Rs Aggarwal 2020 2021 for Class 7 Maths Chapter 20 - Mensuration

Answer:

(i) Length = 24.5 m
Breadth = 18 m
   
   ∴ Area of the rectangle = Length $\times$ Breadth
                                 = 24.5 m $\times$ 18 m
                                 = 441 m²
              
  (ii) Length = 12.5 m
Breadth = 8 dm = (8 $\times$ 10) = 80 cm = 0.8 m     [since 1 dm = 10 cm and 1 m = 100 cm]
      
      ∴ Area of the rectangle = Length $\times$ Breadth
                                    = 12.5 m $\times$ 0.8 m
                                    = 10 m²

Answer:

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, 48 m will be one side of the triangle and the diagonal, which is 50 m, will be the hypotenuse.
According to the Pythagoras theorem:
(Hypotenuse)² = (Base)² + (Perpendicular)²
Perpendicular = $\sqrt{{\left(\mathrm{Hypotenuse}\right)}^2-(\mathrm{Base}{)}^2}$
Perpendicular = $\sqrt{{\left(50\right)}^2-{\left(48\right)}^2}=\sqrt{2500-2304}=\sqrt{196}=14$ m
∴ Other side of the rectangular plot = 14 m
Length = 48m
Breadth = 14m

∴ Area of the rectangular plot = 48 m $\times$ 14 m = 672 m²
Hence, the area of a rectangular plot is 672 m².

Answer:

Let the length of the field be 4x m.
Breadth = 3x m
∴ Area of the field = (4x $\times$ 3x) m² = 12x² m²
But it is given that the area is 1728 m².
∴ 12x² = 1728
⇒ x² = $\left(\frac{1728}{12}\right)$ = 144
⇒ x = $\sqrt{144}$ = 12
∴ Length = (4 $\times$ 12) m = 48 m
Breadth = (3 $\times$ 12) m =36 m
∴ Perimeter of the field = 2(l + b) units
                                       = 2(48 + 36) m = (2 $\times$ 84) m = 168 m
∴ Cost of fencing = Rs (168 $\times$ 30) = Rs 5040

Answer:

Area of the rectangular field = 3584 m²
Length of the rectangular field = 64 m
Breadth of the rectangular field = $\left(\frac{\mathrm{Area}}{\mathrm{Length}}\right)$ = $\left(\frac{3584}{64}\right)$ m = 56 m
Perimeter of the rectangular field = 2 (length + breadth)
                                                      = 2(64+ 56) m = (2 $\times$ 120) m = 240 m
Distance covered by the boy = 5 $\times$ Perimeter of the rectangular field
                                              = 5 $\times$ 240 = 1200 m

The boy walks at the rate of 6 km/hr.
or
Rate = $\left(\frac{6\times 1000}{60}\right)$ m/min = 100 m/min.

∴ Required time to cover a distance of 1200 m = $\left(\frac{1200}{100}\right)$ min = 12 min
Hence, the boy will take 12 minutes to go five times around the field.

Answer:

Given:
Length of the verandah = 40 m = 400 dm    [since 1 m = 10 dm ]
Breadth of the verandah = 15 m = 150 dm
∴ Area of the verandah= (400 $\times$ 150) dm² = 60000 dm²

Length of a stone = 6 dm
Breadth of a stone = 5 dm
∴ Area of a stone = (6 $\times$ 5) dm² = 30 dm²

∴ Total number of stones needed to pave the verandah = $\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{verandah}}{\mathrm{Area} \mathrm{of} \mathrm{each} \mathrm{stone}}$

                                                                                    = $\left(\frac{60000}{30}\right)$ = 2000

Answer:

Area of the carpet = Area of the room
                              = (13 m $\times$ 9 m) = 117 m²

Now, width of the carpet = 75 cm    (given)
                                         = 0.75 m      [since 1 m = 100 cm]

Length of the carpet = $\left(\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{carpet}}{\mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{carpet}}\right)$ = $\left(\frac{117}{0.75}\right)$ m = 156 m
Rate of carpeting = Rs 105 per m
∴ Total cost of carpeting = Rs (156 $\times$105) = Rs 16380
Hence, the total cost of carpeting the room is Rs 16380.

Answer:

Given:
Length of the room = 15 m
Width of the carpet = 75 cm = 0.75 m          (since 1 m = 100 cm)

Let the length of the carpet required for carpeting the room be x m.
Cost of the carpet = Rs. 80 per m
∴ Cost of x m carpet = Rs. (80 $\times$ x) = Rs. (80x)
Cost of carpeting the room = Rs. 19200
∴ 80x = 19200 ⇒ x = $\left(\frac{19200}{80}\right)$ = 240
Thus, the length of the carpet required for carpeting the room is 240 m.
Area of the carpet required for carpeting the room = Length of the carpet $\times$ Width of the carpet
                                                                           = ( 240 $\times$ 0.75) m² = 180 m²
Let the width of the room be b m.
Area to be carpeted = 15 m $\times$ b m = 15b m²
∴ 15b m² = 180 m²
⇒ b = $\left(\frac{180}{15}\right)$ m = 12 m
Hence, the width of the room is 12 m.

Answer:

Total cost of fencing a rectangular piece = Rs. 9600 
Rate of fencing = Rs. 24
∴ Perimeter of the rectangular field = $\left(\frac{\mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{fencing}}{\mathrm{Rate} \mathrm{of} \mathrm{fencing}}\right)$ m = $\left(\frac{9600}{24}\right)$ m = 400 m

Let the length and breadth of the rectangular field be 5x and 3x, respectively.
Perimeter of the rectangular land =  2(5x + 3x) = 16x
But the perimeter of the given field is 400 m.
∴ 16x = 400
 x = $\left(\frac{400}{16}\right)$ = 25
Length of the field = (5 $\times$ 25) m = 125 m
Breadth of the field = (3 $\times$ 25) m = 75 m

Answer:

Length of the diagonal of the room = $\sqrt{{\mathrm{l}}^2+{\mathrm{b}}^2+{\mathrm{h}}^2}$
                                                         = $\sqrt{{\left(10\right)}^2+{\left(10\right)}^2+{\left(5\right)}^2}$ m
                                                         = $\sqrt{100+100+25}$m
                                                         = $\sqrt{225}$m = 15 m

Hence, length of the largest pole that can be placed in the given hall is 15 m.

Answer:

Side of the square = 8.5 m
∴ Area of the square = (Side)²
                               = (8.5 m)²
                                = 72.25 m²

Answer:

(i) Diagonal of the square = 72 cm
     ∴ Area of the square = $\left[\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right]$ sq. unit
                                    = $\left[\frac{1}{2}\times {\left(72\right)}^2\right]$ cm²
                                    = 2592 cm²

(ii)Diagonal of the square = 2.4 m
     ∴ Area of the square = $\left[\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right]$ sq. unit
                                    = $\left[\frac{1}{2}\times {\left(2.4\right)}^2\right]$ m²
                                    = 2.88 m²

Answer:

We know:
Area of a square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$ sq. units
Diagonal of the square = $\sqrt{2\times \mathrm{Area} \mathrm{of} \mathrm{square}}$ units
                                  = $\left(\sqrt{2\times 16200}\right)$m = 180 m
∴ Length of the diagonal of the square = 180 m

                             

Answer:

Area of the square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$ sq. units
Given:
Area of the square field = $\frac{1}{2}$ hectare
                                  = $\left(\frac{1}{2}\times 10000\right)$ m² = 5000 m²             [since 1 hectare = 10000 m² ]

Diagonal of the square = $\sqrt{2\times \mathrm{Area} \mathrm{of} the \mathrm{square}}$ 

                                  = $\left(\sqrt{2\times 5000}\right)$m = 100 m

∴ Length of the diagonal of the square field = 100 m

Answer:

Area of the square plot = 6084 m²
Side of the square plot = $\left(\sqrt{\mathrm{Area}}\right)$
                                         = $\left(\sqrt{6084}\right)$ m
                                         = $\left(\sqrt{78\times 78}\right)$m = 78 m

∴ Perimeter of the square plot = 4 $\times$ side = (4 $\times$ 78) m = 312 m
312 m wire is needed to go along the boundary of the square plot once.

Required length of the wire that can go four times along the boundary = 4 $\times$ Perimeter of the square plot
                                                                                                           = (4 $\times$ 312) m = 1248 m

Answer:

Side of the square = 10 cm
Length of the wire = Perimeter of the square = 4 $\times$ Side = 4 $\times$ 10 cm = 40 cm
Length of the rectangle (l) = 12 cm
Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

⇒ 2(l + b) = 40
⇒ 2(12 + b) = 40
⇒ 24 + 2b = 40
⇒ 2b = 40 - 24 = 16
⇒ b = $\left(\frac{16}{2}\right)$ cm = 8 cm
∴ Breadth of the rectangle = 8 cm
Now, Area of the square = (Side)² = (10 cm $\times$ 10 cm) = 100 cm²
Area of the rectangle = l $\times$ b = (12 cm $\times$ 8 cm) = 96 cm²

Hence, the square encloses more area.
It encloses 4 cm² more area.

Answer:

Given:
Length = 50 m
Breadth = 40 m
Height = 10 m

Area of the four walls  = {2h(l + b)} sq. unit
                                   = {2 $\times$ 10 $\times$ (50 + 40)}m²
                                   = {20 $\times$ 90} m² = 1800 m²
Area of the ceiling = l $\times$ b = (50 m $\times$ 40 m) = 2000 m²
∴ Total area to be white washed = (1800 + 2000) m² = 3800 m²
Rate of white washing = Rs 20/sq. metre
∴ Total cost of white washing = Rs (3800 $\times$ 20) = Rs 76000

Answer:

Let the length of the room be l m.
Given:
Breadth of the room = 10 m
Height of the room = 4 m
Area of the four walls = [2(l + b)h] sq units.
                              = 168 m²
∴ 168 = [2(l + 10) $\times$ 4]
⇒ 168 = [8l + 80]
⇒ 168 - 80 = 8l
⇒  88 = 8l
⇒ l = $\left(\frac{88}{8}\right)$ m = 11 m
∴ Length of the room = 11 m

Answer:

Given:
Length of the room = 7.5 m
Breadth of the room = 3.5 m
Area of the four walls = [2(l + b)h] sq. units.
                              = 77 m²
∴ 77 = [2(7.5 + 3.5)h]
⇒ 77 = [(2 $\times$ 11)h]
⇒ 77 = 22h
⇒ h =  $\left(\frac{77}{22}\right)$ m = $\left(\frac{7}{2}\right)$ m = 3.5 m
∴ Height of the room = 3.5 m

Answer:

Let the breadth of the room be x m.
Length of the room = 2x m
Area of the four walls = {2(l + b) $\times$ h} sq. units
           120 m²  = {2(2x + x) $\times$ 4} m²  
⇒ 120 = {8 $\times$ 3x }
⇒ 120 = 24x
⇒ x =$\left(\frac{120}{24}\right)$ = 5
∴ Length of the room = 2x = (2 $\times$ 5) m = 10 m
Breadth of the room = x = 5 m
∴ Area of the floor = l $\times$ b = (10 m $\times$ 5 m) = 50 m²

Answer:

Length = 8.5 m
Breadth = 6.5 m
Height = 3.4 m

Area of the four walls = {2(l + b) $\times$ h} sq. units
                          = {2(8.5 + 6.5) $\times$ 3.4}m² = {30 $\times$ 3.4} m² = 102 m²
Area of one door = (1.5 $\times$ 1) m² = 1.5 m²
∴ Area of two doors = (2 $\times$ 1.5) m² = 3 m²
Area of one window = (2 $\times$ 1) m² = 2 m²
 ∴ Area of two windows = (2 $\times$ 2) m² = 4 m²
Total area of  two doors and two windows = (3 + 4) m²
                                                               = 7 m²
Area to be painted = (102 - 7) m² = 95 m²
Rate of painting = Rs 160 per m²
Total cost of painting = Rs (95 $\times$ 160) = Rs 15200

Answer:

Let PQRS be the given grassy plot and ABCD be the inside boundary of the path.

Length = 75 m
Breadth = 60 m
Area of the plot = (75 $\times$ 60) m² = 4500 m²
Width of the path = 2 m
∴ AB = (75 - 2 $\times$ 2) m = (75 - 4) m =71 m
 AD = (60 - 2 $\times$ 2) m = (60 - 4) m = 56 m
Area of rectangle ABCD = (71 x 56) m² = 3976 m²
Area of the path = (Area of PQRS - Area of ABCD)
                           = (4500 - 3976) m² = 524 m²
Rate of constructing the path = Rs 125 per m²
∴ Total cost of constructing the path = Rs (524 $\times$ 125) = Rs 65,500

Answer:

Let PQRS be the given rectangular plot and ABCD be the inside boundary of the path.

Length = 95 m
Breadth = 72 m
Area of the plot = (95 $\times$ 72) m² = 6,840 m²
Width of the path = 3.5 m
∴ AB = (95 - 2 $\times$ 3.5) m = (95 - 7) m = 88 m
AD = (72 - 2 $\times$ 3.5) m = (72 - 7) m = 65 m
Area of the grassy rectangle plot ABCD = (88 $\times$ 65) m² = 5,720 m²
Area of the path = (Area PQRS - Area ABCD)
                           = (6840 - 5720) m² = 1,120 m²
Rate of constructing the path = Rs. 80 per m²
∴ Total cost of  constructing the path = Rs. (1,120 $\times$ 80) = Rs. 89,600
Rate of laying the grass on the plot ABCD = Rs. 40 per m²
∴ Total cost of laying the grass on the plot = Rs. (5,720 $\times$ 40) = Rs. 2,28,800
∴ Total expenses involved = Rs. ( 89,600 + 2,28,800) = Rs. 3,18,400

Answer:

Let ABCD be the saree and EFGH be the part of saree without border.

Length, AB= 5 m
Breadth, BC = 1.3 m
Width of the border of the saree = 25 cm = 0.25 m

∴ Area of ABCD = 5 m $\times$ 1.3 m = 6.5 m²

Length, GH = {5 -( 0.25 + 0.25} m = 4.5 m
Breadth, FG = {1.3 - 0.25 + 0.25} m = 0.8 m
∴ Area of EFGH = 4.5 m $\times$ .8 m = 3.6 m²

Area of the border = Area of ABCD − Area of EFGH
                              =  6.5 m²  − 3.6 m²
                              = 2.9 m² = 29000 cm²     [since 1 m² = 10000 cm²]
Rate of printing the border = Rs 1 per 10 cm²
∴ Total cost of printing the border = Rs $\left(\frac{1\times 29000}{10}\right)$
                                                       = Rs 2900

Answer:

Length, EF = 38 m
Breadth, FG = 25 m

∴ Area of EFGH = 38 m $\times$  25 m = 950 m²

Length, AB = (38  + 2.5  + 2.5 ) m = 43 m
Breadth, BC = ( 25 + 2.5 + 2.5 ) m = 30 m
∴ Area of ABCD = 43 m $\times$ 30 m = 1290 m²

Area of the path = Area of ABCD − Area of PQRS
                          = 1290 m² − 950 m²
                         = 340 m²
Rate of gravelling the path = Rs 120 per m²

∴ Total cost of gravelling the path = Rs (120 $\times$ 340)
                                                    = Rs 40800

Answer:

Let EFGH denote the floor of the room.
The white region represents the floor of the 1.25 m verandah.

Length, EF = 9.5 m
Breadth, FG = 6 m

∴ Area of EFGH = 9.5 m  $\times$  6 m = 57 m²

Length, AB = (9.5  + 1.25  + 1.25 ) m = 12 m
Breadth, BC = ( 6 + 1.25 + 1.25 ) m = 8.5 m
∴ Area of ABCD = 12 m $\times$ 8.5 m = 102 m²

Area of the verandah = Area of ABCD − Area of EFGH
                                  = 102 m² − 57 m²
                                  = 45 m²
Rate of cementing the verandah = Rs 80 per m²

∴ Total cost of cementing the verandah = Rs ( 80 $\times$ 45)
                                                            = Rs 3600

Answer:

Side of the flower bed = 2 m 80 cm = 2.80 m      [since 100 cm = 1 m]

∴ Area of the square flower bed = (Side)² = (2.80 m )² = 7.84 m²
Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm
                                                                        = (2.80 + 0.3 + 0.3) m = 3.4 m
Area of the enlarged flower bed with the digging strip = (Side )² = (3.4 )² = 11.56 m²

∴ Increase in the area of the flower bed = 11.56 m² − 7.84 m²
                                                            = 3.72 m²

Answer:

Let the length and the breadth of the park be 2x m and x m, respectively.
Perimeter of the park = 2(2x + x) = 240 m
⇒ 2(2x + x) = 240
⇒ 6x = 240
⇒ x = $\left(\frac{240}{6}\right)$ m =40 m
∴ Length of the park = 2x = (2 $\times$ 40) = 80 m
Breadth = x = 40 m
Let PQRS be the given park and ABCD be the inside boundary of the path.

Length = 80 m
Breadth = 40 m
Area of the park = (80 $\times$ 40) m² = 3200 m²
Width of the path = 2 m
∴ AB = (80 - 2 $\times$ 2) m = (80 - 4) m =76 m
 AD = (40 - 2 $\times$ 2) m = (40 - 4) m = 36 m
Area of the rectangle ABCD = (76 $\times$ 36) m² = 2736 m²
Area of the path = (Area of PQRS - Area of ABCD)
                           = (3200 - 2736) m² = 464 m²
Rate of paving the path = Rs. 80 per m²
∴ Total cost of  paving the path = Rs. (464 $\times$ 80) = Rs. 37,120

Answer:

Length of the hall, PQ = 22 m
Breadth of the hall, QR = 15.5 m

∴ Area of the school hall PQRS = 22 m $\times$ 15.5 m = 341 m²
Length of the carpet, AB = 22 m − ( 0.75 m + 0.75 m) = 20.5 m         [since 100 cm = 1 m]
Breadth of the carpet, BC = 15.5 m − ( 0.75 m + 0.75 m) = 14 m
∴ Area of the carpet ABCD = 20.5 m $\times$ 14 m = 287 m²
Area of the strip = Area of the school hall (PQRS) − Area of the carpet (ABCD)
                           = 341 m² − 287 m²
                           = 54 m²

Area of 1 m length of the carpet = 1 m $\times$ 0.82 m = 0.82 m²

∴ Length of the carpet whose area is 287 m² = 287 m² ÷ 0.82 m² = 350 m
Cost of the 350 m long carpet = Rs 60 $\times$ 350 = Rs 21000

Answer:

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let a side of the lawn (AB) be x m.
Area of the square lawn = x²
Length, PQ = (x m + 2.5 m + 2.5 m) = (x + 5) m
∴ Area of PQRS = (x + 5)² = (x² + 10x + 25) m²

Area of the path = Area of PQRS − Area of the square lawn (ABCD)
⇒ 165 = x² + 10x + 25 − x²
⇒ 165 = 10x + 25
⇒ 165 − 25 = 10x
⇒ 140 = 10x
∴ x = 140 ÷ 10 = 14
∴ Side of the lawn = 14 m

∴ Area of the lawn = (Side)² = (14 m)² = 196 m²

Answer:

Area of the path = 305 m²

Let the length of the park be 5x m and the breadth of the park be 2x m.

∴ Area of the rectangular park = 5x $\times$ 2x = 10x² m²
Width of the path = 2.5 m
Outer length, PQ = 5x m + 2.5 m + 2.5 m = (5x + 5) m
Outer breadth, QR = 2x + 2.5 m + 2.5 m = (2x + 5) m
Area of PQRS = (5x + 5) $\times$ (2x + 5)  = (10x² + 25x + 10x + 25) = (10x² + 35x + 25) m²
∴ Area of the path = [(10x² + 35x + 25) − 10x² ] m²
⇒  305 = 35x + 25
⇒ 305 − 25 = 35x  
⇒ 280 = 35x
⇒ x = 280 ÷ 35 = 8

∴ Length of the park = 5x = 5 $\times$ 8 = 40 m
Breadth of the park = 2x = 2 $\times$ 8 = 16 m

Answer:

Let ABCD be the rectangular park.
Let EFGH and IJKL be the two rectangular roads with width 5 m.

Length of the rectangular park, AD = 70 m 
  Breadth of the rectangular park, CD = 50 m
∴ Area of the rectangular park = Length $\times$ Breadth = 70 m $\times$ 50 m = 3500 m²
Area of road EFGH = 70 m $\times$ 5 m = 350 m²
Area of road IJKL = 50 m $\times$ 5 m = 250 m²

Clearly, area of MNOP is common to both the two roads.

∴ Area of MNOP = 5 m $\times$ 5 m = 25 m²

Area of the roads = Area (EFGH) + Area (IJKL) − Area (MNOP)
                            = (350  + 250 ) m²− 25 m² = 575 m²
It is given that the cost of constructing the roads is Rs. 120/m².

Cost of constructing 575 m² area of the roads = Rs. (120 × 575)
                                                                    = Rs. 69000

Answer:

Let ABCD be the rectangular field and PQRS and KLMN be the two rectangular roads with width 2 m and 2.5 m, respectively.

Length of the rectangular field, CD = 115 cm
Breadth of the rectangular field, BC = 64 m
∴ Area of the rectangular lawn ABCD = 115 m $\times$ 64 m = 7360 m²
Area of the road PQRS = 115 m $\times$ 2 m = 230 m²
Area of the road KLMN = 64 m $\times$ 2.5 m = 160 m²

Clearly, the area of EFGH is common to both the two roads.

∴ Area of EFGH = 2 m $\times$ 2.5 m = 5 m²

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
                            = (230 m² + 160 m²) − 5 m² = 385 m²

Rate of gravelling the roads = Rs 60 per m²
∴ Total cost of gravelling the roads = Rs (385 $\times$ 60)
                                                     = Rs 23,100

Answer:

Let ABCD be the rectangular field and KLMN and PQRS be the two rectangular roads with width 2.5 m and 2 m, respectively.

Length of the rectangular field CD = 50 cm
Breadth of the rectangular field BC = 40 m
∴ Area of the rectangular field ABCD = 50 m $\times$ 40 m = 2000 m²
Area of road KLMN = 40 m $\times$ 2.5 m = 100 m²
Area of road PQRS = 50 m $\times$ 2 m = 100 m²

Clearly, area of EFGH is common to both the two roads.

∴ Area of EFGH = 2.5 m $\times$ 2 m = 5 m²

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
                                = (100 m² + 100 m²) − 5 m² = 195 m²

Area of the remaining portion of the field = Area of the rectangular field (ABCD) − Area of the roads
                                                                   = (2000 − 195) m²
                                                                                  = 1805 m²

Answer:

(i) Complete the rectangle as shown below:
    
    Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGH] sq. units
                                              = [(43 m $\times$ 27 m) - {(43 - 2 $\times$ 1.5) m x (27 - 1 $\times$ 2) m}]
                                              = [(43 m $\times$ 27 m) - {40 m $\times$ 25 m}]
                                              = 1161 m² - 1000 m²
                                              = 161 m²

(ii) Complete the rectangle as shown below:
     
     Area of the shaded region = [Area of square ABCD - {(Area of EFGH) + (Area of IJKL) - (Area of MNOP)}] sq. units
                                               = [(40 $\times$ 40) - {(40 $\times$ 2) + (40 $\times$ 3) - (2 $\times$ 3)}] m²
                                               = [1600 - {(80 + 120 - 6)] m²
                                               = [1600 - 194] m²
                                               = 1406 m²

Answer:

(i) Complete the rectangle as shown below:
    
Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGD] sq. units
                                          = [(AB $\times$ BC) - (DG $\times$ GF)] m²
                                          = [(24 m $\times$ 19 m) - {(24 - 4) m $\times$ 16.5 m} ]
                                          = [(24 m $\times$ 19 m) - (20 m $\times$ 16.5) m]
                                          = (456 - 330) m² = 126 m²

(ii) Complete the rectangle by drawing lines as shown below:
     
Area of the shaded region ={(12 $\times$ 3) + (12 $\times$ 3) + (5$\times$ 3) + {(15 - 3 - 3) $\times$3)} cm²
                                    = { 36 + 36 + 15 + 27} cm²
                                    = 114 cm²

Answer:

Divide the given figure in four parts shown below:

Given:
Width of each part = 0.5 m

Now, we have to find the length of each part.

Length of part I = 3.5 m
Length of part II = (3.5 - 0.5 - 0.5) m = 2.5 m
Length of part III = (2.5 - 0.5 - 0.5) = 1.5 m
Length of part IV = (1.5 - 0.5 - 0.5) = 0.5 m
∴ Area of the shaded region = [Area of part (I) + Area of part (II) + Area of part (III) + Area of part (IV)] sq. units
                                        = [(3.5 $\times$ 0.5) + (2.5 $\times$ 0.5) + ( 1.5 $\times$ 0.5) + (0.5 $\times$ 0.5)] m²
                                        = [1.75 + 1.25 + 0.75 + 0.25] m²
                                        = 4 m²

Answer:

Base = 32 cm
Height = 16.5 cm

∴ Area of the parallelogram = Base $\times$ Height
                                          =  32 cm $\times$ 16.5 cm
                                          = 528 cm²

Answer:

Base = 1 m 60 cm = 1.6 m             [since 100 cm = 1 m]
Height = 75 cm = 0.75 m

∴ Area of the parallelogram = Base $\times$ Height
                                              = 1.6 m $\times$ 0.75 m
                                              = 1.2 m²

Answer:

(i) Base = 14 dm = (14 $\times$ 10) cm = 140 cm                  [since 1 dm  = 10 cm]
     Height = 6.5 dm = (6.5 $\times$ 10) cm = 65 cm
    
     Area of the parallelogram  = Base $\times$ Height
                                                = 140 cm $\times$ 65 cm
                                                = 9100 cm²

(ii) Base = 14 dm = (14 $\times$ 10) cm                            [since 1 dm = 10 cm and 100 cm  = 1 m]
              = 140 cm = 1.4 m            
      Height = 6.5 dm = (6.5 $\times$ 10) cm
                  = 65 cm = 0.65 m
    
 ∴ Area of the parallelogram = Base $\times$ Height
                                                = 1.4 m $\times$ 0.65 m
                                                = 0.91 m²

Answer:

Area of the given parallelogram = 54 cm²
Base of the given parallelogram = 15 cm
∴ Height of the given parallelogram = $\frac{\mathrm{Area}}{\mathrm{Base}}$ = $\left(\frac{54}{15}\right)$ cm = 3.6 cm

Answer:

Base of the parallelogram = 18 cm
Area of the parallelogram = 153 cm²
∴ Area of the parallelogram = Base $\times$ Height
⇒ Height = $\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{parallelogram}}{\mathrm{Base}}$ = $\left(\frac{153}{18}\right)$ cm = 8.5 cm
Hence, the distance of the given side from its opposite side is 8.5 cm.

Answer:

Base, AB = 18 cm
Height, AL = 6.4 cm
∴ Area of the parallelogram ABCD = Base $\times$ Height
                                                    = (18 cm $\times$ 6.4 cm) = 115.2 cm²                    ... (i)

Now, taking BC as the base:
Area of the parallelogram ABCD = Base $\times$ Height
                                                      = (12 cm $\times$ AM)                          ... (ii)
From equation (i) and (ii):
12 cm $\times$ AM = 115.2 cm²
⇒ AM = $\left(\frac{115.2}{12}\right)$cm
= 9.6 cm

Answer:

ABCD is a parallelogram with side AB of length 15 cm and the corresponding altitude AE of length 4 cm.
The adjacent side AD is of length 8 cm and the corresponding altitude is CF.

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases.

∴ AD $\times$ CF = AB $\times$ AE
⇒ 8 cm $\times$ CF = 15 cm $\times$4 cm

⇒ CF =$\left(\frac{15\times 4}{8}\right)$ cm = $\left(\frac{15}{2}\right)$ cm = 7.5 cm
Hence, the distance between the shorter sides is 7.5 cm.

Answer:

Let the base of the parallelogram be x cm.
Then, the height of the parallelogram will be $\frac{1}{3}$x cm.
It is given that the area of the parallelogram is 108 cm².

Area of a parallelogram =  Base $\times$ Height
                     ∴ 108 cm² = x $\times$ $\frac{1}{3}$x
                        108 cm² = $\frac{1}{3}$x²
⇒ x² = (108 $\times$ 3) cm² = 324 cm²
⇒ x² = (18 cm)²
⇒ x = 18 cm

∴ Base = x = 18 cm
Height = $\frac{1}{3}$x = $\left(\frac{1}{3}\times 18\right)$ cm
            = 6 cm

Answer:

Let the height of the parallelogram be x cm.
Then, the base of the parallelogram will be 2x cm.
It is given that the area of the parallelogram is 512 cm².

Area of a parallelogram =  Base $\times$ Height
                     ∴ 512 cm² = 2x $\times$ x
                        512 cm² = 2x²
⇒ x² =$\left(\frac{512}{2}\right)$ cm² = 256 cm²
⇒ x² = (16 cm)²
⇒ x = 16 cm

∴ Base = 2x = 2 $\times$ 16
             = 32 cm
Height = x = 16 cm

Answer:

A rhombus is a special type of a parallelogram.

The area of a parallelogram is given by the product of its base and height.
∴ Area of the given rhombus = Base × Height
(i) Area of the rhombus = 12 cm $\times$ 7.5 cm = 90 cm²

(ii) Base = 2 dm = (2 $\times$ 10) = 20 cm    [since 1 dm = 10 cm]
     Height = 12.6 cm
      ∴ Area of the rhombus = 20 cm $\times$ 12.6 cm = 252 cm²

Answer:

(i)
     Length of one diagonal = 16 cm
     Length of the other diagonal = 28 cm
     ∴ Area of the rhombus = $\frac{1}{2}$ $\times$ (Product of the diagonals)
                                           = $\left(\frac{1}{2}\times 16\times 28\right)$ cm² = 224 cm²
(ii)
      Length of one diagonal = 8 dm 5 cm = (8 $\times$ 10 + 5) cm = 85 cm              [since 1 dm = 10 cm]
      Length of the other diagonal = 5 dm 6 cm = (5 $\times$ 10 + 6) cm = 56 cm
      ∴ Area of the rhombus = $\frac{1}{2}$ $\times$ (Product of the diagonals)
                                            = $\left(\frac{1}{2}\times 85\times 56\right)$ cm²
                                            = 2380 cm²
    

Answer:

Let ABCD be the rhombus, whose diagonals intersect at O.

 AB = 20 cm and AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.

Therefore, ΔAOB is a right angled triangle, right angled at O. 

Here, OA =$\frac{1}{2}\mathrm{AC}$ = 12 cm
AB = 20 cm

By Pythagoras theorem:
(AB)² = (OA)² + (OB)²
⇒ (20)² = (12)² + (OB)²
⇒ (OB)² = (20)² − (12)²
⇒ (OB)² = 400 − 144 = 256
⇒ (OB)² = (16)²
⇒ OB = 16 cm
∴ BD = 2 $\times$ OB = 2 $\times$ 16 cm = 32 cm

∴  Area of the rhombus ABCD = $\left(\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}\right)$ cm²
                                                  = $\left(\frac{1}{2}\times 24\times 32\right)$ cm²
                                                  = 384 cm²

Answer:

Area of a rhombus =  $\frac{1}{2}$ $\times$ (Product of the diagonals)
Given:
Length of one diagonal = 19.2 cm
Area of the rhombus = 148.8 cm²

∴ Length of the other diagonal = $\left(\frac{148.8\times 2}{19.2}\right)$ cm = 15.5 cm

Answer:

Perimeter of the rhombus = 56 cm
Area of the rhombus = 119 cm²
Side of the rhombus = $\frac{\mathrm{Perimeter}}{4}$ = $\left(\frac{56}{4}\right)$ cm = 14 cm
Area of a rhombus = Base $\times$ Height

∴  Height of the rhombus = $\frac{\mathrm{Area}}{\mathrm{Base}}$= $\left(\frac{119}{14}\right)$ cm
                                          = 8.5 cm

Answer:

Given:
Height of the rhombus = 17.5 cm
Area of the rhombus = 441 cm²

We know:
Area of a rhombus = Base $\times$ Height

∴  Base of the rhombus =$\frac{\mathrm{Area}}{\mathrm{Height}}$ = $\left(\frac{441}{17.5}\right)$ cm = 25.2 cm
Hence, each side of a rhombus is 25.2 cm.

Answer:

Area of a triangle = $\frac{1}{2}$ $\times$ Base $\times$ Height
                          = $\left(\frac{1}{2}\times 24.8\times 16.5\right)$ cm² = 204.6 cm²
Given:
Area of the rhombus = Area of the triangle
Area of the rhombus = 204.6 cm²

Area of the rhombus = $\frac{1}{2}$ $\times$ (Product of the diagonals)
Given:
Length of one diagonal = 22 cm 
∴ Length of the other diagonal = $\left(\frac{204.6\times 2}{22}\right)$ cm
                                                  = 18.6 cm

                          

Answer:

We know:
Area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$
(i) Base = 42 cm
Height = 25 cm
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 42\times 25\right)$ cm² = 525 cm²
(ii) Base = 16.8 m    
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m]
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 16.8\times 0.75\right)$ m² = 6.3 m²
(iii) Base = 8 dm = (8 $\times$ 10) cm = 80 cm     [since 1 dm = 10 cm]
       Height = 35 cm
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 80\times 35\right)$ cm² = 1400 cm²

Answer:

Height of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Base}}$
Here, base = 16 cm and area = 72 cm²

∴ Height = $\left(\frac{2\times 72}{16}\right)$ cm = 9 cm

Answer:

Height of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Base}}$
Here, base = 28 m and area = 224 m²

∴ Height = $\left(\frac{2\times 224}{28}\right)$ m = 16 m

Answer:

Base of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Height}}$
Here, height = 12 cm and area = 90 cm²

∴ Base = $\left(\frac{2\times 90}{12}\right)$ cm = 15 cm

Answer:

Total cost of cultivating the field = Rs. 14580

Rate of cultivating the field = Rs. 1080 per hectare

Answer:

Let the length of the other leg be h cm.
Then, area of the triangle = $\left(\frac{1}{2}\times 14.8\times h\right)$ cm² = (7.4 h) cm²
But it is given that the area of the triangle is 129.5 cm².
∴ 7.4h = 129.5
⇒ h = $\left(\frac{129.5}{7.4}\right)$ = 17.5 cm
∴ Length of the other leg = 17.5 cm

Answer:

Here, base = 1.2 m and hypotenuse = 3.7 m

In the right angled triangle:

Perpendicular = $\sqrt{(H\mathrm{ypotenuse}{)}^2-(B\mathrm{ase}{)}^2}$

                        $$\begin{gathered} =\sqrt{{\left(3.7\right)}^2-{\left(1.2\right)}^2} \\ =\sqrt{13.69-1.44} \\ =\sqrt{12.25} \\ =3.5 \end{gathered}$$
 Area = $\left(\frac{1}{2}\times \mathrm{base}\times \mathrm{perpendicular}\right)$ sq. units
          = $\left(\frac{1}{2}\times 1.2\times 3.5\right)$ m²
∴ Area of the right angled triangle = 2.1 m²

Answer:

In a right angled triangle, if one leg is the base, then the other leg is the height.
Let the given legs be 3x and 4x, respectively.
Area of the triangle = $\left(\frac{1}{2}\times 3x\times 4x\right)$ cm²
⇒ 1014 = (6x²)
⇒ 1014 = 6x²
⇒ x² = $\left(\frac{1014}{6}\right)$ = 169
⇒ x = $\sqrt{169}$ = 13
∴ Base = (3 $\times$ 13) = 39 cm
Height = (4 $\times$ 13) = 52 cm

Answer:

Consider a right-angled triangular scarf (ABC).
Here, ∠B= 90°
BC = 80 cm
AC = 1 m = 100 cm

Now, AB² + BC² = AC²
⇒ AB² = AC² - BC² = (100)² - (80)²
            = (10000 - 6400) = 3600
 ⇒ AB = $\sqrt{3600}$  = 60 cm
Area of the scarf ABC = $\left(\frac{1}{2}\times BC\times AB\right)$ sq. units
                               = $\left(\frac{1}{2}\times 80\times 60\right)$ cm²
                               = 2400 cm² = 0.24 m²     [since 1 m² = 10000 cm²]
Rate of the cloth = Rs 250 per m²
∴ Total cost of the scarf = Rs (250 $\times$ 0.24) = Rs 60
Hence, cost of the right angled scarf is Rs 60.
                            

Answer:

(i) Side of the equilateral triangle = 18 cm
     Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$  sq. units
                                                      =  $\frac{\sqrt{3}}{4}{\left(18\right)}^2$ cm² = $\left(\sqrt{3}\times 81\right)$ cm²
                                                      = (1.73 $\times$ 81) cm² = 140.13 cm²

(ii) Side of the equilateral triangle = 20 cm
     Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$  sq. units
                                                      =  $\frac{\sqrt{3}}{4}{\left(20\right)}^2$ cm² = $\left(\sqrt{3}\times 100\right)$ cm²
                                                      = (1.73 $\times$ 100) cm² = 173 cm²

Answer:

It is given that the area of an equilateral triangle is $16\sqrt{3}$ cm².

We know:
Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2$ sq. units

∴ Side of the equilateral triangle = $\left[\sqrt{\left(\frac{4\times \mathrm{Area}}{\sqrt{3}}\right)}\right]$ cm
                                                      =   $\left[\sqrt{\left(\frac{4\times 16\sqrt{3}}{\sqrt{3}}\right)}\right]$cm = $\left(\sqrt{4\times 16}\right)$cm = $\left(\sqrt{64}\right)$cm = 8 cm

Hence, the length of the equilateral triangle is 8 cm.

Answer:

Let the height of the triangle be h cm.
Area of the triangle = $\left(\frac{1}{2}\times \mathrm{Base} \times \mathrm{Height}\right)$ sq. units
                          = $\left(\frac{1}{2}\times 24\times h\right)$ cm²

Let the side of the equilateral triangle be a cm.
Area of the equilateral triangle = $\left(\frac{\sqrt{3}}{4}{a}^2\right)$ sq. units
                                            = $\left(\frac{\sqrt{3}}{4}\times 24\times 24\right)$ cm² = $\left(\sqrt{3}\times 144\right)$ cm²
∴ $\left(\frac{1}{2}\times 24\times h\right)$ = $\left(\sqrt{3}\times 144\right)$
⇒ 12 h = $\left(\sqrt{3}\times 144\right)$
⇒ h = $\left(\frac{\sqrt{3}\times 144}{12}\right)=\left(\sqrt{3}\times 12\right)=\left(1.73\times 12\right)=20.76$ cm
∴ Height of the equilateral triangle = 20.76 cm

Answer:

(i) Let a = 13 m, b = 14 m and c = 15 m
     s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{13+14+15}{2}\right)=\left(\frac{42}{2}\right)$m = 21 m
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
                               = $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$m²
                               =   $\sqrt{21\times 8\times 7\times 6}$ m²
                               = $\sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}$ m²
                               = (2 $\times$ 2 $\times$ 3 $\times$ 7) m²
                               = 84 m²

(ii) Let a = 52 cm, b = 56 cm and c = 60 cm
      s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{52+56+60}{2}\right)=\left(\frac{168}{2}\right)$ cm = 84 cm
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
                               = $\sqrt{84\left(84-52\right)\left(84-56\right)\left(84-60\right)}$cm²
                               =   $\sqrt{84\times 32\times 28\times 24}$ cm²
                               = $\sqrt{12\times 7\times 4\times 8\times 4\times 7\times 3\times 8}$ cm²
                               = $\sqrt{2\times 2\times 3\times 7\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 3\times 2\times 2\times 2}$ cm²
                               = (2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 3 $\times$ 7) m²
                               = 1344 cm²

(iii) Let a = 91 m, b = 98 m and c = 105 m
      s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{91+98+105}{2}\right)=\left(\frac{294}{2}\right)$ m = 147 m
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
                               = $\sqrt{147\left(147-91\right)\left(147-98\right)\left(147-105\right)}$m²
                               =  $\sqrt{147\times 56\times 49\times 42}$ m²
                               = $\sqrt{3\times 49\times 8\times 7\times 49\times 6\times 7}$ m²
                               = $\sqrt{3\times 7\times 7\times 2\times 2\times 2\times 7\times 7\times 7\times 2\times 3\times 7}$ m²
                               = ( 2 $\times$ 2 $\times$ 3 $\times$ 7 $\times$ 7 $\times$ 7) m²
                               = 4116 m²

Answer:

Let a = 33 cm, b = 44 cm and c = 55 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{33+44+55}{2}\right)$ cm = $\left(\frac{132}{2}\right)$ cm = 66 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                                      = $\sqrt{66\left(66-33\right)\left(66-44\right)\left(66-55\right)}$ cm²
                                      =  $\sqrt{66\times 33\times 22\times 11}$ cm²
                                      = $\sqrt{6\times 11\times 3\times 11\times 2\times 11\times 11}$ cm²
                                      = (6 $\times$ 11 $\times$ 11) cm² = 726 cm²

Let the height on the side measuring 44 cm be h cm.
Then, Area  = $\frac{1}{2}\times b\times h$
⇒ 726 cm² = $\frac{1}{2}\times 44\times h$
⇒ h = $\left(\frac{2\times 726}{44}\right)$ cm = 33 cm.
∴  Area of the triangle = 726 cm²
Height corresponding to the side measuring 44 cm = 33 cm

Answer:

Let a = 13x cm, b = 14x cm and c = 15x cm
Perimeter of the triangle = 13x + 14x + 15x = 84 (given)
⇒ 42x = 84
⇒ x = $\frac{84}{42}=2$
∴ a = 26 cm , b = 28 cm and c = 30 cm
               
s = $\frac{a+b+c}{2}$ = $\left(\frac{26+28+30}{2}\right)$cm = $\left(\frac{84}{2}\right)$cm = 42 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                               = $\sqrt{42\left(42-26\right)\left(42-28\right)\left(42-30\right)}$ cm²
                               =  $\sqrt{42\times 16\times 14\times 12}$ cm²
                               = $\sqrt{6\times 7\times 4\times 4\times 2\times 7\times 6\times 2}$ cm²
                               = (2 $\times$4 $\times$ 6 $\times$ 7) cm² = 336 cm²
Hence, area of the given triangle is 336 cm².

Answer:

Let a = 42 cm, b = 34 cm and c = 20 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{42+34+20}{2}\right)$ cm = $\left(\frac{96}{2}\right)$ cm = 48 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                               = $\sqrt{48\left(48-42\right)\left(48-34\right)\left(48-20\right)}$ cm²
                               =  $\sqrt{48\times 6\times 14\times 28}$ cm²
                               = $\sqrt{6\times 2\times 2\times 2\times 6\times 14\times 2\times 14}$ cm²
                               = (2 $\times$ 2 $\times$ 6 $\times$ 14) cm² = 336 cm²

Let the height on the side measuring 42 cm be h cm.
Then, Area  = $\frac{1}{2}\times b\times h$
⇒ 336 cm² = $\frac{1}{2}\times 42\times h$
⇒ h = $\left(\frac{2\times 336}{42}\right)$ cm = 16 cm
∴ Area of the triangle = 336 cm²
Height corresponding to the side measuring 42 cm = 16 cm

Answer:

Let each of the equal sides be a cm.
b = 48 cm
a = 30 cm
Area of the triangle = $\left\{\frac{1}{2}\times b\times \sqrt{a^2-\frac{b^2}{4}}\right\}$ sq. units
                                = $\left\{\frac{1}{2}\times 48\times \sqrt{{\left(30\right)}^2-\frac{{\left(48\right)}^2}{4}}\right\}$ cm² = $\left(24\times \sqrt{900-\frac{2304}{4}}\right)$ cm²
                                = $\left(24\times \sqrt{900-576}\right)$ cm² = $\left(24\times \sqrt{324}\right)$ cm² = (24 $\times$ 18) cm² = 432 cm²
∴ Area of the triangle = 432 cm²

Answer:

Let each of the equal sides be a cm.
a + a + 12 = 32 ⇒ 2a = 20  ⇒ a = 10
∴ b = 12 cm and a = 10 cm
Area of the triangle = $\left\{\frac{1}{2}\times b\times \sqrt{a^2-\frac{b^2}{4}}\right\}$ sq. units
                                = $\left\{\frac{1}{2}\times 12\times \sqrt{100-\frac{144}{4}}\right\}$ cm² = $\left(6-\sqrt{100-36}\right)$ cm²
                                = $\left(6\times \sqrt{64}\right)$ cm² = (6 $\times$ 8) cm²
                                = 48 cm²

Answer:

We have:
AC = 26 cm, DL = 12.8 cm and BM = 11.2 cm

Area of ΔADC = $\frac{1}{2}$ $\times$ AC $\times$ DL
                          = $\frac{1}{2}$ $\times$ 26 cm $\times$ 12.8 cm = 166.4 cm²
Area of ΔABC = $\frac{1}{2}$ $\times$ AC $\times$ BM
                         = $\frac{1}{2}$ $\times$ 26 cm $\times$ 11.2 cm = 145.6 cm²

∴ Area of the quadrilateral ABCD = Area of ΔADC  + Area of ΔABC
                                              = (166.4 + 145.6) cm²
                                                        = 312 cm²

Answer:

First, we have to find the area of ΔABC and ΔACD.
For ΔACD:
Let a = 30 cm, b = 40 cm and c = 50 cm
     s = $\left(\frac{a+b+c}{2}\right)=\left(\frac{30+40+50}{2}\right)=\left(\frac{120}{2}\right)=60$ cm
∴  Area of triangle ACD = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$  sq. units
                                        = $\sqrt{60\left(60-30\right)\left(60-40\right)\left(60-50\right)}$ cm²
                                        =  $\sqrt{60\times 30\times 20\times 10}$ cm²
                                        = $\sqrt{360000}$ cm²
                                        = 600 cm²
For ΔABC:
Let a = 26 cm, b = 28 cm and c = 30 cm
     s = $\left(\frac{a+b+c}{2}\right)=\left(\frac{26+28+30}{2}\right)=\left(\frac{84}{2}\right)=42$ cm
∴  Area of triangle ABC = $\sqrt{s\left(s-a\right)\left(s-b\left(s-c\right)\right)}$  sq. units
                                        = $\sqrt{42\left(42-26\right)\left(42-28\right)\left(42-30\right)}$ cm²
                                        =  $\sqrt{42\times 16\times 14\times 12}$ cm²
                                        = $\sqrt{2\times 3\times 7\times 2\times 2\times 2\times 2\times 2\times 7\times 3\times 2\times 2}$ cm²
                                        = (2 $\times$2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 7) cm²
                                        = 336 cm²
∴ Area of the given quadrilateral ABCD =  Area of ΔACD + Area of ΔABC
                                                                           = (600 + 336) cm² = 936 cm²

Answer:

Area of the rectangle = AB $\times$ BC
                                   = 36 m $\times$ 24 m
                                   = 864 m²
Area of the triangle  = $\frac{1}{2}$ $\times$ AD $\times$ FE
                                 = $\frac{1}{2}$ $\times$ BC $\times$ FE       [since AD = BC]
                                 = $\frac{1}{2}$ $\times$ 24 m $\times$ 15 m
                                 = 12 m $\times$15 m = 180 m²
∴ Area of the shaded region = Area of the rectangle − Area of the triangle
                                              = (864 − 180) m²
                                                         =  684 m²

Answer:

Join points PR and SQ.
These two lines bisect each other at point O.

Here, AB = DC = SQ = 40 cm
AD = BC =RP = 25 cm

Also, OP = OR = $\frac{RP}{2}=\frac{25}{2}$ = 12.5 cm
From the figure we observe:
Area of ΔSPQ = Area of ΔSRQ
∴ Area of the shaded region   = 2 $\times$ (Area of ΔSPQ)
                                                       = 2 $\times$ ($\frac{1}{2}$$\times$ SQ $\times$OP)
                                                       = 2 $\times$ ($\frac{1}{2}$ $\times$ 40 cm $\times$ 12.5 cm)
                                                       = 500 cm²

Answer:

(i) Area of rectangle ABCD = (10 cm x 18 cm) = 180 cm²
    
    Area of triangle I = $\left(\frac{1}{2}\times 6\times 10\right)$ cm² = 30 cm²
    Area of triangle II = $\left(\frac{1}{2}\times 8\times 10\right)$ cm² = 40 cm²
    ∴ Area of the shaded region = {180 - ( 30 + 40)} cm² = { 180 - 70}cm² = 110 cm²

(ii) Area of square ABCD = (Side)² = (20 cm)² = 400 cm²
     
     Area of triangle I = $\left(\frac{1}{2}\times 10\times 20\right)$ cm² = 100 cm²
    Area of triangle II = $\left(\frac{1}{2}\times 10\times 10\right)$ cm² = 50 cm²
    Area of triangle III = $\left(\frac{1}{2}\times 10\times 20\right)$ cm² = 100 cm²
   ∴ Area of the shaded region = {400 - ( 100 + 50 + 100)} cm² = { 400 - 250}cm² = 150 cm²

Answer:

Let ABCD be the given quadrilateral and let BD be the diagonal such that BD is of the length 24 cm.
Let AL ⊥ BD and CM ⊥ BD
Then, AL = 5 cm and CM = 8 cm
Area of the quadrilateral ABCD = (Area of ΔABD + Area of ΔCBD)
                                              =  $\left[\left(\frac{1}{2}\times BD\times AL\right)+\left(\frac{1}{2}\times BD\times CM\right)\right]$ sq. units
                                              =  $\left[\left(\frac{1}{2}\times 24\times 5\right)+\left(\frac{1}{2}\times 24\times 8\right)\right]$ cm²
                                              = ( 60 + 96) cm² = 156 cm²

∴ Area of the given quadrilateral = 156 cm

Answer:

(i) Here, r = 28 cm
     ∴ Circumference = 2π r
                                  = $\left(2\times \frac{22}{7}\times 28\right)$cm
                                  =  176 cm
    Hence, the circumference of the given circle is 176 cm.

(ii) Here, r = 1.4 m
      ∴ Circumference = 2π r
                                   = $\left(2\times \frac{22}{7}\times 1.4\right)$ m
                                   =  $\left(2\times 22\times 0.2\right)$ m = 8.8 m
    Hence, the circumference of the given circle is 8.8 m.

Answer:

(i) Here, d = 35 cm
     Circumference = 2π r
                              = $\left(\mathrm{\pi }d\right)$    [since 2r = d]
                              =  $\left(\frac{22}{7}\times 35\right)$ cm = (22 $\times$ 5) = 110 cm
     Hence, the circumference of the given circle is 110 cm.

(ii) Here, d = 4.9 m
      Circumference =2π r
                              = $\left(\mathrm{\pi }d\right)$    [since 2r = d]
                              =  $\left(\frac{22}{7}\times 4.9\right)$ m = (22 $\times$ 0.7) = 15.4 m
      Hence, the circumference of the given circle is 15.4 m.

Answer:

Here, r = 15 cm
∴ Circumference = $2\mathrm{\pi r}$
                             = ( 2 $\times$ 3.14 $\times$ 15) cm
                             = 94.2 cm
Hence, the circumference of the given circle is 94.2 cm

Answer:

Circumference of the given circle = 57.2 cm
∴ C = 57.2 cm
Let the radius of the given circle be r cm.
C = $2\mathrm{\pi r}$
⇒ r = $\frac{\mathrm{C}}{2\mathrm{\pi }}$ cm
⇒ r = $\left(\frac{57.2}{2}\times \frac{7}{22}\right)$ cm = 9.1 cm
Thus, radius of the given circle is 9.1 cm.

Answer:

Circumference of the given circle = 63.8 m
∴ C = 63.8 m
Let the radius of the given circle be r cm.
C = $2\mathrm{\pi r}$
⇒ r = $\frac{\mathrm{C}}{2\mathrm{\pi }}$
⇒ r = $\left(\frac{63.8}{2}\times \frac{7}{22}\right)$m =10.15 m
∴ Diameter of the given circle = 2r = (2 $\times$ 10.15) m = 20.3 m

Answer:

Let the radius of the given circle be r cm.
Then, its circumference = $2\mathrm{\pi r}$

Given:
(Circumference) - (Diameter) = 30 cm     
∴ ($2\mathrm{\pi r}$ - 2r ) =  30
⇒ $2r\left(\mathrm{\pi }-1\right)=30$
⇒ $2r\left(\frac{22}{7}-1\right)=30$
⇒ $2r\times \frac{15}{7}=30$
⇒ $r=\left(30\times \frac{7}{30}\right)=7$
∴ Radius of the given circle = 7 cm

Answer:

Let the radii of the given circles be 5x and 3x, respectively.
Let their circumferences be C₁ and C_2, respectively.

C₁ = $2\times \pi \times 5x=10\mathrm{\pi }x$

C₂ = $2\times \pi \times 3x=6\mathrm{\pi }x$
∴ $\frac{C_1}{C_2}=\frac{10\pi x}{6\pi x}=\frac{5}{3}$
⇒ C₁:C₂ = 5:3
Hence, the ratio of the circumference of the given circle is 5:3.

Answer:

Radius of the circular field, r = 21 m.
Distance covered by the cyclist = Circumference of the circular field
                                             = $2\mathrm{\pi r}$
                                             = $\left(2\times \frac{22}{7}\times 21\right)$ m = 132 m
Speed of the cyclist = 8 km per hour = $\frac{8000 \mathrm{m}}{(60\times 60) \mathrm{s}}=\left(\frac{8000}{3600}\right)\mathrm{m}/\mathrm{s}$ = $\left(\frac{20}{9}\right)\mathrm{m}/\mathrm{s}$
                                                                                                
Time taken by the cyclist to cover the field = $\frac{\mathrm{Distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{cyclist}}{\mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{cyclist}}$
                                                                     = $\left[\frac{132}{\left({\displaystyle \frac{20}{9}}\right)}\right]\mathrm{s}$
                                                                     = $\left(\frac{132\times 9}{20}\right)\mathrm{s}$
                                                                     = 59.4 s

Answer:

Let the inner and outer radii of the track be r metres and R metres, respectively.

Then, $2\mathrm{\pi r}$ = 528
$2\mathrm{\pi R}$ = 616
⇒ $2\times \frac{22}{7}\times r=528$ 
$2\times \frac{22}{7}\times R=616$
⇒ r = $\left(528\times \frac{7}{44}\right)=84$ 
R = $\left(616\times \frac{7}{44}\right)=98$
⇒ (R - r) = (98 - 84) m = 14 m
Hence, the width of the track is 14 m.

Answer:

Let the inner and outer radii of the track be r metres and (r + 10.5) metres, respectively.

Inner circumference = 330 m
∴ $2\mathrm{\pi r}=330$ ⇒ $2\times \frac{22}{7}\times r=330$
                      ⇒ r = $\left(330\times \frac{7}{44}\right)=52.5 \mathrm{m}$
Inner radius of the track = 52.5 m
∴ Outer radii of the track = (52.5 + 10.5) m = 63 m


∴ Circumference of the outer circle = $\left(2\times \frac{22}{7}\times 63\right) \mathrm{m}=396 \mathrm{m}$
Rate of fencing = Rs. 20 per metre
∴ Total cost of fencing the outer circle = Rs. (396 $\times$ 20) = Rs. 7920

Answer:

We know that the concentric circles are circles that form within each other, around a common centre point.
Radius of the inner circle, r = 98 cm
∴  Circumference of the inner circle = $2\mathrm{\pi r}$
                                                         = $\left(2\times \frac{22}{7}\times 98\right)$ cm = 616 cm

Radius of the outer circle, R = 1 m 26 cm = 126 cm              [since 1 m = 100 cm]
∴  Circumference of the outer circle = $2\mathrm{\pi R}$
                                                           = $\left(2\times \frac{22}{7}\times 126\right)$ cm = 792 cm
∴ Difference in the lengths of the circumference of the circles = (792 - 616) cm = 176 cm
Hence, the circumference of the second circle is 176 cm larger than that of the first circle.

Answer:

Length of the wire = Perimeter of the equilateral triangle
                              = 3 $\times$ Side of the equilateral triangle = (3 $\times$ 8.8) cm = 26.4 cm
Let the wire be bent into the form of a circle of radius r cm.
Circumference of the circle = 26.4 cm
⇒ $2\mathrm{\pi r}=26.4$
⇒ $2\times \frac{22}{7}\times r=26.4$
⇒ r = $\left(\frac{26.4\times 7}{2\times 22}\right)$ cm = 4.2 cm

∴ Diameter = 2r = (2 × 4.2) cm = 8.4 cm
Hence, the diameter of the ring is 8.4 cm.

Answer:

Circumference of the circle = Perimeter of the rhombus
                                           = 4  × Side of the rhombus = (4  ×  33) cm = 132 cm

∴ Circumference of the circle = 132 cm
⇒ $2\mathrm{\pi r}=132$
⇒ $2\times \frac{22}{7}\times r=132$
⇒ r = $\left(\frac{132\times 7}{2\times 22}\right)$cm = 21 cm
Hence, the radius of the circle is 21 cm.

Answer:

Length of the wire = Perimeter of the rectangle
                               = 2(l + b) = 2  × (18.7 + 14.3) cm = 66 cm

Let the wire be bent into the form of a circle of radius r cm.

Circumference of the circle = 66 cm

⇒ $2\mathrm{\pi r}=66$
⇒ $\left(2\times \frac{22}{7}\times r\right)=66$
⇒ r = $\left(\frac{66\times 7}{2\times 22}\right)$ cm = 10.5 cm

Hence, the radius of the circle formed is 10.5 cm.

Answer:

It is given that the radius of the circle is 35 cm.
Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = $2\mathrm{\pi r}$ = $\left(2\times \frac{22}{7}\times 35\right)$ cm = 220 cm
Let the wire be bent into the form of a square of side a cm.
Perimeter of the square = 220 cm
⇒ 4a = 220
⇒ a = $\left(\frac{220}{4}\right)$cm = 55 cm
Hence, each side of the square will be 55 cm.

Answer:

Given:
Diameter of the well (d) = 140 cm.
Radius of the well (r) = $\left(\frac{140}{2}\right)$cm = 70 cm

Let the radius of the outer circle (including the stone parapet) be R cm.
Length of the outer edge of the parapet = 616 cm
⇒ $2\mathrm{\pi R}=616$
⇒ $\left(2\times \frac{22}{7}\times R\right)=616$
⇒ R = $\left(\frac{616\times 7}{2\times 22}\right)$ cm = 98 cm

Now, width of the parapet = {Radius of the outer circle (including the stone parapet) - Radius of the well}
                                           = {98 -70} cm = 28 cm
Hence, the width of the parapet is 28 cm.

Answer:

It may be noted that in one rotation, the bus covers a distance equal to the circumference of the wheel.
Now, diameter of the wheel = 98 cm
∴ Circumference of the wheel = $\mathrm{\pi d}$ = $\left(\frac{22}{7}\times 98\right)$ cm = 308 cm
Thus, the bus travels 308 cm in one rotation.

∴ Distance covered by the bus in 2000 rotations = (308  × 2000) cm
                                                                          = 616000 cm 
                                                                          = 6160 m          [since 1 m = 100 cm]

Answer:

It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Diameter of the wheel = 70 cm
∴ Circumference of the wheel = $\mathrm{\pi d}$ = $\left(\frac{22}{7}\times 70\right)$ cm = 220 cm
Thus, the cycle covers 220 cm in one revolution.

∴ Distance covered by the cycle in 250 revolutions = (220  × 250) cm
                                                                               = 55000 cm 
                                                                               = 550 m              [since 1 m = 100 cm]

Hence, the cycle will cover 550 m in 250 revolutions.

Answer:

Diameter of the wheel = 77 cm
⇒ Radius of the wheel = $\left(\frac{77}{2}\right)$ cm
Circumference of the wheel = $2\mathrm{\pi r}$
                                              = $\left(2\times \frac{22}{7}\times \frac{77}{2}\right)$cm = (22  × 11) cm = 242 cm
                                                                                                        = $\left(\frac{242}{100}\right)$m = $\left(\frac{121}{50}\right)$m
Distance covered by the wheel in 1 revolution = $\left(\frac{121}{50}\right)$ m
Now, $\left(\frac{121}{50}\right)$ m is covered by the car in 1 revolution.
(121  × 1000) m will be covered by the car in $\left(1\times \frac{50}{121}\times 121\times 1000\right)$ revolutions, i.e. 50000 revolutions.
∴ Required number of revolutions = 50000

Answer:

It may be noted that in one revolution, the bicycle covers a distance equal to the circumference of the wheel.
Total distance covered by the bicycle in 5000 revolutions = 11 km

⇒ 5000 × Circumference of the wheel = 11000 m                [since 1 km = 1000 m]

Circumference of the wheel = $\left(\frac{11000}{5000}\right)$ m =2.2 m = 220 cm               [since 1 m = 100 cm]

Circumference of the wheel = $\mathrm{\pi }\times \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{wheel}$
⇒ 220 cm = $\frac{22}{7}\times \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{wheel}$
⇒ Diameter of the wheel = $\left(\frac{220\times 7}{22}\right)$ cm = 70 cm
Hence, the circumference of the wheel is 220 cm and its diameter is 70 cm.

Answer:

Length of the hour hand (r)= 4.2 cm.
Distance covered by the hour hand in 12 hours = $2\mathrm{\pi r}$ = $\left(2\times \frac{22}{7}\times 4.2\right)$ cm = 26.4 cm

∴ Distance covered by the hour hand in 24 hours = (2  × 26.4) = 52.8 cm
Length of the minute hand (R)= 7 cm
Distance covered by the minute hand in 1 hour = $2\mathrm{\pi R}$ = $\left(2\times \frac{22}{7}\times 7\right)$ cm = 44 cm

∴ Distance covered by the minute hand in 24 hours = (44  × 24) cm = 1056 cm

∴ Sum of the distances covered by the tips of both the hands in 1 day = (52.8 + 1056) cm
                                                                                                                = 1108.8 cm

Answer:

(i) Given:
r = 21 cm
     
∴ Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 21\times 21\right)$ cm² = $\left(22\times 3\times 21\right)$ cm² = 1386 cm²

(ii) Given:
r = 3.5 m
     
Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 3.5\times 3.5\right)$ m² = $\left(22\times 0.5\times 3.5\right)$ m² = 38.5 m²

Answer:

(i) Given:
d = 28 cm ⇒ r = $\left(\frac{d}{2}\right)$ = $\left(\frac{28}{2}\right)$ cm = 14 cm
Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 14\times 14\right)$ cm² = $\left(22\times 2\times 14\right)$ cm² = 616 cm²

(ii) Given:
r = 1.4 m ⇒ r = $\left(\frac{d}{2}\right)$ = $\left(\frac{1.4}{2}\right)$m = 0.7 m
      Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 0.7\times 0.7\right)$ m² = $\left(22\times 0.1\times 0.7\right)$ m² = 1.54 m²

Answer:

Let the radius of the circle be r cm.
Circumference = $\left(2\mathrm{\pi r}\right)$cm
∴ $\left(2\mathrm{\pi r}\right)$ = 264
⇒ $\left(2\times \frac{22}{7}\times r\right)=264$
⇒ r = $\left(\frac{264\times 7}{2\times 22}\right)$ = 42
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 42\times 42\right)$ cm²
                                = 5544 cm²

Answer:

Let the radius of the circle be r m.
Then, its circumference will be $\left(2\mathrm{\pi r}\right)$m.
∴ $\left(2\mathrm{\pi r}\right)$ = 35.2
⇒ $\left(2\times \frac{22}{7}\times r\right)=35.2$
⇒ r = $\left(\frac{35.2\times 7}{2\times 22}\right)$ = 5.6
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 5.6\times 5.6\right)$ m² = 98.56 m²

Answer:

Let the radius of the circle be r cm.
Then, its area will be ${\mathrm{\pi r}}^2$ cm².
∴ ${\mathrm{\pi r}}^2$ = 616
⇒ $\left(\frac{22}{7}\times r\times r\right)$ = 616
⇒ r² = $\left(\frac{616\times 7}{22}\right)$ = 196
⇒ r = $\sqrt{196}$ = 14
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ cm
                                                  = $\left(2\times \frac{22}{7}\times 14\right)$ cm = 88 cm

Answer:

Let the radius of the circle be r m.
Then, area = ${\mathrm{\pi r}}^2$ m²
∴ ${\mathrm{\pi r}}^2$ = 1386
⇒ $\left(\frac{22}{7}\times r\times r\right)$ = 1386
⇒ r² = $\left(\frac{1386\times 7}{22}\right)$ = 441
⇒ r = $\sqrt{441}$ = 21
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ m
                                                  = $\left(2\times \frac{22}{7}\times 21\right)$ m = 132 m

Answer:

Let r₁ and r₂ be the radii of the two given circles and A₁ and A₂ be their respective areas.

$\frac{r_1}{r_2}=\frac{4}{5}$
∴ $\frac{A_1}{A_2}=\frac{\mathrm{\pi }{r_1}^2}{\mathrm{\pi }{r_2}^2}=\frac{{r_1}^2}{{r_2}^2}={\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{4}{5}\right)}^2=\frac{16}{25}$
Hence, the ratio of the areas of the given circles is 16:25.

Answer:

If the horse is tied to a pole, then the pole will be the central point and the area over which the horse will graze will be a circle. The string by which the horse is tied will be the radius of the circle.
Thus,
Radius of the circle (r) = Length of the string = 21 m

Now, area of the circle = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 21\times 21\right)$ m² = 1386 m²
∴ Required area = 1386 m²

Answer:

Let a be one side of the square.
Area of the square = 121 cm²                (given)
                 ⇒ a² = 121
                 ⇒ a = 11 cm   (since 11  × 11 = 121)
Perimeter of the square = 4  × side = 4a = (4  × 11) cm = 44 cm
Length of the wire = Perimeter of the square
                               = 44 cm
The wire is bent in the form of a circle.
Circumference of a circle = Length of the wire
∴ Circumference of a circle = 44 cm
⇒ $2\mathrm{\pi r}=44$
⇒ $\left(2\times \frac{22}{7}\times r\right)=44$
⇒ r = $\left(\frac{44\times 7}{2\times 22}\right)$= 7 cm
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 7\times 7\right)$ cm²
                                 = 154 cm²

Answer:

It is given that the radius of the circle is 28 cm.

Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = $2\mathrm{\pi r}=\left(2\times \frac{22}{7}\times 28\right)$ cm = 176 cm
Let the wire be bent into the form of a square of side a cm.

Perimeter of the square = 176 cm

⇒ 4a = 176
⇒ a = $\left(\frac{176}{4}\right)$cm = 44 cm
Thus, each side of the square is 44 cm.

Area of the square = (Side)² = (a)² = (44 cm)²
                                                        = 1936 cm²
∴ Required area of the square formed = 1936 cm²

Answer:

Area of the acrylic sheet = 34 cm  × 24 cm = 816 cm²
Given that the diameter of a circular button is 3.5 cm.
∴ Radius of the circular button (r)= $\left(\frac{3.5}{2}\right)$ cm = 1.75 cm
∴ Area of 1 circular button = ${\mathrm{\pi r}}^2$
                                                 = $\left(\frac{22}{7}\times 1.75\times 1.75\right)$ cm²
                                                 = 9.625 cm²
∴ Area of 64 such buttons = (64  × 9.625) cm² = 616 cm²
Area of the remaining acrylic sheet = (Area of the acrylic sheet - Area of 64 circular buttons)
                                                                    = (816 - 616) cm² = 200 cm²

Answer:

Area of the rectangular ground = 90 m  × 32 m = (90  × 32) m² = 2880 m²
Given:
Radius of the circular tank (r) = 14 m
∴ Area covered by the circular tank = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 14\times 14\right)$ m²
                                                               = 616 m²

∴ Remaining portion of the rectangular ground for turfing = (Area of the rectangular ground - Area covered by the circular tank)
                                                                                          = (2880 - 616) m² = 2264 m²
Rate of turfing = Rs 50 per sq. metre
∴ Total cost of turfing the remaining ground = Rs (50  × 2264) = Rs 1,13,200

Answer:

Area of each of the four quadrants is equal to each other with radius 7 cm.

Area of the square ABCD = (Side)² = (14 cm)² = 196 cm²
Sum of the areas of the four quadrants = $\left(4\times \frac{1}{4}\times \frac{22}{7}\times 7\times 7\right)$ cm²
                                                        = 154 cm²
∴ Area of the shaded portion  = Area of square ABCD - Areas of the four quadrants
                                          = (196 - 154) cm²
                                          = 42 cm²

Answer:

Let ABCD be the rectangular field.

Here, AB = 60 m
BC = 40 m

Let the horse be tethered to corner A by a 14 m long rope.

Then, it can graze through a quadrant of a circle of radius 14 m.
∴ Required area of the field = $\left(\frac{1}{4}\times \frac{22}{7}\times 14\times 14\right)$ m² = 154 m²
Hence, horse can graze 154 m² area of the rectangular field.

Answer:

Diameter of the big circle = 21 cm
Radius = $\left(\frac{21}{2}\right)$ cm = 10.5 cm
∴ Area of the bigger circle = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 10.5\times 10.5\right)$ cm²
                                          = 346.5 cm²


Diameter of circle I  = $\frac{2}{3}$ of the diameter of the bigger circle
                                 = $\frac{2}{3}$ of 21 cm = $\left(\frac{2}{3}\times 21\right)$ cm = 14 cm
Radius of circle I (r₁) = $\left(\frac{14}{2}\right)$ cm = 7 cm
∴ Area of circle I = ${{\mathrm{\pi r}}_1}^2$ = $\left(\frac{22}{7}\times 7\times 7\right)$ cm²
                                         = 154 cm²

Diameter of circle II  = $\frac{1}{3}$ of the diameter of the bigger circle
                                   = $\frac{1}{3}$ of 21 cm = $\left(\frac{1}{3}\times 21\right)$ cm = 7 cm
Radius of circle II (r₂) = $\left(\frac{7}{2}\right)$ cm = 3.5 cm
∴ Area of circle II = ${{\mathrm{\pi r}}_2}^2$ = $\left(\frac{22}{7}\times 3.5\times 3.5\right)$ cm²
                                          = 38.5 cm²

∴ Area of the shaded portion = {Area of the bigger circle - (Sum of the areas of circle I and II)}
                                      = {346.5 - (154 + 38.5)} cm²
                                      = {346.5 - 192.5} cm²
                                              = 154 cm²
Hence, the area of the shaded portion is 154 cm²

Answer:

Let ABCD be the rectangular plot of land that measures 8 m by 6 m.
∴ Area of the plot = (8 m  × 6 m) = 48 m²
Area of the four flower beds = $\left(4\times \frac{1}{4}\times \frac{22}{7}\times 2\times 2\right)$ m² = $\left(\frac{88}{7}\right)$ m²
Area of the circular flower bed in the middle of the plot = ${\mathrm{\pi r}}^2$
                                                                                       = $\left(\frac{22}{7}\times 2\times 2\right)$ m² = $\left(\frac{88}{7}\right)$ m²

Area of the remaining part = $\left\{48-\left(\frac{88}{7}+\frac{88}{7}\right)\right\}$ m²
                                           = $\left\{48-\frac{176}{7}\right\}$ m²
                                           = $\left\{\frac{336-176}{7}\right\}$ m² = $\left(\frac{160}{7}\right)$ m² = 22.86 m²
∴ Required area of the remaining plot = 22.86 m²

Answer:

(c) 192 cm²

Let ABCD be the rectangular plot.
Then, AB = 16 cm
          AC = 20 cm

Let BC = x cm
From right triangle ABC:
AC² = AB² + BC²
⇒ (20)² = (16)² + x²
⇒ x² = (20)² - (16)² ⇒ {400 - 256} = 144
⇒ x = $\sqrt{144}$ = 12
∴ BC = 12 cm
∴ Area of the plot = (16 × 12) cm² = 192 cm²

Answer:

(b) 72 cm²

Given:
Diagonal of the square = 12 cm
∴ Area of the square = $\left\{\frac{1}{2}\times {\left(\mathrm{Diagonal}\right)}^2\right\}$ sq. units.
                                  = $\left\{\frac{1}{2}\times {\left(12\right)}^2\right\}$ cm²
                                  = 72 cm²

Answer:

(b) 20 cm

Area of the square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$ sq. units.
Area of the square field = 200 cm²

Diagonal of a square = $\sqrt{2\times \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{square}}$ 
                                  = $\left(\sqrt{2\times 200}\right)$ cm = $\left(\sqrt{400}\right)$ cm = 20 cm
∴ Length of the diagonal of the square = 20 cm

Answer:

(a) 100 m

Area of the square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$sq. units.
Given:
Area of square field = 0.5 hectare
                                 = $\left(0.5\times 10000\right)$m²                     [since 1 hectare = 10000 m²]
                                 = 5000 m²                  

Diagonal of a square = $\sqrt{2\times \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{square}}$ 
                                  = $\left(\sqrt{2\times 5000}\right)$m = 100 m
Hence, the length of the diagonal of a square field is 100 m.

Answer:

(c) 90 m

Let the breadth of the rectangular field be x m.
Length = 3x m
Perimeter of the rectangular field = 2(l + b)
⇒ 240 = 2( x + 3x)
⇒ 240 = 2(4x)
⇒ 240 = 8x     ⇒ x = $\left(\frac{240}{8}\right)=30$
∴ Length of the field = 3x = (3 × 30) m = 90 m

Answer:

(d) 56.25%

Let the side of the square be a cm.
Area of the square = (a)² cm²
Increased side = (a + 25% of a) cm
                        = $\left(a+\frac{25}{100}a\right)$ cm = $\left(a+\frac{1}{4}a\right)\mathrm{cm}=\left(\frac{5}{4}a\right)$ cm
Area of the square = ${\left(\frac{5}{4}a\right)}^{2}c{\mathrm{m}}^2=\left(\frac{25}{16}{a}^2\right)$ cm²
Increase in the area = $\left[\left(\frac{25}{16}{a}^2\right)-a^2\right]$ cm²= $\left(\frac{25a^2-16a^2}{16}\right)$ cm² = $\left(\frac{9a^2}{16}\right)$ cm²
% increase in the area = $\frac{\mathrm{Increased} \mathrm{area}}{\mathrm{Old} \mathrm{area}}\times 100$
                              = $\left[\frac{\left({\displaystyle \frac{9}{16}}{a}^2\right)}{a^2}\times 100\right]$ = $\left(\frac{9\times 100}{16}\right)=56.25$

Answer:

(b) 1:2

Let the side of the square be a.
Length of its diagonal = $\sqrt{2}a$
∴ Required ratio = $\frac{a^2}{{\left(\sqrt{2}a\right)}^2}=\frac{a^2}{2a^2}=\frac{1}{2}=1:2$

Answer:

(c) A > B

We know that a square encloses more area even though its perimeter is the same as that of the rectangle.

∴ Area of a square  > Area of a rectangle

Answer:

(b) 13500 m²

Let the length of the rectangular field be 5x.
Breadth = 3x
Perimeter of the field = 2(l + b) = 480 m          (given)
⇒ 480 = 2(5x + 3x)  ⇒ 480 = 16x
⇒ x = $\frac{480}{16}$ = 30
∴ Length = 5x = (5 × 30) = 150 m
Breadth = 3x = (3 × 30) = 90 m
∴ Area of the rectangular park = 150 m × 90 m = 13500 m²

Answer:

(a) 6 m

Total cost of carpeting = Rs 6000
Rate of carpeting = Rs 50 per m
∴ Length of the carpet = $\left(\frac{6000}{50}\right)$ m = 120 m
∴ Area of the carpet = $\left(120\times \frac{75}{100}\right)$ m² =  90 m²     [since 75 cm = $\frac{75}{100} \mathrm{m}$]
Area of the floor = Area of the carpet = 90 m²
∴ Width of the room = $\left(\frac{\mathrm{Area}}{\mathrm{Length}}\right)=\left(\frac{90}{15}\right) \mathrm{m}=6 \mathrm{m}$

Answer:

(a) 84 cm²

Let a = 13 cm, b = 14 cm and c = 15 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{13+14+15}{2}\right)$ cm = 21 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                               = $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$ cm²
                               =  $\sqrt{21\times 8\times 7\times 6}$ cm²
                               = $\sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}$ cm²
                               = (2 × 2 × 3 × 7) cm²
                               = 84 cm²

Answer:

(b) 48 m²

Base = 12 m
Height = 8 m
Area of the triangle = $\left(\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\right)$ sq. units
                          = $\left(\frac{1}{2}\times 12\times 8\right)$ m²
                          = 48 m²

Answer:

(b) 4 cm

Area of the equilateral triangle = $4\sqrt{3}$ cm²

We know:
Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2$ sq. units
∴ Side of the equilateral triangle = $\left[\sqrt{\left(\frac{4\times \mathrm{Area}}{\sqrt{3}}\right)}\right]$ cm
                                                    =   $\left[\sqrt{\left(\frac{4\times 4\sqrt{3}}{\sqrt{3}}\right)}\right]$cm = $\left(\sqrt{4\times 4}\right)$ cm = $\left(\sqrt{16}\right)$cm = 4 cm

Answer:

(c) $16\sqrt{3}$ cm²

It is given that one side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$ sq. units
                                                      =  $\frac{\sqrt{3}}{4}{\left(8\right)}^2$ cm²
                                                      = $\left(\frac{\sqrt{3}}{4}\times 64\right)$ cm² = $16\sqrt{3}$ cm²

Answer:

(b) $2\sqrt{3}$ cm²

Let ΔABC be an equilateral triangle with one side of the length a cm.
Diagonal of an equilateral triangle = $\frac{\sqrt{3}}{2}a$ cm
⇒ $\frac{\sqrt{3}}{2}a=\sqrt{6}$
⇒ a = $\frac{\sqrt{6}\times 2}{\sqrt{3}}=\frac{\sqrt{3}\times \sqrt{2}\times 2}{\sqrt{3}}=2\sqrt{2}$ cm
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{a}^2$
                                                 = $\frac{\sqrt{3}}{4}{\left(2\sqrt{2}\right)}^2$ cm² = $\left(\frac{\sqrt{3}}{4}\times 8\right)$ cm² = $2\sqrt{3}$ cm²

Answer:

(b) 72 cm²

Base of the parallelogram = 16 cm
Height of the parallelogram = 4.5 cm
∴ Area of the parallelogram = Base × Height
                                              = (16 × 4.5) cm²  = 72 cm²

Answer:

(b) 216 cm²

Length of one diagonal = 24 cm
 Length of the other diagonal = 18 cm
     ∴ Area of the rhombus = $\frac{1}{2}$ × (Product of the diagonals)
                                           = $\left(\frac{1}{2}\times 24\times 18\right)$ cm² = 216 cm²

Answer:

(c) 154 cm²

Let the radius of the circle be r cm.
Circumference = $2\mathrm{\pi r}$

(Circumference) - (Radius) = 37
∴ $\left(2\mathrm{\pi r} -\mathrm{r}\right)=37$
⇒ $r\left(2\mathrm{\pi }-1\right)=37$
⇒ r = $\frac{37}{\left(2\mathrm{\pi }-1\right)}$ = $\frac{37}{\left(2\times {\displaystyle \frac{22}{7}}-1\right)}=\frac{37}{\left({\displaystyle \frac{44}{7}}-1\right)}=\frac{37}{\left({\displaystyle \frac{44-7}{7}}\right)}=\left(\frac{37\times 7}{37}\right)=7$
∴ Radius of the given circle is 7 cm.
∴ Area = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 7\times 7\right)$ cm² = 154 cm²

Answer:

(c) 54 m²

Given:
Perimeter of the floor = 2(l + b) = 18 m      
Height of the room = 3 m        
                    
∴ Area of the four walls = {2(l + b) × h}
                                        = Perimeter × Height
                                        = 18 m × 3 m = 54 m²

Answer:

(a) 200 m

Area of the floor of a room = 14 m × 9 m = 126 m²

Width of the carpet = 63 cm = 0.63 m                (since 100 cm = 1 m)

∴ Required length of the carpet = $\frac{\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{of} \mathrm{a} \mathrm{room}}{\mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{carpet}}$
                                                = $\left(\frac{126}{0.63}\right) \mathrm{m}=200 \mathrm{m}$

Answer:

(c) 120 cm²

Let the length of the rectangle be x cm and the breadth be y cm.
Area of the rectangle = xy cm²
Perimeter of the rectangle = 2( x + y) = 46 cm          (given)
⇒ 2( x + y) = 46
⇒ ( x + y) = $\left(\frac{46}{2}\right)$ cm = 23 cm

Diagonal of the rectangle = $\sqrt{x^2+y^2}$ = 17 cm
⇒  $\sqrt{x^2+y^2}$ = 17

Squaring both the sides, we get:
⇒ x² + y² = (17)²
⇒ x² + y² = 289

 Now, (x² + y²) = ( x + y)² - 2xy
⇒ 2xy = ( x + y)² - (x² + y²)
           = (23)² - 289
           = 529 - 289 = 240
∴ xy = $\left(\frac{240}{2}\right)$ cm² = 120 cm²

Answer:

(b) 3:1

Let a side of the first square be a cm and that of the second square be b cm.
Then, their areas will be a² and b², respectively.
Their perimeters will be 4a and 4b, respectively.

According to the question:
$\frac{a^2}{b^2}=\frac{9}{1}$⇒ ${\left(\frac{a}{b}\right)}^2=\frac{9}{1}={\left(\frac{3}{1}\right)}^2$⇒ $\frac{a}{b}=\frac{3}{1}$

∴ Required ratio of the perimeters = $\frac{4a}{4b}=\frac{4\times 3}{4\times 1}=\frac{3}{1}$= 3:1

Answer:

(d) 4:1

Let the diagonals be 2d and d.
Area of the square = $\left[\frac{1}{2}\times {\left(\mathrm{Diagonal}\right)}^2\right]$ sq. units
 Required ratio = $\frac{A_1}{A_2}=\frac{{\displaystyle \frac{1}{2}}{\left(2d\right)}^2}{{\displaystyle \frac{1}{2}}{\left(d\right)}^2}=\frac{4d^2}{d^2}=\frac{4}{1}=4:1$

Answer:

(c) 49 m

Let the width of the rectangle be x m.

Given:
Area of the rectangle = Area of the square
⇒ Length × Width = Side × Side
⇒ (144 × x) = 84 × 84
∴ Width (x) = $\left(\frac{84\times 84}{144}\right)$ m = 49 m
Hence, width of the rectangle is 49 m.

Answer:

(d) $4:\sqrt{3}$

Let one side of the square and that of an equilateral triangle be the same, i.e. a units.
Then, Area of the square = (Side)² = (a)²
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$ = $\frac{\sqrt{3}}{4}{\left(\mathrm{a}\right)}^2$
∴ Required ratio = $\frac{a^2}{{\displaystyle \frac{\sqrt{3}}{4}}{a}^2}=\frac{4}{\sqrt{3}}=4:\sqrt{3}$

Answer:

(a) $\sqrt{\mathrm{\pi }}:1$

Let the side of the square be x cm and the radius of the circle be r cm.
Area of the square = Area of the circle
⇒ (x)² = ${\mathrm{\pi r}}^2$
∴ Side of the square (x) = $\sqrt{\mathrm{\pi }}r$
Required ratio = $\frac{\mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{square}}{\mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{circle}}$
                        = $\frac{x}{r}=\frac{\sqrt{\mathrm{\pi }}r}{r}=\frac{\sqrt{\mathrm{\pi }}}{1}=\sqrt{\mathrm{\pi }}:1$