Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 15 - Perimeter And Area Of Plane Figures

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Page No 683:

Answer:

Given: $base = 24 cm, correponding height = 14.5 cm$
Area of a triangle = $\frac{1}{2} \times base \times coresponding height$

$= \frac{1}{2} \times 24 \times 14.5 = 174 {cm}^{2}$

Page No 683:

Answer:

Let the sides of the triangle be a = 20 cm, b = 34 cm and c = 42 cm.
Let s be the semi-perimeter of the triangle.
$s = \frac{1}{2} (a + b + c) s = \frac{1}{2} (20 + 34 + 42) s = 48 cm$

$Area of t h e triangle = \sqrt{s (s - a) (s - b) (s - c)} \Rightarrow \sqrt{48 (48 - 20) (48 - 34) (48 - 42)} \Rightarrow \sqrt{48 \times 28 \times 14 \times 6} \Rightarrow \sqrt{112896} \Rightarrow 336 {cm}^{2}$

Length of the longest side is 42 cm.

Area of a triangle = $\frac{1}{2} \times b \times h$
$\Rightarrow 336 = \frac{1}{2} \times 42 \times h \Rightarrow 672 = 42 h \Rightarrow \frac{672}{42} = h \Rightarrow h = 16 cm$

The height corresponding to the longest side is 16 cm.

Page No 683:

Answer:

Let the sides of triangle be a = 18 cm, b = 24 cm and c = 30 cm.
Let s be the semi-perimeter of the triangle.
$s = \frac{1}{2} (a + b + c) s = \frac{1}{2} (18 + 24 + 30) s = 36 cm$

$Area of a t riangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{36 (36 - 18) (36 - 24) (36 - 30)} = \sqrt{36 \times 18 \times 12 \times 6} = \sqrt{46656} = 216 {cm}^{2}$
The smallest side is 18 cm long. This is the base.

Now, area of a triangle = $\frac{1}{2} \times b \times h$
$\Rightarrow 216 = \frac{1}{2} \times 18 \times h \Rightarrow 216 = 9 h \Rightarrow \frac{216}{9} = h \Rightarrow h = 24 cm$

The height corresponding to the smallest side is 24 cm.

Page No 683:

Answer:

Let the sides of a triangle be 5x m ,12x m and 13x m.

Since, perimeter is the sum of all the sides,
$5 x + 12 x + 13 x = 150 \Rightarrow 30 x = 150 o r, x = \frac{150}{30} = 5$

The lengths of the sides are:
$a = 5 \times 5 = 25 m b = 12 \times 5 = 60 m c = 13 \times 5 = 65 m Semiperimeter (s) of the triangle = \frac{Perimeter}{2} = \frac{25 + 60 + 65}{2} = \frac{150}{2} = 75 m Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{75 (75 - 25) (75 - 60) (75 - 65)} = \sqrt{75 \times 50 \times 15 \times 10} = \sqrt{562500} = 750 m^{2}$

Page No 683:

Answer:

$Let the sides of the triangular field be 25 x, 17 x and 12 x . As, perimeter = 540 m \Rightarrow 25 x + 17 x + 12 x = 540 \Rightarrow 54 x = 540 \Rightarrow x = \frac{540}{54} \Rightarrow x = 10 So, the sides are 250 m, 170 m and 120 m . Now, semi - perimeter, s = \frac{250 + 170 + 120}{2} = \frac{540}{2} = 270 m So, area of the field = \sqrt{270 (270 - 250) (270 - 170) (270 - 120)} = \sqrt{270 \times 20 \times 100 \times 150} = \sqrt{3^{3} \times 10 \times 2 \times 10 \times 10^{2} \times 3 \times 5 \times 10} = 3^{2} \times 10^{3} = 9000 m^{2} Also, the cost of ploughing the field = \frac{9000 \times 40}{100} = ₹ 3, 600$

Page No 683:

Answer:

The perimeter of a right-angled triangle = 40 cm
Therefore , a+b+c= 40 cm
Hypotenuse = 17 cm
Therefore, c = 17 cm
a+b+c= 40 cm
$\Rightarrow$ a+b+17 = 40
$\Rightarrow$ a+b = 23
$\Rightarrow$ b = 23 $-$ a...........(i)
Now, using Pythagoras' theorem, we have:
$a^{2} + b^{2} = c^{2} \Rightarrow a^{2} + (23 - a)^{2} = 17^{2} \Rightarrow a^{2} + 529 - 46 a + a^{2} = 289 \Rightarrow 2 a^{2} - 46 a + 529 - 289 = 0 \Rightarrow 2 a^{2} - 46 a + 240 = 0 \Rightarrow a^{2} - 23 a + 120 = 0 \Rightarrow (a - 15) (a - 8) = 0 \Rightarrow a = 15 o r a = 8$

Substituting the value of a=15, in equation(i) we get:
b = 23-a
= 23 - 15
= 8 cm

If we had chosen $a = 8 cm$ , then, $b = 23 - 8 = 15 cm$

In any case,
$Area of a t riangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times 8 \times 15 = 60 {cm}^{2}$

Page No 683:

Answer:

Given:
Area of the triangle = $60 {cm}^{2}$
Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.
$a - b = 7 cm$
a = 7 + b.......(1)
$Area of triangle = \frac{1}{2} \times b \times h$
$\Rightarrow 60 = \frac{1}{2} \times b \times (7 + b) \Rightarrow 120 = 7 b + b^{2} \Rightarrow b^{2} + 7 b - 120 = 0 \Rightarrow (b + 15) (b - 8) = 0 \Rightarrow b = - 15 or 8$

Side of a triangle cannot be negative.
Therefore, b = 8 cm.

Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm

Now, a = 15 cm, b = 8 cm

Now, in the given right triangle, we have to find third side.
$(Hyp)^{2} = (F i r s t side)^{2} + (S e c o n d side)^{2} \Rightarrow {Hyp}^{2} = 8^{2} + 15^{2} \Rightarrow {Hyp}^{2} = 64 + 225 \Rightarrow {Hyp}^{2} = 289 \Rightarrow Hyp = 17 cm$
So, the third side is 17 cm.

Perimeter of a triangle = $a + b + c$ .
Therefore, required perimeter of the triangle $= 15 + 8 + 17 = 40 cm$ .

Page No 683:

Answer:

Given:
Area of triangle = $24 {cm}^{2}$
Let the sides be a and b, where a is the height and b is the base of triangle.

$a - b = 2 cm$
a = 2 + b.......(1)

$Area of triangle = \frac{1}{2} \times b \times h$

$\Rightarrow 24 = \frac{1}{2} \times b \times (2 + b) \Rightarrow 48 = b + \frac{1}{2} b^{2} \Rightarrow 48 = 2 b + b^{2} \Rightarrow b^{2} + 2 b - 48 = 0 \Rightarrow (b + 8) (b - 6) = 0 \Rightarrow b = - 8 or 6$
Side of a triangle cannot be negative.

Therefore, b = 6 cm.

Substituting the value of b = 6 cm in equation(1), we get:
a = 2+6 = 8 cm

Now, a = 8 cm, b = 6 cm

In the given right triangle we have to find third side. Using the relation

$(Hyp)^{2} = (Oneside)^{2} + (Otherside)^{2} \Rightarrow {Hyp}^{2} = 8^{2} + 6^{2} \Rightarrow {Hyp}^{2} = 64 + 36 \Rightarrow {Hyp}^{2} = 100 \Rightarrow Hyp = 10 cm$

So, the third side is 10 cm.

So, perimeter of the triangle = a + b + c
= 8+6+10
=24 cm

Page No 683:

Answer:

$(i) The area of the equilateral triangle = \frac{\sqrt{3}}{4} \times {side}^{2} = \frac{\sqrt{3}}{4} \times 10^{2} = \frac{\sqrt{3}}{4} \times 100 = 25 \sqrt{3} {cm}^{2} or 25 \times 1.732 = 43.3 {cm}^{2} So, the area of the triangle is 25 \sqrt{3} {cm}^{2} or 43.3 {cm}^{2} . (ii) As, area of the equilateral triangle = 25 \sqrt{3} {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 25 \sqrt{3} \Rightarrow \frac{1}{2} \times 10 \times Height = 25 \sqrt{3} \Rightarrow 5 \times Height = 25 \sqrt{3} \Rightarrow Height = \frac{25 \sqrt{3}}{5} = 5 \sqrt{3} or height = 5 \times 1.732 = 8.66 m ∴ The height of the triangle is 5 \sqrt{3} cm or 8.66 cm .$

Page No 683:

Answer:

Let the side of the equilateral triangle be x cm.

$As, the area of an equilateral triangle = \frac{\sqrt{3}}{4} {(side)}^{2} = \frac{x^{2} \sqrt{3}}{4} Also, the area of the triangle = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times x \times 6 = 3 x So, \frac{x^{2} \sqrt{3}}{4} = 3 x \Rightarrow \frac{x \sqrt{3}}{4} = 3 \Rightarrow x = \frac{12}{\sqrt{3}} \Rightarrow x = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow x = \frac{12 \sqrt{3}}{3} \Rightarrow x = 4 \sqrt{3} cm Now, area of the equilateral triangle = 3 x = 3 \times 4 \sqrt{3} = 12 \sqrt{3} = 12 \times 1.73 = 20.76 {cm}^{2}$

Page No 683:

Answer:

Area of equilateral triangle = $36 \sqrt{3} {cm}^{2}$

Area of equilateral triangle = $(\frac{\sqrt{3}}{4} \times a^{2})$ , where a is the length of the side.
$\Rightarrow 36 \sqrt{3} = \frac{\sqrt{3}}{4} \times a^{2} \Rightarrow 144 = a^{2} \Rightarrow a = 12 cm$

Perimeter of a triangle = 3a
$= 3 \times 12 = 36 cm$

Page No 683:

Answer:

Area of the equilateral triangle = $81 \sqrt{3} {cm}^{2}$
Area of an equilateral triangle = $(\frac{\sqrt{3}}{4} \times a^{2})$ , where a is the length of the side.
$\Rightarrow 81 \sqrt{3} = \frac{\sqrt{3}}{4} \times a^{2} \Rightarrow 324 = a^{2} \Rightarrow a = 18 cm$
Height of triangle = $\frac{\sqrt{3}}{2} \times a$

$= \frac{\sqrt{3}}{2} \times 18 = 9 \sqrt{3} cm$

Page No 683:

Answer:

Base = 48 cm
Hypotenuse =50 cm
First we will find the height of the triangle; let the height be 'p'.
$\Rightarrow {(Hypotenuse)}^{2} = {(b a s e)}^{2} + p^{2} \Rightarrow 50^{2} = 48^{2} + p^{2} \Rightarrow p^{2} = 50^{2} - 48^{2} \Rightarrow p^{2} = (50 - 48) (50 + 48) \Rightarrow p^{2} = 2 \times 98 \Rightarrow p^{2} = 196 \Rightarrow p = 14 cm$

Area of the triangle= $\frac{1}{2} \times base \times height$

$= \frac{1}{2} \times 48 \times 14 = 336 {cm}^{2}$

Page No 683:

Answer:

Hypotenuse = 65 cm
Base = 60 cm
In a right-angled triangle,
${(Hypotenuse)}^{2} = {(B ase)}^{2} + {(P erpendicular)}^{2} \Rightarrow {(65)}^{2} = {(60)}^{2} + {(perpendicular)}^{2} \Rightarrow {(65)}^{2} - {(60)}^{2} = {(perpendicular)}^{2} \Rightarrow {(perpendicular)}^{2} = (65 - 60) (65 + 60) \Rightarrow {(perpendicular)}^{2} = 5 \times 125 \Rightarrow {(perpendicular)}^{2} = 625 \Rightarrow perpendicular = 25 cm$

Area of triangle = $\frac{1}{2} \times base \times perpendicular$

$= \frac{1}{2} \times 60 \times 25 = 750 {cm}^{2}$

Page No 684:

Answer:

Given: Radius = 8 cm
Height = 6 cm
Area=?
In a right-angled triangle, the centre of the circumcircle is the mid-point of the hypotenuse.

$Hypotenuse = 2 \times (radius of circumcircle) for a right triangle = 2 \times 8 = 16 cm So, h ypotenuse = 16 cm$
Now, base = 16 cm and height = 6 cm

Area of the triangle = $\frac{1}{2} \times b a s e \times h e i g h t$

$= \frac{1}{2} \times 16 \times 6 = 48 {cm}^{2}$

Page No 684:

Answer:

In a right isosceles triangle, $base = height = a$

Therefore,
$Area of the triangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times a \times a = \frac{1}{2} a^{2}$

Further, given that area of isosceles right triangle = 200 cm²

$\Rightarrow \frac{1}{2} a^{2} = 200 \Rightarrow a^{2} = 400 o r, a = \sqrt{400} = 20 cm$

In an isosceles right triangle, two sides are equal ('a') and the third side is the hypotenuse, i.e. 'c'

Therefore, c = $\sqrt{a^{2} + a^{2}}$

    $= \sqrt{2 a^{2}} = a \sqrt{2} = 20 \times 1.41 = 28.2 cm$

Perimeter of the triangle = $a + a + c$
                           $= 20 + 20 + 28.2$

                            = 68.2 cm

The length of the hypotenuse is 28.2 cm and the perimeter of the triangle is 68.2 cm.

Page No 684:

Answer:

Given :
Base = 80 cm
Area = 360 ${cm}^{2}$
Area of an isosceles triangle = $(\frac{1}{4} b \sqrt{4 a^{2} - b^{2}})$

$\Rightarrow 360 = \frac{1}{4} \times 80 \sqrt{4 a^{2} - 80^{2}} \Rightarrow 360 = 20 \sqrt{4 a^{2} - 6400} \Rightarrow 18 = 2 \sqrt{a^{2} - 1600} \Rightarrow 9 = \sqrt{a^{2} - 1600}$
Squaring both the sides, we get:
$\Rightarrow 81 = a^{2} - 1600 \Rightarrow a^{2} = 1681 \Rightarrow a = 41 cm$
Perimeter $= (2 a + b)$
$= [2 (41) + 80] = 82 + 80 = 162 cm$
So, the perimeter of the triangle is 162 cm.

Page No 684:

Answer:

Let the height of the triangle be h cm.
Each of the equal sides measures $a = (h + 2) cm$ and b = 12 cm (base).

Now,
Area of the triangle = Area of the isosceles triangle
$\frac{1}{2} \times base \times height = \frac{1}{4} \times b \sqrt{4 a^{2} - b^{2}}$

$\Rightarrow \frac{1}{2} \times 12 \times h = \frac{1}{4} \times 12 \times \sqrt{4 (h + 2)^{2} - 144} \Rightarrow 6 h = 3 \sqrt{4 h^{2} + 16 h + 16 - 144} \Rightarrow 2 h = \sqrt{4 h^{2} + 16 h + 16 - 144} On squaring both the sides, we get : \Rightarrow 4 h^{2} = 4 h^{2} + 16 h + 16 - 144 \Rightarrow 16 h - 128 = 0 \Rightarrow h = 8$

Area of the triangle = $\frac{1}{2} \times b \times h$
$= \frac{1}{2} \times 12 \times 8 = 48 {cm}^{2}$

Page No 684:

Answer:

Let:
Length of each of the equal sides of the isosceles right-angled triangle = a = 10 cm
And,
Base = Height = a
$Area of isosceles right - angled triangle = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 10 \times 10 = 50 {cm}^{2}$

The hypotenuse of an isosceles right-angled triangle can be obtained using Pythagoras' theorem.

If h denotes the hypotenuse, we have:

$h^{2} = a^{2} + a^{2} \Rightarrow h = 2 a^{2} \Rightarrow h = \sqrt{2} a \Rightarrow h = 10 \sqrt{2} cm$

∴ Perimeter of the isosceles right-angled triangle = $2 a + \sqrt{2} a$
$= 2 \times 10 + 1.41 \times 10 = 20 + 14.1 = 34.1 cm$

Page No 684:

Answer:

Given:
Side of equilateral triangle ABC = 10 cm
BD = 8 cm
Area of $equilateral ∆ A B C = \frac{\sqrt{3}}{4} a^{2}$ (where a = 10 cm)
$Area of equilateral △ A B C = \frac{\sqrt{3}}{4} \times 10^{2} = 25 \sqrt{3} = 25 \times 1.732 = 43.30 {cm}^{2}$

$In the right ∆ B D C, we have : B C^{2} = B D^{2} + C D^{2} \Rightarrow 10^{2} = 8^{2} + C D^{2} \Rightarrow C D^{2} = 10^{2} - 8^{2} \Rightarrow C D^{2} = 36 \Rightarrow C D = 6$

Area of triangle $△$ BCD = $\frac{1}{2} \times b \times h$

                $= \frac{1}{2} \times 8 \times 6 = 24 {cm}^{2}$

Area of the shaded region = $Area of ∆ A B C - Area of ∆ B D C$
   = 43.30 $-$ 24
           = 19.3 cm²

Page No 691:

Answer:

$As, perimeter = 80 m \Rightarrow 2 (length + breadth) = 80 \Rightarrow 2 (length + 16) = 80 \Rightarrow 2 \times length + 32 = 80 \Rightarrow 2 \times length = 80 - 32 \Rightarrow length = \frac{48}{2} ∴ length = 24 m Now, the area of the plot = length \times breadth = 24 \times 16 = 384 m^{2}$

So, the length of the plot is 24 m and its area is 384 m².

Page No 691:

Answer:

Let the breadth of the rectangular park be $b$ .
∴ Length of the rectangular park $= l = 2 b$
Perimeter = 840 m

$\Rightarrow 840 = 2 (l + b) \Rightarrow 840 = 2 (2 b + b) \Rightarrow 840 = 2 (3 b) \Rightarrow 840 = 6 b \Rightarrow b = 140 m Thus, we have : l = 2 b = 2 \times 140 = 280 m$

$Area = l \times b = 280 \times 140 = 39200 m^{2}$

Page No 692:

Answer:

One side of the rectangle = 12 cm
Diagonal of the rectangle = 37 cm

The diagonal of a rectangle forms the hypotenuse of a right-angled triangle. The other two sides of the triangle are the length and the breadth of the rectangle.

Now, using Pythagoras' theorem, we have:

${(one side)}^{2} + {(other side)}^{2} = {(hypotenuse)}^{2}$
$\Rightarrow {(12)}^{2} + {(other side)}^{2} = {(37)}^{2} \Rightarrow 144 + {(other side)}^{2} = 1369 \Rightarrow {(other side)}^{2} = 1369 - 144 \Rightarrow {(other side)}^{2} = 1225 \Rightarrow other side = \sqrt{1225} \Rightarrow other side = 35 cm$

Thus, we have:
Length = 35 cm
Breadth = 12 cm
$Area of the rectangle = 35 \times 12 = 420 {cm}^{2}$

Page No 692:

Answer:

Area of the rectangular plot = 462 m²
Length (l) = 28 m
$Area of a rectangle = Length (l) x Breadth (b) \Rightarrow 462 = 28 x b \Rightarrow b = 16.5 m$

Perimeter of the plot = $2 (l + b)$
$= 2 (28 + 16.5) = 2 \times 44.5 = 89 m$

Page No 692:

Answer:

Let the length and breadth of the rectangular lawn be 5x m and 3x m, respectively.

Given:
Area of the rectangular lawn = $3375 m^{2}$

$\Rightarrow 3375 = 5 x \times 3 x \Rightarrow 3375 = 15 x^{2} \Rightarrow \frac{3375}{15} = x^{2} \Rightarrow 225 = x^{2} \Rightarrow x = 15$

Thus, we have:

$l = 5 x = 5 \times 15 = 75 m b = 3 x = 3 \times 15 = 45 m$

$Perimeter of the rectangular lawn = 2 (l + b) = 2 (75 + 45) = 2 (120) = 240 m$

Cost of fencing 1 m lawn = Rs 65
∴ Cost of fencing 240 m lawn = $240 \times 65 = Rs 15, 600$

Page No 692:

Answer:

$As, the area of the floor = length \times breadth = 16 \times 13.5 = 216 m^{2} And, the width of the carpet = 75 m So, the length of the carpet required = \frac{Area of the floor}{Width of the carpet} = \frac{216}{75} = 2.88 m Now, the cost of the carpet required = 2.88 \times 60 = ₹ 172.80$

Hence, the cost of covering the floor with carpet is ₹172.80.

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

Page No 692:

Answer:

The length and breadth of floor of a rectangular hall is 24 m and 18 m respectively.
The area of rectangular floor is $24 \times 18 = 432 m^{2} .$
The length and breadth of carpet is 2.5 m and 80 cm or 0.8 m respectively.
The area of carpet is $2.5 \times 0.8 = 2 m^{2} .$

So, the number of such carpets required to cover the floor of rectangular hall is = $\frac{the area of floor}{the area of carpet} = \frac{432}{2} = 216 carpets .$

Page No 692:

Answer:

$Area of the verandah = Length \times Breadth = 36 \times 15 = 540 m^{2}$
Length of the stone = 6 dm = 0.6 m
Breadth of the stone = 5 dm = 0.5 m
$Area of one stone = 0.6 \times 0.5 = 0.3 m^{2}$

$Number of stones r equired = \frac{Area of the verendah}{Area of the stone} = \frac{540}{0.3} = 1800$

Thus, 1800 stones will be required to pave the verandah.

Page No 692:

Answer:

Area of the rectangle = 192 cm²
Perimeter of the rectangle = 56 cm

$Perimeter = 2 (length + breadth) \Rightarrow 56 = 2 (l + b) \Rightarrow l + b = 28 \Rightarrow l = 28 - b$

$Area = length \times breadth \Rightarrow 192 = (28 - b) x b \Rightarrow 192 = 28 b - b^{2} \Rightarrow b^{2} - 28 b + 192 = 0 \Rightarrow (b - 16) (b - 12) = 0 \Rightarrow b = 16 or 12$

Thus, we have;
$l = 28 - b \Rightarrow l = 28 - 12 \Rightarrow l = 16$

We will take length as 16 cm and breath as 12 cm because length is greater than breadth by convention.

Page No 692:

Answer:

The field is planted with grass, with 2.5 m uncovered on its sides.

The field is shown in the given figure.

Thus, we have;
Length of the area planted with grass = $35 - (2.5 + 2.5) = 35 - 5 = 30 m$

Width of the area planted with grass = $18 - (2.5 + 2.5) = 18 - 5 = 13 m$

Area of the rectangular region planted with grass = $30 \times 13 = 390 m^{2}$

Page No 692:

Answer:

The plot with the gravel path is shown in the figure.

Area of the rectangular plot = $l \times b$
Area of the rectangular plot = $125 \times 78 = 9750 m^{2}$
Length of the park including the path = 125 + 6 = 131 m
Breadth of the park including the path = 78 + 6 = 84 m
Area of the plot including the path
$= 131 \times 84 = 11004 m^{2}$

Area of the path = $11004 - 9750$
= 1254 m²
Cost of gravelling 1 m²of the path = Rs 75
∴ Cost of gravelling 1254 m²of the path = $1254 \times 75$
= Rs 94050

Page No 692:

Answer:

(i) Area of the rectangular field = $54 \times 35 = 1890 m^{2}$

Let the width of the path be x m. The path is shown in the following diagram:

Length of the park excluding the path = (54 $-$ 2x) m
Breadth of the park excluding the path = (35 $-$ 2x ) m

Thus, we have:
Area of the path = 420 m²
$\Rightarrow 420 = 54 \times 35 - (54 - 2 x) (35 - 2 x) \Rightarrow 420 = 1890 - (1890 - 70 x - 108 x + 4 x^{2}) \Rightarrow 420 = - 4 x^{2} + 178 x \Rightarrow 4 x^{2} - 178 x + 420 = 0 \Rightarrow 2 x^{2} - 89 x + 210 = 0 \Rightarrow 2 x^{2} - 84 x - 5 x + 210 = 0 \Rightarrow 2 x (x - 42) - 5 (x - 42) = 0 \Rightarrow (x - 42) (2 x - 5) = 0 \Rightarrow x - 42 = 0 or 2 x - 5 = 0 \Rightarrow x = 42 or x = 2.5$

The width of the path cannot be more than the breadth of the rectangular field.
∴ x = 2.5 m

Thus, the path is 2.5 m wide.

(ii) Let the width of the border be x m.
The length and breadth of the carpet are 8 m and 5 m, respectively.
Area of the carpet = $8 \times 5 = 40 m^{2}$
Length of the carpet without border = $(8 - 2 x)$
Breadth of carpet without border = $(5 - 2 x)$
Area of the border = 12 m²
Area of the carpet without border = $(8 - 2 x) (5 - 2 x)$
$Thus, we have : 12 = 40 - [(8 - 2 x) (5 - 2 x)] \Rightarrow 12 = 40 - (40 - 26 x + 4 x^{2}) \Rightarrow 12 = 26 x - 4 x^{2}$

$\Rightarrow 26 x - 4 x^{2} = 12 \Rightarrow 4 x^{2} - 26 x + 12 = 0 \Rightarrow 2 x^{2} - 13 x + 6 = 0$

$\Rightarrow (2 x - 1) (x - 6) = 0 \Rightarrow 2 x - 1 = 0 and x - 6 = 0 \Rightarrow x = \frac{1}{2} and x = 6$

Because the border cannot be wider than the entire carpet, the width of the carpet is $\frac{1}{2}$ m, i.e., 50 cm.

Page No 692:

Answer:

Let the length and breadth of the garden be 9x m and 5x m, respectively,
Now,
Area of the garden = $(9 x \times 5 x) = 45 x^{2}$
Length of the garden excluding the path = ( $9 x - 7)$
Breadth of the garden excluding the path = $(5 x - 7)$
Area of the path = $45 x^{2} - [(9 x - 7) (5 x - 7)]$
$\Rightarrow 1911 = 45 x^{2} - [45 x^{2} - 63 x - 35 x + 49] \Rightarrow 1911 = 45 x^{2} - 45 x^{2} + 63 x + 35 x - 49 \Rightarrow 1911 = 98 x - 49 \Rightarrow 1960 = 98 x \Rightarrow x = \frac{1960}{98} \Rightarrow x = 20$
Thus, we have:
Length = $9 x = 20 \times 9 = 180 m$
Breadth = $5 x = 5 \times 20 = 100 m$

Page No 692:

Answer:

Width of the room left uncovered = 0.25 m
Now,
Length of the room to be carpeted = $4.9 - (0.25 + 0.25) = 4.9 - 0.5 = 4.4 m$
Breadth of the room be carpeted = $3.5 - (0.25 + 0.25) = 3.5 - 0.5 = 3 m$

Area to be carpeted = $4.4 \times 3 = 13.2 m^{2}$

Breadth of the carpet = 80 cm = 0.8 m
We know:
Area of the room = Area of the carpet

$Length of the carpet = \frac{Area of the room}{Breadth of the carpet} = \frac{13.2}{0.8} = 16.5 m$

Cost of 1 m carpet = Rs 80
Cost of 16.5 m carpet = $80 \times 16.5 = Rs 1, 320$

Page No 692:

Answer:

It is given that the dimensions of rectangular park is 50 m × 40 m.
∴ Area of the rectangular park = 50 × 40 = 2000 m²
Area of the grass surrounding the pond = 1184 m²
Now,
Area of the rectangular pond
= Area of the rectangular park − Area of the grass surrounding the rectangular pond
= 2000 − 1184
= 816 m²
Let the uniform width of the surrounding grass be x.
∴ Length of the rectangular pond = (50 − 2x) m
Breadth of the rectangular pond = (40 − 2x) m
Now,
Area of rectangular pond = 816 m²
∴ (50 − 2x) × (40 − 2x) = 816
⇒ 2000 − 80x − 100x + 4x² = 816
⇒ 4x² − 180x + 2000 − 816 = 0
⇒ 4x² − 180x + 1184 = 0
⇒ x² − 45x + 296 = 0
⇒ x² − 37x − 8x + 296 = 0
⇒ x(x − 37) − 8(x − 37) = 0
⇒ (x − 8)(x − 37) = 0
⇒ x − 8 = 0 or x − 37 = 0
⇒ x = 8 or x = 37
For x = 37,
Length of rectangular pond = 50 − 2 × 37 = −24 m, which is not possible
So, x ≠ 37
Therefore, x = 8.
When x = 8,
Length of the rectangular pond = 50 − 2 × 8 = 50 − 16 = 34 m
Breadth of the rectangular pond = 40 − 2 × 8 = 40 − 16 = 24 m.

Page No 692:

Answer:

The length and breadth of the lawn are 80 m and 64 m, respectively.
The layout of the roads is shown in the figure below:

Area of the road ABCD = $80 \times 5 = 400 m^{2}$
Area of the road PQRS = $64 \times 5 = 320 m^{2}$
Clearly, the area EFGH is common in both the roads.
Area EFGH = $5 \times 5 = 25 m^{2}$
Area of the roads = $400 + 320 - 25$
= $695 m^{2}$
Given:
Cost of gravelling 1 m² area = Rs 40
∴ Cost of gravelling 695 m² area = $695 \times 40$
= Rs 27,800

Page No 693:

Answer:

The room has four walls to be painted.

$Area of these walls = 2 (l \times h) + 2 (b \times h) = (2 \times 14 \times 6.5) + (2 \times 10 \times 6.5) = 312 m^{2}$

Now,
Area of the two doors = $(2 \times 2.5 \times 1.2) = 6 m^{2}$

Area of the four windows = $(4 \times 1.5 \times 1) = 6 m^{2}$

The walls have to be painted; the doors and windows are not to be painted.

∴ Total area to be painted $= 312 - (6 + 6) = 300 m^{2}$
Cost for painting 1 m² = Rs 35
Cost for painting 300 m² = $300 \times 35 = Rs 10, 500$

Page No 693:

Answer:

$As, the rate of covering the floor = ₹ 25 per m^{2} And, the cost of covering the floor = ₹ 2700 So, the area of the floor = \frac{2700}{25} \Rightarrow length \times breadth = 108 \Rightarrow 12 \times breadth = 108 \Rightarrow breadth = \frac{108}{12} ∴ breadth = 9 m Also, As, the rate of painting the four walls = ₹ 30 per m^{2} And, the cost of painting the four walls = ₹ 7560 So, the area of the four walls = \frac{7560}{30} \Rightarrow 2 (length + breadth) height = 252 \Rightarrow 2 (12 + 9) height = 252 \Rightarrow 2 (21) height = 252 \Rightarrow 42 \times height = 252 \Rightarrow height = \frac{252}{42} ∴ height = 6 m$

So, the dimensions of the room are $12 m \times 9 m \times 6 m$ .

Page No 693:

Answer:

Area of the square = $\frac{1}{2} \times {Diagonal}^{2}$

       = $\frac{1}{2} \times 24 \times 24$
             = $288 m^{2}$
Now, let the side of the square be x m.
Thus, we have:
$Area = {Side}^{2} \Rightarrow 288 = x^{2} \Rightarrow x = 12 \sqrt{2} \Rightarrow x = 16.92$

Perimeter = $4 \times Side$
   $= 4 \times 16.92 = 67.68 m$

Thus, the perimeter of the square plot is 67.68 m.

Page No 693:

Answer:

Area of the square = 128 cm²
$Area = \frac{1}{2} d^{2} (where d is a diagonal of the square) \Rightarrow 128 = \frac{1}{2} d^{2} \Rightarrow d^{2} = 256 \Rightarrow d = 16 cm$

Now,
Area = ${Side}^{2}$
$\Rightarrow 128 = {Side}^{2} \Rightarrow Side = 11.31 cm$

Perimeter = 4(Side)
$= 4 (11.31) = 45.24 cm$
Thus, the perimeter of the square is 45.24 cm.

Page No 693:

Answer:

We know that 1 hectare = 10000 m².
So, Area of square field = 8 hectares = 80000 m².
$\Rightarrow \frac{1}{2} {(diagonal)}^{2} = 80000 \Rightarrow {(diagonal)}^{2} = 160000 \Rightarrow diagonal = \sqrt{160000} = 400 m .$
Now, distance to be travelled = 400 m
speed is given to be 4 km/h = $4 \times \frac{1000 m}{60 \min} = \frac{4000}{60} m / \min = \frac{200}{3} m / \min$
Therefore, $time taken = \frac{distance}{speed} = \frac{400}{\frac{200}{3}} = \frac{400 \times 3}{200} = 6 minutes .$

Page No 693:

Answer:

$As, the rate of the harvesting = ₹ 900 per hectare And, the cost of harvesting = ₹ 8100 So, the area of the square field = \frac{8100}{900} = 9 hectare \Rightarrow the area = 90000 m^{2} (As, 1 hectare = 10000 m^{2}) \Rightarrow {(side)}^{2} = 90000 \Rightarrow side = \sqrt{90000} So, side = 300 m Now, perimeter of the field = 4 \times side = 4 \times 300 = 1200 m Since, the rate of putting the fence = ₹ 18 per m So, the cost of putting the fence = 1200 \times 18 = ₹ 21, 600$

Page No 693:

Answer:

Cost of fencing the lawn = Rs 28000
Let l be the length of each side of the lawn. Then, the perimeter is 4l.
We know:
$Cost = Rate \times Perimeter \Rightarrow 28000 = 14 \times 4 l$

$\Rightarrow 28000 = 56 l O r, l = \frac{28000}{56} = 500 m$

Area of the square lawn $= 500 \times 500 = 250000 m^{2}$

Cost of mowing 100 m² of the lawn = Rs 54
Cost of mowing 1 m² of the lawn $= Rs \frac{54}{100}$

∴ Cost of mowing 250000 m² of the lawn $= \frac{250000 \times 54}{100} = Rs 135000$

Page No 693:

Answer:

$We have, BD = 24 cm, AL = 9 cm, CM = 12 cm, AL ⊥ BD and CM ⊥ BD Area of the quadrilateral = ar (∆ ABD) + ar (∆ BCD) = \frac{1}{2} \times BD \times AL + \frac{1}{2} \times BD \times CM = \frac{1}{2} \times 24 \times 9 + \frac{1}{2} \times 24 \times 12 = 108 + 144 = 252 {cm}^{2}$

So, the area of the quadrilateral ABCD is 252 cm².

Page No 693:

Answer:

$∆$ BDC is an equilateral triangle with side a= 26 cm.
Area of $∆ B D C = \frac{\sqrt{3}}{4} a^{2}$
$= \frac{\sqrt{3}}{4} \times 26^{2} = \frac{1.73}{4} \times 676 = 292.37 {cm}^{2}$

By using Pythagoras' theorem in the right-angled triangle $∆$ DAB, we get:
$A D^{2} + A B^{2} = B D^{2} \Rightarrow 24^{2} + A B^{2} = 26^{2} \Rightarrow A B^{2} = 26^{2} - 24^{2} \Rightarrow A B^{2} = 676 - 576 \Rightarrow A B^{2} = 100 \Rightarrow A B = 10 cm$

Area of $∆ A B D = \frac{1}{2} \times b \times h$
$= \frac{1}{2} \times 10 \times 24 = 120 {cm}^{2}$

Area of the quadrilateral = Area of $∆$ BCD + Area of $∆$ ABD
   $= 292.37 + 120$
           = 412.37 cm²

Perimeter of the quadrilateral = AB + BC + CD + AD
                                                 = 24 + 10 + 26 + 26
                                                 = 86 cm

Page No 693:

Answer:

In the right-angled $∆$ ACB:

$A B^{2} = B C^{2} + A C^{2} \Rightarrow 17^{2} = B C^{2} + 15^{2} \Rightarrow 17^{2} - 15^{2} = B C^{2} \Rightarrow 64 = B C^{2} \Rightarrow B C = 8 cm$

Perimeter = $A B + B C + C D + A D$
     = $17 + 8 + 12 + 9$
   = 46 cm

Area of $∆ A B C = \frac{1}{2} (b \times h)$

   = $\frac{1}{2} (8 \times 15)$
= $60 {cm}^{2}$
$In ∆ A D C : A C^{2} = A D^{2} + C D^{2} So, ∆ A D C is a right - angled triangle at D .$

Area of $∆ A D C = \frac{1}{2} \times b \times h$
    $= \frac{1}{2} \times 9 \times 12 = 54 {cm}^{2}$

∴ Area of the quadrilateral = Area of $∆$ ABC + Area of $∆$ ADC
   = 60 + 54
   = 114 cm²

Page No 693:

Answer:

Area of $∆$ ABD = $\sqrt{s (s - a) (s - b) (s - c)}$
$s = \frac{1}{2} (a + b + c) s = \frac{42 + 20 + 34}{2} s = 48 cm$

Area of $∆$ ABD = $\sqrt{48 (48 - 42) (48 - 20) (48 - 34)}$

      $= \sqrt{48 \times 6 \times 28 \times 14} = \sqrt{112896} = 336 {cm}^{2}$

Area of $∆$ BDC = $\sqrt{s (s - a) (s - b) (s - c)}$
$s = \frac{1}{2} (a + b + c) s = \frac{21 + 20 + 29}{2} s = 35 cm$

Area of $∆$ BDC = $\sqrt{35 (35 - 29) (35 - 20) (35 - 21)}$

$= \sqrt{35 \times 6 \times 15 \times 14} = \sqrt{44100} = 210 {cm}^{2}$

∴ Area of quadrilateral ABCD = Area of $∆$ ABD + Area of $∆$ BDC
       = 336 + 210
        = 546 cm²

Page No 694:

Answer:

Given:
Base = 25 cm
Height = 16.8 cm
∴ Area of the parallelogram $= Base \times Height = 25 cm \times 16.8 cm = 420 {cm}^{2}$

Page No 694:

Answer:

Longer side = 32 cm
Shorter side = 24 cm
Let the distance between the shorter sides be x cm.
Area of a parallelogram = $Longer side \times Distance between the longer sides$
= $Shorter side \times Distance between the shorter sides$
or, $32 \times 17.4 = 24 \times x$

or, $x = \frac{32 \times 17.4}{24} = 23.2 cm$

∴ Distance between the shorter sides = 23.2 cm

Page No 694:

Answer:

Area of the parallelogram = 392 m²
Let the base of the parallelogram be b m.
Given:
Height of the parallelogram is twice the base.
∴ Height = 2b m
Area of a parallelogram = $Base x Height$
$\Rightarrow 392 = b \times 2 b \Rightarrow 392 = 2 b^{2} \Rightarrow \frac{392}{2} = b^{2} \Rightarrow 196 = b^{2} \Rightarrow b = 14$
∴ Base = 14 m
Altitude = $2 \times Base = 2 \times 14 = 28 m$

Page No 694:

Answer:

Parallelogram ABCD is made up of congruent $∆$ ABC and $∆$ ADC.

Area of triangle ABC = $\sqrt{s (s - a) (s - b) (s - c)}$ (Here, s is the semiperimeter.)

Thus, we have:

$s = \frac{a + b + c}{2} s = \frac{34 + 20 + 42}{2} s = 48 cm$

$Area of ∆ A B C = \sqrt{48 (48 - 34) (48 - 20) (48 - 42)} = \sqrt{48 \times 14 \times 28 \times 6} = 336 {cm}^{2}$

Now,
Area of the parallelogram = $2 \times Area of ∆ A B C$
$= 2 \times 336 = 672 {cm}^{2}$

Page No 694:

Answer:

Area of the rhombus = $\frac{1}{2} \times d_{1 \times} d_{2}$ , where d₁ and d₂ are the lengths of the diagonals.

      $= \frac{1}{2} \times 30 \times 16 = 240 {cm}^{2}$

Side of the rhombus = $\frac{1}{2} \sqrt{{d_{1}}^{2} + {d_{2}}^{2}}$

   $= \frac{1}{2} \sqrt{30^{2} + 16^{2}} = \frac{1}{2} \sqrt{1156} = \frac{1}{2} \times 34 = 17 cm$

Perimeter of the rhombus = $4 a$
      = $4 \times 17$
= 68 cm

Page No 694:

Answer:

Perimeter of a rhombus = 4a (Here, a is the side of the rhombus.)

$\Rightarrow 60 = 4 a \Rightarrow a = 15 cm$

(i) Given:
One of the diagonals is 18 cm long.
$d_{1} = 18 cm$

Thus, we have:
$Side = \frac{1}{2} \sqrt{{d_{1}}^{2} + {d_{2}}^{2}} \Rightarrow 15 = \frac{1}{2} \sqrt{18^{2} + {d_{2}}^{2}} \Rightarrow 30 = \sqrt{18^{2} + {d_{2}}^{2}} Squaring both sides, we get : \Rightarrow 900 = 18^{2} + {d_{2}}^{2} \Rightarrow 900 = 324 + {d_{2}}^{2} \Rightarrow {d_{2}}^{2} = 576 \Rightarrow d_{2} = 24 cm$

∴ Length of the other diagonal = 24 cm

(ii) Area of the rhombus $= \frac{1}{2} d_{1} \times d_{2}$
$= \frac{1}{2} \times 18 \times 24 = 216 {cm}^{2}$

Page No 694:

Answer:

i) Area of a rhombus $= \frac{1}{2} \times d_{1} \times d_{2}$ , where d₁ and d₂ are the lengths of the diagonals.
$\Rightarrow 480 = \frac{1}{2} \times 48 \times d_{2} \Rightarrow d_{2} = \frac{480 \times 2}{48} \Rightarrow d_{2} = 20 cm$

∴ Length of the other diagonal = 20 cm

(ii) Side = $\frac{1}{2} \sqrt{{d_{1}}^{2} + {d_{2}}^{2}}$

$= \frac{1}{2} \sqrt{48^{2} + 20^{2}} = \frac{1}{2} \sqrt{2304 + 400} = \frac{1}{2} \sqrt{2704} = \frac{1}{2} \times 52 = 26 cm$

∴ Length of the side of the rhombus = 26 cm

(iii) Perimeter of the rhombus = $4 \times Side$
$= 4 \times 26 = 104 cm$

Page No 694:

Answer:

$Area of the trapezium = \frac{1}{2} \times (sum of the parallel sides) \times distance between the parallel sides = \frac{1}{2} \times (12 + 9) \times 8 = 21 \times 4 = 84 {cm}^{2}$

So, the area of the trapezium is 84 cm².

Page No 694:

Answer:

Area of the canal = $640 m^{2}$
Area of trapezium = $\frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$
$\Rightarrow 640 = \frac{1}{2} \times (10 + 6) \times h \Rightarrow \frac{1280}{16} = h \Rightarrow h = 80 m$

Therefore, the depth of the canal is 80 m.

Page No 694:

Answer:

Draw $D E ∥ B C$ and DL perpendicular to AB.
The opposite sides of quadrilateral DEBC are parallel. Hence, DEBC is a parallelogram.
∴ DE = BC = 13 m
Also,
$A E = (A B - E B) = (A B - D C) = (25 - 11) = 14 m$
For $∆ D A E$ :
Let:
AE = a =14 m
DE = b = 13 m
DA = c =15 m

Thus, we have:

$s = \frac{a + b + c}{2} s = \frac{14 + 13 + 15}{2} = 21 m$

Area of $∆ D A E = \sqrt{s (s - a) (s - b) (s - c)}$
$= \sqrt{21 \times (21 - 14) \times (21 - 13) \times (21 - 15)} = \sqrt{21 \times 7 \times 8 \times 6} = \sqrt{7056} = 84 m^{2}$

Area of $∆ D A E = \frac{1}{2} \times A E \times D L$
$\Rightarrow 84 = \frac{1}{2} \times 14 \times D L \Rightarrow \frac{84 \times 2}{14} = D L \Rightarrow D L = 12 m$

Area of trapezium = $\frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$
$= \frac{1}{2} \times (11 + 25) \times 12 = \frac{1}{2} \times 36 \times 12 = 216 m^{2}$

Page No 697:

Answer:

(d) 30 m

Let the length of the rectangle be x m.
∴ Breadth of the rectangle $= (x - 5) m$

$Area = x (x - 5) = x^{2} - 5 x \Rightarrow x^{2} - 5 x = 750$
$\Rightarrow x^{2} - 5 x - 750 = 0 \Rightarrow x^{2} - 30 x + 25 x - 750 = 0 \Rightarrow x (x - 30) + 25 (x - 30) = 0 \Rightarrow (x + 25) (x - 30) = 0 \Rightarrow x + 25 = 0 and x - 30 = 0 \Rightarrow x = - 25 and x = 30$

Length cannot be negative.
∴ Length = x = 30 m

Page No 697:

Answer:

(b) 2520 m²

Let the breadth of the field be x m.
∴ Length = (x + 23) m

Now,
Perimeter $= 2 (Length + Breadth) = 2 (x + x + 23) = (4 x + 46) m$
Thus, we have:
$4 x + 46 = 206 \Rightarrow 4 x = 206 - 46 = 160 \Rightarrow x = \frac{160}{4} = 40$
∴ Breadth = x = 40 m
Length = x + 23 $= 40 + 23 = 63 m$
Area $= Length \times Breadth = 63 \times 40 = 2520 m^{2}$

Page No 697:

Answer:

(a) 108 m²

Length of the rectangular field = 12 m
Diagonal = 15 m

${Diagonal}^{2} = {Length}^{2} + {Breadth}^{2}$
$Breadth = \sqrt{{Diagonal}^{2} - {Length}^{2}} = \sqrt{15^{2} - 12^{2}} = \sqrt{225 - 144} = 9 m$

∴ Area of the field $= Length \times Breadth = 12 \times 9 = 108 m^{2}$

Page No 697:

Answer:

$We have, Width of the carpet = 75 cm = 0.75 m and length of the room = 15 m Length of the carpet = \frac{Cost of carpeting}{Rate of carpeting} = \frac{8400}{70} = 120 m Now, area of the carpet required for carpeting = 120 \times 0.75 m^{2} \Rightarrow area of the floor = 120 \times 0.75 m^{2} \Rightarrow 15 \times breadth = 120 \times 0.75 \Rightarrow breadth = \frac{120 \times 0.75}{15} So, breadth = 6 m$

Hence, the correct answer is option (c).

Page No 697:

Answer:

(d) 64 cm

Let the breadth of the rectangle be x cm.
∴ Length of the rectangle = 3x cm
We know:
$Diagonal = \sqrt{(Length)^{2} + (Breadth)^{2}} \Rightarrow 8 \sqrt{10} = \sqrt{x^{2} + (3 x)^{2}} \Rightarrow 8 \sqrt{10} = \sqrt{x^{2} + 9 x^{2}} \Rightarrow 8 \sqrt{10} = x \sqrt{10} \Rightarrow x = 8$

Now,
Breadth of the rectangle = x = 8 cm
Length of the rectangle = 3x = 24 cm
Perimeter of the rectangle $= 2 (Length + Breadth) = 2 (8 + 24) = 64 cm$

Page No 697:

Answer:

(d) 4% decrease

Let:
Length $= x$
breadth $= y$
Area $= x y$

Now,
New length $= x + 20 % x = x + \frac{1}{5} x = \frac{6}{5} x$

New breadth $= y - 20 % y = y - \frac{1}{5} y = \frac{4}{5} y$

New area $= \frac{6}{5} x \times \frac{4}{5} y = \frac{24}{25} x y$

Difference in the areas $= x y - \frac{24}{25} x y = \frac{1}{25} x y$

Difference in percentage $= [(\frac{\frac{1}{25} x y}{x y}) \times 100] % = 4 %$

Page No 697:

Answer:

(a) 264 m²

Length of the ground including the path $= 80 + 2 = 82 m$
Breadth of the ground including the path $= 50 + 2 = 52 m$

Total area (including the path) $= Length \times Breadth = 82 \times 52 = 4264 m^{2}$

Area of the field $= 80 \times 50 = 4000 m^{2}$

Area of the path $= 4264 - 4000 = 264 m^{2}$

Page No 697:

Answer:

(b) 100 cm²

A diagonal of a square forms the hypotenuse of a right-angled triangle with base and height equal to side a.

${Diagonal}^{2} = a^{2} + a^{2} \Rightarrow {Diagonal}^{2} = 2 a^{2} \Rightarrow a = \frac{1}{\sqrt{2}} Diagonal = \frac{1}{\sqrt{2}} \times 10 \sqrt{2} = 10 cm$

∴ Area of the square $= a^{2} = 10 \times 10 = 100 {cm}^{2}$

Page No 697:

Answer:

(d) 110 m

Let the diagonal of the square field be d m.

In case of a square field, $d^{2} = 2 a^{2}$ , where a is the side of the square field.
Now,
$Area of a square field = a^{2}$
$d^{2} = 2 a^{2} \Rightarrow d^{2} = 2 \times A r e a o f t h e s q u a r e f i e l d \Rightarrow d = \sqrt{2 \times Area of the square field}$

∴ $d = \sqrt{2 \times 6050} = \sqrt{12100} = 110$

Page No 697:

Answer:

(c) 100 m

Disclaimer :- The length cannot be in hectare So we used is as area of the square.
Area of the square field $= 0.5 \times 10000 = 5000 m^{2}$

The diagonal divides the square into two isosceles right-angled triangles.

Using Pythagoras' theorem, we have:

${Diagonal}^{2} = a^{2} + a^{2} = 2 a^{2}$
Area of a square = $a^{2}$

∴ Diagonal $= \sqrt{2 area} = \sqrt{2 \times 5000} = \sqrt{10000} = 100 m$

Page No 697:

Answer:

(b) 12 cm

Area of an equilateral triangle $= \frac{\sqrt{3}}{4} a^{2}$ (where a is the length of the side)
Thus, we have:
$4 \sqrt{3} = \frac{\sqrt{3}}{4} a^{2}$
$\Rightarrow a^{2} = 16 \Rightarrow a = 4 cm$

Perimeter of the equilateral triangle = 3a $= 3 \times 4 = 12 cm$

Page No 697:

Answer:

(c) $16 \sqrt{3} {cm}^{2}$

Let the side of the equilateral triangle be a.
Given:
a = 8 cm
Now,
Area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 8 \times 8 = 16 \sqrt{3} {cm}^{2}$

Page No 698:

Answer:

$As, area of an equilateral triangle = \frac{\sqrt{3}}{4} \times {(side)}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = \frac{\sqrt{3}}{4} \times {(6 \sqrt{3})}^{2} \Rightarrow \frac{1}{2} \times 6 \sqrt{3} \times Height = \frac{\sqrt{3}}{4} \times 36 \times 3 \Rightarrow 3 \sqrt{3} \times Height = 27 \sqrt{3} \Rightarrow Height = \frac{27 \sqrt{3}}{3 \sqrt{3}} ∴ Height = 9 cm$

Hence, the correct option is (b).

Page No 698:

Answer:

$Let the side of the equilateral triangle be x . As, area of the equilateral triangle = \frac{\sqrt{3}}{4} \times {(side)}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = \frac{\sqrt{3}}{4} \times x^{2} \Rightarrow \frac{1}{2} \times x \times 3 \sqrt{3} = \frac{x^{2} \sqrt{3}}{4} \Rightarrow \frac{3 x \sqrt{3}}{2} = \frac{x^{2} \sqrt{3}}{4} \Rightarrow \frac{3 \sqrt{3}}{2} = \frac{x \sqrt{3}}{4} \Rightarrow x = \frac{3 \sqrt{3} \times 4}{2 \sqrt{3}} \Rightarrow x = 6 cm Now, the area of the triangle = \frac{\sqrt{3}}{4} \times 6^{2} = \frac{\sqrt{3}}{4} \times 36 = 9 \sqrt{3} {cm}^{2}$

Hence, the correct answer is option (c).

Page No 698:

Answer:

$Let the base of the triangle be 3 x and its height be 4 x . As, the area of the triangle = 216 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 216 \Rightarrow \frac{1}{2} \times 3 x \times 4 x = 216 \Rightarrow 6 x^{2} = 216 \Rightarrow x^{2} = \frac{216}{6} \Rightarrow x^{2} = 36 \Rightarrow x = \sqrt{36} \Rightarrow x = 6 cm So, the height of the triangle = 4 \times 6 = 24 cm$

Hence, the correct answer is option (b).

Page No 698:

Answer:

$As, the sides of the triangle are 20 m, 21 m and 29 m So, the semi - perimeter = \frac{20 + 21 + 29}{2} = 35 m Now, the area of the triangular field = \sqrt{35 (35 - 20) (35 - 21) (35 - 29)} = \sqrt{35 \times 15 \times 14 \times 6} = \sqrt{7 \times 5 \times 5 \times 3 \times 7 \times 2 \times 2 \times 3} = 2 \times 3 \times 5 \times 7 = 210 m^{2} ∴ The cost of cultivating the field = 210 \times 9 = ₹ 1890$

Hence, the correct answer is option (c).

Page No 698:

Answer:

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

Let:
Length of the side of the square = Length of the side of the equilateral triangle = a unit

Now,

Area of the square $= a \times a = a^{2} {unit}^{2}$

Area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^{2} {unit}^{2}$

$Ratio of areas = \frac{Area of the square}{Area of the equilateral triangle} = \frac{a^{2}}{\frac{\sqrt{3}}{4} a^{2}} = \frac{4}{\sqrt{3}}$

Page No 698:

Answer:

(b) $\frac{49 \sqrt{3}}{4} {cm}^{2}$

Area of a circle $= π r^{2}$
$\Rightarrow 154 = π r^{2} \Rightarrow r = \sqrt{\frac{154 \times 7}{22}} = \sqrt{7 \times 7} = 7 cm$

The radius of the circle is equal to the side of the equilateral triangle.

∴ r = a (Here, a is the side of the equilateral triangle.)

$a = 7 cm$

∴ Area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 7 \times 7 = \frac{49 \sqrt{3}}{4} {cm}^{2}$

Page No 698:

Answer:

$We have, BD = 20 cm \Rightarrow BO = \frac{BD}{2} = \frac{20}{2} = 10 cm As, area of the rhombus ABCD = 480 {cm}^{2} \Rightarrow \frac{1}{2} \times AC \times BD = 480 \Rightarrow \frac{1}{2} \times AC \times 20 = 480 \Rightarrow AC \times 10 = 480 \Rightarrow AC = \frac{480}{10} \Rightarrow AC = 48 cm \Rightarrow AO = \frac{AC}{2} = \frac{48}{2} = 24 cm Now, in ∆ AOB, Using Pythagoras theorem, {AB}^{2} = {AO}^{2} + {BO}^{2} = 24^{2} + 10^{2} = 576 + 100 \Rightarrow {AB}^{2} = 676 \Rightarrow AB = \sqrt{676} ∴ AB = 26 cm$

Hence, the correct answer is option (c).

Page No 698:

Answer:

$WE have, AB = BC = CD = DA = 20 cm and BD = 24 cm Also, BO = \frac{BD}{2} = \frac{24}{2} = 12 cm In ∆ AOB, Using Pythagoras theorem {AO}^{2} = {AB}^{2} - {BO}^{2} = 20^{2} - 12^{2} = 400 - 144 \Rightarrow {AO}^{2} = 256 \Rightarrow AO = \sqrt{256} \Rightarrow AO = 16 cm \Rightarrow AC = 2 AO = 2 \times 16 = 32 cm Now, the area of the rhombus ABCD = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 32 \times 24 = 384 {cm}^{2}$

Hence, the correct answer is option (d).

Page No 701:

Answer:

(b) 114 sq cm

Using Pythagoras' theorem in $∆$ ABC, we get:
$A C^{2} = A B^{2} + B C^{2} \Rightarrow A B = \sqrt{A C^{2} - B C^{2}} = \sqrt{17^{2} - 15^{2}} = 8 cm$

$Area of ∆ A B C = \frac{1}{2} \times A B \times B C = \frac{1}{2} \times 8 \times 15 = 60 {cm}^{2}$

$Area of ∆ B C D = \frac{1}{2} \times B D \times C D = \frac{1}{2} \times 12 \times 9 = 54 {cm}^{2}$

∴ Area of quadrilateral ABCD $= Ar (∆ A B C) + Ar (∆ B C D) = 54 + 60 = 114 {cm}^{2}$

Page No 701:

Answer:

(a) 306 m²

In the given figure, AECD is a rectangle.

Length AE = Length CD = 28 m

Now,
$B E = A B - A E = 40 - 28 = 12 m$

Also,
AD = CE = 9 m
$Area of trapezium = \frac{1}{2} \times Sum of parallel sides \times Distance between them = \frac{1}{2} \times (D C + A B) \times C E = \frac{1}{2} \times (28 + 40) \times 9 = \frac{1}{2} \times 68 \times 9 = 306 m^{2}$

In the given figure, if DA is perpendicular to AE, then it can be solved, otherwise it cannot be solved.

Page No 701:

Answer:

(c) 12.5 cm

Let the sides of the triangle be 12x cm, 14x cm and 25x cm.
Thus, we have:
$Perimeter = 12 x + 14 x + 25 x \Rightarrow 25.5 = 51 x \Rightarrow x = \frac{25.5}{51} = 0.5$

∴ Greatest side of the triangle $= 25 x = 25 \times 0.5 = 12.5 cm$

Page No 701:

Answer:

(c) 52 cm²

$Area of trapezium = \frac{1}{2} (Sum of parallel sides) \times Distance between them = \frac{1}{2} \times (9.7 + 6.3) \times 6.5 = 8 \times 6.5 = 52.0 {cm}^{2}$

Page No 701:

Answer:

Given:
Side of the equilateral triangle = 10 cm
Thus, we have:
$Area of the equilateral triangle = \frac{\sqrt{3}}{4} {side}^{2} = \frac{\sqrt{3}}{4} \times 10 \times 10 = 25 \times 1.732 = 43.3 {cm}^{2}$

Page No 701:

Answer:

Area of an isosceles triangle:
$= \frac{1}{4} b \sqrt{4 a^{2} - b^{2}} (Where a is the length of the equal sides and b is the base) = \frac{1}{4} \times 24 \sqrt{4 {(13)}^{2} - 24^{2}} = 6 \sqrt{4 \times 169 - 576} = 6 \sqrt{676 - 576} = 6 \sqrt{100} = 6 \times 10 = 60 {cm}^{2}$

Page No 701:

Answer:

Let the rectangle ABCD represent the hall.

Using the Pythagorean theorem in the right-angled triangle ABC, we have:
${Diagonal}^{2} = {Length}^{2} + {Breadth}^{2}$

$\Rightarrow Breadth = \sqrt{{Diagonal}^{2} - {Length}^{2}} = \sqrt{26^{2} - 24^{2}} = \sqrt{676 - 576} = \sqrt{100} = 10 m$

∴ Area of the hall $= Length \times Breadth = 24 \times 10 = 240 m^{2}$

Page No 701:

Answer:

The diagonal of a square forms the hypotenuse of an isosceles right triangle. The other two sides are the sides of the square of length a cm.

Using Pythagoras' theorem, we have:

${Diagonal}^{2} = a^{2} + a^{2} = 2 a^{2} \Rightarrow Diagonal = \sqrt{2} a$

$Diagonal of the square = \sqrt{2} a \Rightarrow 24 = \sqrt{2} a$

$\Rightarrow$ a $= \frac{24}{\sqrt{2}}$
Area of the square $= {Side}^{2} = {(\frac{24}{\sqrt{2}})}^{2} = \frac{24 \times 24}{2} = 288 {cm}^{2}$

Page No 701:

Answer:

$Area of the rhombus = \frac{1}{2} (Product of diagonals) = \frac{1}{2} (48 \times 20) = 480 {cm}^{2}$

Page No 701:

Answer:

To find the area of the triangle, we will first find the semiperimeter of the triangle.

Thus, we have:

$s = \frac{1}{2} (a + b + c) = \frac{1}{2} (42 + 34 + 20) = \frac{1}{2} \times 96 = 48 cm$

Now,

$Area of the triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{48 (48 - 42) (48 - 34) (48 - 20)} = \sqrt{48 \times 6 \times 14 \times 28} = \sqrt{112896} = 336 {cm}^{2}$

Page No 701:

Answer:

Let the length and breadth of the lawn be $5 x m and 3 x m$ , respectively.

Now,

Area of the lawn $= 5 x \times 3 x = 5 x^{2}$
$\Rightarrow 15 x^{2} = 3375 \Rightarrow x = \sqrt{\frac{3375}{15}} \Rightarrow x = \sqrt{225} = 15$

$Length = 5 x = 5 \times 15 = 75 m Breadth = 3 x = 3 \times 15 = 45 m$

∴ Perimeter of the lawn $= 2 (Length + Breadth) = 2 (75 + 45) = 2 \times 120 = 240 m$
Total cost of fencing the lawn at Rs 20 per metre $= 240 \times 20 = Rs 4800$

Page No 701:

Answer:

Given:
Sides are 20 cm each and one diagonal is of 24 cm.
The diagonal divides the rhombus into two congruent triangles, as shown in the figure below.

We will now use Hero's formula to find the area of triangle ABC.
First, we will find the semiperimeter.
$s = \frac{1}{2} (a + b + c) = \frac{1}{2} (20 + 20 + 24) = \frac{64}{2} = 32 m$

$Area of ∆ ABC = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{32 (32 - 20) (32 - 20) (32 - 24)} = \sqrt{32 \times 12 \times 12 \times 8} = \sqrt{36864} = 192 {cm}^{2}$

Now,
Area of the rhombus $= 2 \times Area of triangle ABC = 192 \times 2 = 384 {cm}^{2}$

Page No 702:

Answer:

We will divide the trapezium into a triangle and a parallelogram.

Difference in the lengths of parallel sides $= 25 - 11 = 14 cm$
We can represent this in the following figure:

Trapezium ABCD is divided into parallelogram AECD and triangle CEB.

Consider triangle CEB.

In triangle CEB, we have:

E B = 25 - 11 = 14 cm

Using Hero's theorem, we will first evaluate the semiperimeter of triangle CEB and then evaluate its area.

Semiperimeter,

s = \frac{1}{2} (a + b + c) = \frac{1}{2} (15 + 13 + 14) = \frac{42}{2} = 21 cm

Area of triangle C E B = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{21 (21 - 15) (21 - 13) (21 - 14)} = \sqrt{21 \times 6 \times 8 \times 7} = \sqrt{7056} = 84 {cm}^{2}

Also,

Area of triangle CEB

= \frac{1}{2} (Base \times Height)

Height of triangle CEB

= \frac{Area \times 2}{Base} = \frac{84 \times 2}{14} = 12 cm

Consider parallelogram AECD.

Area of parallelogram AECD

= Height \times Base = A E \times C F = 12 \times 11 = 132 {cm}^{2}

Area of trapezium ABCD

= Ar (∆ B E C) + Ar (parallelogram A E C D) = 132 + 84 = 216 {cm}^{2}

Page No 702:

Answer:

The diagonal of a parallelogram divides it into two congruent triangles. Also, the area of the parallelogram is the sum of the areas of the triangles.

We will now use Hero's formula to calculate the area of triangle ABC.

$Semiperimeter, s = \frac{1}{2} (34 + 20 + 42) = \frac{1}{2} (96) = 48 cm$

$Area of ∆ ABC = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{48 (48 - 42) (48 - 34) (48 - 20)} = \sqrt{48 \times 6 \times 14 \times 28} = \sqrt{112896} = 336 {cm}^{2}$

Area of the parallelogram $= 2 \times Area ∆ ABC = 2 \times 336 = 672 {cm}^{2}$

Page No 702:

Answer:

Given:
Cost of fencing = Rs 2800
Rate of fencing = Rs 14

Now,
Perimeter $= \frac{Total cost}{Rate} = \frac{2800}{14} = 200 m$

Because the lawn is square, its perimeter $is 4 a, where a is the side of the square)$ .
$\Rightarrow 4 a = 200 \Rightarrow a = \frac{200}{4} = 50 m$

Area of the lawn = ${Side}^{2} = 50^{2} = 2500 m^{2}$

Cost for mowing the lawn per 100 m²= Rs 54

Cost for mowing the lawn per 1 m² $= Rs \frac{54}{100}$

Total cost for mowing the lawn per 2500 m² $= \frac{54}{100} \times 2500 = Rs 1350$

Page No 702:

Answer:

Quadrilateral ABCD is divided into triangles $∆$ ABD and $∆$ BCD.
We will now use Hero's formula.
For $∆$ ABD:
$Semiperimeter, s = \frac{1}{2} (42 + 20 + 34) = \frac{96}{2} = 48 cm$

$Area of ∆ ABD = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{48 (48 - 42) (48 - 34) (48 - 20)} = \sqrt{48 \times 6 \times 14 \times 28} = \sqrt{112896} = 336 {cm}^{2}$

For $∆$ BCD:
$s = \frac{1}{2} (20 + 21 + 29) = \frac{70}{2} = 35 cm$

$Area of ∆ BCD = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{35 (35 - 20) (35 - 21) (35 - 29)} = \sqrt{35 \times 15 \times 14 \times 6} = \sqrt{44100} = 210 {cm}^{2}$

Thus, we have:
Area of quadrilateral ABCD $= Ar (∆ A B D) + Ar (B D C) = 336 + 210 = 546 {cm}^{2}$

Page No 702:

Answer:

Area of the rhombus $= \frac{1}{2} (Product of diagonals) = \frac{1}{2} (120 \times 44) = 2640 m^{2}$

Area of the parallelogram $= Base \times Height = 66 \times Height$

Given:
The area of the rhombus is equal to the area of the parallelogram.

$Thus, we have : 66 \times Height = 2640 \Rightarrow Height = \frac{2640}{66} = 40 m$

∴ Corresponding height of the parallelogram = 40 m

Page No 702:

Answer:

Diagonals of a rhombus perpendicularly bisect each other. The statement can help us find a side of the rhombus. Consider the following figure.

ABCD is the rhombus and AC and BD are the diagonals. The diagonals intersect at point O.

We know:
$∠ D O C = 90 °$
$D O = O B = \frac{1}{2} D B = \frac{1}{2} \times 48 = 24 cm$
Similarly,
$A O = O C = \frac{1}{2} A C = \frac{1}{2} \times 20 = 10 cm$

Using Pythagoras' theorem in the right-angled triangle $∆$ DOC, we get:

$D C^{2} = \sqrt{D O^{2} + O C^{2}} = \sqrt{24^{2} + 10^{2}} = \sqrt{576 + 100} = \sqrt{676} = 26 cm$

DC is a side of the rhombus.
We know that in a rhombus, all sides are equal.
∴ Perimeter of ABCD $= 26 \times 4 = 104 cm$

Page No 702:

Answer:

$Area of a parallelogram = Base \times Height ∴ A B \times D E = B C \times D F \Rightarrow D E = \frac{B C \times D F}{A B} = \frac{27 \times 12}{36} = 9 cm$

∴ Distance between the longer sides = 9 cm

Page No 702:

Answer:

The field, which is represented as ABCD, is given below.

The area of the field is the sum of the areas of triangles ABC and ADC.

Area of the triangle ABC $= \frac{1}{2} (AC \times BF) = \frac{1}{2} (128 \times 22.7) = 1452.8 m^{2}$

Area of the triangle ADC $= \frac{1}{2} (AC \times DE) = \frac{1}{2} (128 \times 17.3) = 1107.2 m^{2}$

Area of the field = Sum of the areas of both the triangles $= 1452.8 + 1107.2 = 2560 m^{2}$

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