Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 17 - Construction Of Quadrilaterals

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Rs Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 17 Construction Of Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Construction Of Quadrilaterals are extremely popular among Class 8 students for Maths Construction Of Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw $AB = 4.2 cm$ .
Step 2: With A as the centre and radius equal to $8 cm$ , draw an arc.
Step 3: With B as the centre and radius equal to $6 cm$ , draw another arc, cutting the previous arc at C.
Step 4: Join BC.
Step 5: With A as the centre and radius equal to $5 cm,$ draw an arc.
Step 6: With C as the centre and radius equal to $5.2 cm$ , draw another arc, cutting the previous arc at D.
Step 7: Join AD and CD.

Thus, ABCD is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw $PQ = 5.4 cm$ .
Step 2: With P as the centre and radius equal to $4 cm$ , draw an arc.
Step 3: With Q as the centre and radius equal to $4.6 cm$ , draw another arc, cutting the previous arc at R.
Step 4: Join QR.
Step 5: With P as the centre and radius equal to $3.5 cm,$ draw an arc.
Step 6: With R as the centre and radius equal to $4.3 cm$ , draw another arc, cutting the previous arc at S.
Step 7: Join PS and RS.

Thus, PQRS is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw $AB = 3.5 cm$ .
Step 2: With B as the centre and radius equal to $5.6 cm$ , draw an arc.
Step 3: With A as the centre and radius equal to $4.5 cm$ , draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With D as the centre and radius equal to $4.5 cm,$ draw an arc.
Step 6: With B as the centre and radius equal to $3.8 cm$ , draw another arc, cutting the previous arc at C.
Step 7: Join BC and CD.

Thus, ABCD is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw $AB = 3.6 cm$ .
Step 2: With B as the centre and radius equal to $4 cm$ , draw an arc.
Step 3: With A as the centre and radius equal to $2.7 cm$ , draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With A as the centre and radius equal to $4.6 cm,$ draw an arc.
Step 6: With B as the centre and radius equal to $3.3 cm$ , draw another arc, cutting the previous arc at C.
Step 7: Join AC, BC and CD.

Thus, ABCD is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw $Q R = 7.5 cm .$
Step 2: With Q as the centre and radius equal to $10 cm$ , draw an arc.
Step 3: With R as the centre and radius equal to $5 cm$ , draw another arc, cutting the previous arc at S.
Step 4: Join QS and RS.
Step 5: With S as the centre and radius equal to $6 cm,$ draw an arc.
Step 6: With R as the centre and radius equal to $6 cm$ , draw another arc, cutting the previous arc at P.
Step 7: Join PS and PR.
Step 8: PQ = 4.9 cm
Thus, PQRS is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw $A B = 3.4 cm .$
Step 2: With B as the centre and radius equal to $4 cm$ , draw an arc.
Step 3: With A as the centre and radius equal to $5.7 cm$ , draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With A as the centre and radius equal to $8 cm,$ draw an arc.
Step 6: With D as the centre and radius equal to $3 cm$ , draw another arc, cutting the previous arc at C.
Step 7: Join AC, CD and BC.

Thus, ABCD is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw AB= $3.5 cm$ .
Step 2: Make $∠ A B C = 120^{\circ}$ .
Step 3: With B as the centre, draw an arc $3.5 cm$ and name that point C.
Step 4: With C as the centre, draw an arc $5.2 cm$ .
Step 5: With A as the centre, draw another arc $5.2 cm$ , cutting the previous arc at D.
Step 6: Join CD and AD.
Thus, $A B C D$ is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw AB= $2.9 c m$
Step 2: Make $∠ A = 70^{\circ}$
Step 3: With A as the centre, draw an arc of $3.4 c m$ . Name that point as D.
Step 4: With D as the centre, draw an arc of $2.7 c m$ .
Step 5: With B as the centre, draw an arc of 3.2 cm, cutting the previous arc at C.
Step 6: Join CD and BC.
Then, $A B C D$ is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw BC= $5 c m$
Step 2: Make $∠ B = 125^{\circ} a n d ∠ C = 60^{\circ}$
Step 3: With B as the centre, draw an arc of $3.5 c m$ . Name that point as A.
Step 4: With C as the centre, draw an arc of $4.6 c m$ . Name that point as D.
Step 5: Join A and D.
Then, $A B C D$ is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw QR= $5.6 c m$
Step 2: Make $∠ Q = 45^{\circ} and ∠ R = 90^{\circ}$
Step 3: With Q as the centre, draw an arc of $6 c m$ . Name that point as P.
Step 4: With R as the centre, draw an arc of $2.7 c m$ . Name that point as S.
Step 6: Join P and S.
Then, $P Q R S$ is the required quadrilateral.

Page No 198:

Answer:

Steps of construction:
Step 1: Draw AB= $5.6 c m$
Step 2: Make $∠ A = 50^{\circ} a n d ∠ B = 105^{\circ}$
Step 3: With B as the centre, draw an arc of $4 c m$ .
Step 3: Sum of all the angles of the quadrilateral is $360^{\circ}$ .
$∠ A + ∠ B + ∠ C + ∠ D = 360^{\circ} 50^{\circ} + 105^{\circ} + ∠ C + 80^{\circ} = 360^{\circ} 235^{\circ} + ∠ C = 360^{\circ} ∠ C = 360^{\circ} - 235^{\circ} ∠ C = 125^{\circ}$
Step 5: With C as the centre, make $∠ C equal to ∠ 125^{\circ}$ .
Step 6: Join C and D.
Step 7: Measure $∠ D = 80^{\circ}$
Then, $A B C D$ is the required quadrilateral.

Page No 199:

Answer:

Steps of construction:
Step 1: Draw PQ= $5 c m$
Step 2:
$∠ P + ∠ Q + ∠ R + ∠ S = 360^{\circ} 100^{\circ} + ∠ Q + 100^{\circ} + 75^{\circ} = 360^{\circ} 275^{\circ} + ∠ Q = 360^{\circ} ∠ Q = 360^{\circ} - 275^{\circ} ∠ Q = 85^{\circ}$
Step 3: Make $∠ P = 100^{\circ} a n d ∠ Q = 85^{\circ}$
Step 3: With Q as the centre, draw an arc of $6.5 c m$ .
Step 4: Make $∠ R = 100^{\circ}$
Step 6: Join R and S.
Step 7: Measure $∠ S = 75^{\circ}$
Then, $P Q R S$ is the required quadrilateral.

Page No 199:

Answer:

Steps of construction:
Step 1: Draw $A B = 4 c m$
Step 2: $M a k e ∠ B = 90^{\circ}$
Step 3: $A C^{2} = A B^{2} + B C^{2} 5^{2} = 4^{2} + B C^{2} 25 - 16 = B C^{2} B C = 3 c m$
With B as the centre, draw an arc equal to 3 cm.
Step 4: Make $∠ C = 90^{\circ}$
Step 5: With A as the centre and radius equal to $5.5 c m$ , draw an arc and name that point as D.
Then, $A B C D$ is the required quadrilateral.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw AB = $5.2 c m$
Step 2: With B as the centre, draw an arc of $4.7 c m$ .
Step 3: With A as the centre, draw another arc of $7.6 c m$ , cutting the previous arc at C.
Step 4: Join A and C.
Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with C as the centre, draw an arc of $5.2 c m$ .
Step 6: With A as the centre, draw another arc of $4.7 c m$ , cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw AB= $4.3 c m$
Step 2: With B as the centre, draw an arc of $6.8 c m$ .
Step 3: With A as the centre, draw another arc of $4 c m$ , cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: We know that the opposite sides of a parallelogram are equal.
Thus, with D as the centre, draw an arc of $4.3 c m$ .
Step 6: With B as the centre, draw another arc of $4 c m$ , cutting the previous arc at C.
Step 7: Join CD and BC.
then, ABCD is the required parallelogram.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw PQ= 4 cm
Step 2: Make $∠ P Q R = 60^{\circ}$
Step 2: With Q as the centre, draw an arc of 6 cm and name that point as R.
Step 3: With R as the centre, draw an arc of 4 cm and name that point as S.
Step 4: Join SR and PS.
Then, PQRS is the required parallelogram.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw BC= $5 c m$
Step 2: Make an $∠ B C D = 120^{\circ}$
Step 2: With C as centre draw an arc of $4.8 cm$ , name that point as D
Step 3: With D as centre draw an arc $5 cm$ , name that point as A
Step 4: With B as centre draw another arc $4.8 cm$ cutting the previous arc at A.
Step 5: Join AD and AB
then, ABCD is a required parallelogram.

Page No 201:

Answer:

We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw AB= $4.4 c m$
Step 2: With A as the centre and radius $2.8 c m$ , draw an arc.
Step 3: With B as the centre and radius $3.5 c m$ , draw another arc, cutting the previous arc at point O.
Step 4: Join OA and OB.
Step 5: Produce OA to C, such that OC= AO. Produce OB to D, such that OB=OD.
Step 5: Join AD, BC, and CD.
Thus, ABCD is the required parallelogram. The other side is 4.5 cm in length.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw AB= 6.5cm
Step 2: Draw a perpendicular at point A. Name that ray as AX. From point A, draw an arc of length 2.5 cm on the ray AX and name that point as L.
Step 3: On point L, make a perpendicular. Draw a straight line YZ passing through L, which is perpendicular to the ray AX.
Step 4: Cut an arc of length 3.4 cm on the line YZ and name it as C.
Step 5: From point C, cut an arc of length 6.5 cm on the line YZ. Name that point as D.
Step 6: Join BC and AD.

Therefore, quadrilateral ABCD is a parallelogram.

The altitude from C measures 2.5 cm in length.

Page No 201:

Answer:

We know that the diagonals of a parallelogram bisect each other.

Steps of construction:
Step 1: Draw AC= $3.8 cm$
Step 2: Bisect AC at O.
Step 3: Make $∠ C O X = 60^{\circ}$
Produce XO to Y.
Step 4:
$O B = \frac{1}{2} (4.6) cm O B = 2.3 cm and O D = \frac{1}{2} (4.6) cm O D = 2.3 cm$
Step 5: Join AB, BC, CD and AD.
Thus, ABCD is the required parallelogram.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw AB = $11 c m$
Step 2: Make $∠ A = 90^{\circ} ∠ B = 90^{\circ}$
Step 3: Draw an arc of 8.5 cm from point A and name that point as D.
Step 4: Draw an arc of 8.5 cm from point B and name that point as C.
Step 5: Join C and D.
Thus, ABCD is the required rectangle.

Page No 201:

Answer:

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB = $6.4 c m$
Step 2: Make $∠ A = 90^{\circ} ∠ B = 90^{\circ}$
Step 3: Draw an arc of length 6.4 cm from point A and name that point as D.
Step 4: Draw an arc of length 6.4 cm from point B and name that point as C.
Step 5: Join C and D.
Thus, ABCD is a required square.

Page No 201:

Answer:

We know that the diagonals of a square bisect each other at right angles.
Steps of construction:
Step 1: Draw AC= $5.8 cm$
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:
: $F r o m O : O B = \frac{1}{2} (5.8) cm = 2.9 cm O D = \frac{1}{2} (5.8) cm = 2.9 cm$
Step 4: Join AB, BC, CD and DA.
ABCD is the required square.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw QR = $3.6 c m$
Step 2: Make $∠ Q = 90^{\circ} ∠ R = 90^{\circ}$
Step 3:
$P R^{2} = P Q^{2} + Q R^{2} 6^{2} = P Q^{2} + 3 . 6^{2} P Q^{2} = 36 - 12.96 P Q^{2} = 23.04 P Q = 4.8 cm$

Step 3: Draw an arc of length 4.8 cm from point Q and name that point as P.
Step 4: Draw an arc of length 6 cm from point R, cutting the previous arc at P.
Step 5: Join PQ
Step 6: Draw an arc of length 4.8 cm from point R.
From point P, draw an arc of length 3.6 cm, cutting the previous arc. Name that point as S.
Step 7: Join P and S.
Thus, PQRS is the required rectangle. The other side is 4.8 cm in length.

Page No 201:

Answer:

We know that the diagonals of a rhombus bisect each other.
.Steps of construction:
Step 1: Draw AC= $6 c m$
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:
$O B = \frac{1}{2} (8) cm O B = 4 cm and O D = \frac{1}{2} (8) cm O D = 4 cm$
Draw an arc of length 4 cm on OX and name that point as B.
Draw an arc of length 4 cm on OY and name that point as D.
Step 4 : Join AB, BC, CD and AD.
Thus, ABCD is the required rhombus, as shown in the figure.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw AB = $4 c m$
Step 2: With B as the centre, draw an arc of $4 cm$ .
Step 3: With A as the centre, draw another arc of $6.5 cm$ , cutting the previous arc at C.
Step 4: Join AC and BC.
Step 5: With C as the centre, draw an arc of 4 cm.
Step 6: With A as the centre, draw another arc of $4 cm$ , cutting the previous arc at D.
Step 7: Join AD and CD.
ABCD is the required rhombus.

Page No 201:

Answer:

Steps of construction:
Step1: Draw AB = $7.2 cm$
Step2: Draw $∠ A B Y = 60 ° ∠ B A X = 120 °$
Sum of the adjacent angles is 180°.
$∠ B A X + ∠ A B Y = 180 ° = > ∠ B A X = 180 ° - 60 ° = 120 °$
Step 3:
$Set off A D (7.2 cm) along A X and B C (7.2 cm) along B Y .$
Step 4: Join C and D.
Then, ABCD is the required rhombus.

Page No 201:

Answer:

Steps of construction:
Step 1: Draw AB= $6 cm$
Step 2: Make $∠ A B X = 75^{\circ}$
Step 3: With B as the centre, draw an arc at $4 c m$ . Name that point as C.
Step 4: $A B ∥ C D$

$∴ ∠ A B X + ∠ B C Y = 180 ° \Rightarrow ∠ B C Y = 180 ° - 75 ° = 105 °$
Make $∠ B C Y = 105 °$
At C, draw an arc of length $3.2 cm$ .
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

Page No 201:

Answer:

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle, $∠ A B X, equal to 60 ° .$
Step3: With B as the centre, draw an arc of $5 cm$ . Name that point as C. Join B and C.
Step4:
$A B ∥ D C ∴ ∠ A B X + ∠ B C Y = 180 ° \Rightarrow ∠ B C Y = 180 ° - 60 ° = 120 °$

Draw an angle, $∠ B C Y, equal to 120 ° .$

Step4: With A as the centre, draw an arc of length $6.5 cm$ , which cuts CY. Mark that point as D.

Step5: Join A and D.

Thus, ABCD is the required trapezium.

Page No 202:

Answer:

( i) Open curve: An open curve is a curve where the beginning and end points are different.
Example: Parabola

(ii) Closed Curve: A curve that joins up so there are no end points.
Example: Ellipse

(iii) Simple closed curve: A closed curve that does not intersect itself.

Page No 202:

Answer:

Let the angles be $(x) °, (2 x) °, (3 x) ° and (4 x) ° .$
Sum of the angles of a quadrilateral is $360^{\circ}$ .
$x + 2 x + 3 x + 4 x = 360 10 x = 360 x = \frac{360}{10} x = 36$

$(2 x) ° = (2 \times 36)^{\circ} = 72^{\circ} (3 x) ° = (3 \times 36)^{\circ} = 108^{\circ} (4 x) ° = (4 \times 36)^{\circ} = 144^{\circ}$

The angles of the quadrilateral are $36^{\circ}, 72^{\circ}, 108^{\circ} and 144^{\circ} .$

Page No 202:

Answer:

$Let the two adjacent angles of the parallelogram b e (2 x) ° and (3 x) ° .$
Sum of any two adjacent angles of a parallelogram is $180^{\circ}$ .

$∴ 2 x + 3 x = 180 \Rightarrow 5 x = 180 \Rightarrow x = 36$

$(2 x) ° = (2 \times 36) ° = 72^{\circ} (3 x) ° = (3 \times 36) ° = 108^{\circ}$

Measures of the angles are $72^{\circ} a n d 108^{\circ}$ .

Page No 202:

Answer:

Let the length be $4 x$ cm and the breadth be $5 x$ cm.
Perimeter of the rectangle =180 $c m$
Perimeter of the rectangle= $2 (l + b)$

$2 (l + b) = 180 \Rightarrow 2 (4 x + 5 x) = 180 \Rightarrow 2 (9 x) = 180 \Rightarrow 18 x = 180 \Rightarrow x = 10$

$∴ Length = 4 x cm = 4 \times 10 = 40 cm Breadth = 5 x cm = 5 \times 10 = 50 cm$

Page No 202:

Answer:

Rhombus is a parallelogram.

Consider:

∆ A O B and ∆ C O D ∠ O A B = ∠ O C D (alternate angle) ∠ O D C = ∠ O B A (alternate angle) ∠ D O C = ∠ A O B (vertically opposite angles) ∆ A O B ≅ C O B ∴ A O = C O O B = O D

Therefore, the diagonals bisects at O.

Now, let us prove that the diagonals intersect each other at right angles.

Consider

∆ C O D a n d ∆ C O B

C D = C B (all sides of a rhombus are equal) C O = C O (common side) O D = O B (p oint O bisects B D)

∴

∆ C O D ≅ ∆ C O B

∴

∠ C O D = ∠ C O B

(corresponding parts of congruent triangles)

Further,

∠ C O D + ∠ C O B = 180 ° (l inear pair)

∴

∠ C O D = ∠ C O B = 90 °

It is proved that the diagonals of a rhombus are perpendicular bisectors of each other.

Page No 202:

Answer:

All the sides of a rhombus are equal in length.
The diagonals of a rhombus intersect at $90^{\circ}$ .
The diagonal and the side of a rhombus form right triangles.

In $△ A O B$ :
$A B^{2} = A O^{2} + O B^{2} = 8^{2} + 6^{2} = 64 + 36 = 100 A B = 10 cm$
Therefore, the length of each side of the rhombus is 10 cm.

Page No 202:

Answer:

(b) 37^o, 143^o, 37^o 143^o

Opposite angles of a parallelogram are equal.
$∴ 3 x - 2 = 50 - x \Rightarrow 3 x + x = 50 + 2 \Rightarrow 4 x = 52 \Rightarrow x = 13$

Therefore, the first and the second angles are:
${(3 x - 2)}^{\circ} = {(2 \times 13 - 2)}^{\circ} = 37^{\circ} {(50 - x)}^{\circ} = {(50 - 13)}^{\circ} = 37^{\circ}$
Sum of adjacent angles in a parallelogram is $180^{\circ}$ .
Adjacent angles = $180^{\circ} - 37^{\circ} = 143^{\circ}$

Page No 202:

Answer:

(d) none of the these

Let the angles be $(x) °, (3 x) °, (7 x) ° and (9 x) °$ .

Sum of the angles of the quadrilateral is $360^{\circ}$ .

$x + 3 x + 7 x + 9 x = 360 20 x = 360 x = 18$

$Angles : (3 x) ° = (3 \times 18) = 54^{\circ} (7 x) ° = (7 \times 18) ° = 126^{\circ} (9 x) ° = (9 \times 18) ° = 162^{\circ}$

Page No 202:

Answer:

(b) 6 cm
Let the breadth of the rectangle be x cm.
Diagonal =10 cm
Length= 8 cm
The rectangle is divided into two right triangles.
$D i a g o n a l^{2} = L e n g t h^{2} + B r e a d t h^{2} 10^{2} = 8^{2} + x^{2} 100 - 64 = x^{2} x^{2} = 36 x = 6 c m$

Breadth of the rectangle = 6 cm

Page No 202:

Answer:

(d) x = 8
All sides of a square are equal.

$P Q = Q R (2 x + 3) = (3 x - 5) = > 2 x - 3 x = - 5 - 3 = > x = 8 cm$

Page No 202:

Answer:

(d) 90°

We know that the opposite sides and the angles in a parallelogram are equal.
Also, its adjacent sides are supplementary, i.e. sum of the sides is equal to 180.
Now, the bisectors of these angles form a triangle, whose two angles are:
$\frac{A}{2} and \frac{B}{2} or \frac{A}{2} = (90 - \frac{A}{2}) Sum of the angles of a triangle is 180 ° . \frac{∠ A}{2} + 90 - \frac{∠ A}{2} + ∠ O = 180^{\circ} ∠ O = 180 - 90 ∠ O = 90^{\circ}$

Hence, the two bisectors intersect at right angles.

Page No 202:

Answer:

(c) 9
Hexagon has six sides.
$Number of diagonals = \frac{n (n - 3)}{2} (where n is the number of side s) = \frac{6 (6 - 3)}{2} = 9$

Page No 202:

Answer:

(b) 8
$Interior angle = \frac{180 (n - 2)}{n} \Rightarrow 135 = \frac{180 (n - 2)}{n} \Rightarrow 135 n = 180 n - 360 \Rightarrow 360 = 180 n - 135 n \Rightarrow n = 8$

It has 8 sides.

Page No 202:

Answer:

(i) Sum of all exterior angles = $360^{\circ}$

(ii) Sum of all interior angles = $(n - 2) \times 180 °$

(iii) Number of diagonals = $\frac{n (n - 3)}{2}$

Page No 202:

Answer:

(i) Sum of all exterior angles of a regular polygon is $360^{\circ}$ .

(ii) Sum of all interior angles of a polygon is $(n - 2) \times 180 °, where n is the number of sides .$

Page No 202:

Answer:

(i) Octagon has 8 sides.
$∴ Interior angle = \frac{180 ° n - 360 °}{n} Int erior angle = \frac{(180 ° \times 8) - 360 °}{8} = 135 °$
(ii) Sum of the interior angles of a regular hexagon = $(6 - 2) \times 180^{\circ} = 720^{\circ}$

(iii) Each exterior angle of a regular polygon is $60^{\circ}$ .
$∴ \frac{360}{60} = 6$
Therefore, the given polygon is a hexagon.

(iv) If the interior angle is $108^{\circ}$ , then the exterior angle will be $72^{\circ}$ . (interior and exterior angles are supplementary)
Sum of the exterior angles of a polygon is 360°.

Let there be n sides of a polygon.

$72 n = 360 n = \frac{360}{72} n = 5$

Since it has 5 sides, the polygon is a pentagon.

(v) A pentagon has 5 diagonals.

$If n is the number of s ides, the number of diagonals = \frac{n (n - 3)}{2} \frac{5 (5 - 3)}{2} = 5$

Page No 203:

Answer:

(i) F
The diagonals of a parallelogram need not be equal in length.

(ii) F
The diagonals of a rectangle are not perpendicular to each other.

(iii) T

(iv) T

Adjacent sides of a kite are equal and this is also true for a rhombus. Additionally, all the sides of a rhombus are equal to each other.

Page No 203:

Answer:

Steps of construction:
Step 1: Take PQ = 4.2 cm
Step 2: $Make ∠ X P Q = 120^{\circ}, ∠ Y Q P = 60^{\circ}$
Step 3: Cut an arc of length 5 cm from point Q. Name that point as R.
Step 4: From P, make an arc of length 6 cm. Name that point as S.
Step 5: Join P and S.
Thus, PQRS is a quadrilateral.

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