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 Meritnationâ€™s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

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

Thus, ABCD is the required quadrilateral.

#### Question 2:

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

Thus, ABCD is the required quadrilateral.

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

Thus, PQRS is the required quadrilateral.

#### Question 3:

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

Thus, PQRS is the required quadrilateral.

Steps of construction:
Step 1: Draw .
Step 2: With B as the centre and radius equal to , draw an arc.
Step 3: With A as the centre and radius equal to , 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  draw an arc.
Step 6: With B as the centre and radius equal to , draw another arc, cutting the previous arc at C.
Step 7: Join BC and CD.

Thus, ABCD is the required quadrilateral.

#### Question 4:

Steps of construction:
Step 1: Draw .
Step 2: With B as the centre and radius equal to , draw an arc.
Step 3: With A as the centre and radius equal to , 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  draw an arc.
Step 6: With B as the centre and radius equal to , draw another arc, cutting the previous arc at C.
Step 7: Join BC and CD.

Thus, ABCD is the required quadrilateral.

Steps of construction:
Step 1: Draw .
Step 2: With B as the centre and radius equal to , draw an arc.
Step 3: With A as the centre and radius equal to , 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  draw an arc.
Step 6: With B as the centre and radius equal to , draw another arc, cutting the previous arc at C.
Step 7: Join AC, BC and CD.

Thus, ABCD is the required quadrilateral.

#### Question 5:

Steps of construction:
Step 1: Draw .
Step 2: With B as the centre and radius equal to , draw an arc.
Step 3: With A as the centre and radius equal to , 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  draw an arc.
Step 6: With B as the centre and radius equal to , draw another arc, cutting the previous arc at C.
Step 7: Join AC, BC and CD.

Thus, ABCD is the required quadrilateral.

Steps of construction:
Step 1: Draw
Step 2: With Q as the centre and radius equal to , draw an arc.
Step 3: With R as the centre and radius equal to , 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  draw an arc.
Step 6: With R as the centre and radius equal to , 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.

#### Question 6:

Steps of construction:
Step 1: Draw
Step 2: With Q as the centre and radius equal to , draw an arc.
Step 3: With R as the centre and radius equal to , 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  draw an arc.
Step 6: With R as the centre and radius equal to , 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.

Steps of construction:
Step 1: Draw
Step 2: With B as the centre and radius equal to , draw an arc.
Step 3: With A as the centre and radius equal to , 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  draw an arc.
Step 6: With D as the centre and radius equal to , draw another arc, cutting the previous arc at C.
Step 7: Join AC, CD and BC.

Thus, ABCD is the required quadrilateral.

#### Question 7:

Steps of construction:
Step 1: Draw
Step 2: With B as the centre and radius equal to , draw an arc.
Step 3: With A as the centre and radius equal to , 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  draw an arc.
Step 6: With D as the centre and radius equal to , draw another arc, cutting the previous arc at C.
Step 7: Join AC, CD and BC.

Thus, ABCD is the required quadrilateral.

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

#### Question 8:

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

Steps of construction:
Step 1: Draw AB$2.9cm$
Step 2: Make $\angle A={70}^{\circ }$
Step 3: With A as the centre, draw an arc of $3.4cm$. Name that point as D.
Step 4: With D as the centre, draw an arc of $2.7cm$.
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, $ABCD$ is the required quadrilateral.

#### Question 9:

Steps of construction:
Step 1: Draw AB$2.9cm$
Step 2: Make $\angle A={70}^{\circ }$
Step 3: With A as the centre, draw an arc of $3.4cm$. Name that point as D.
Step 4: With D as the centre, draw an arc of $2.7cm$.
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, $ABCD$ is the required quadrilateral.

Steps of construction:
Step 1: Draw BC$5cm$
Step 2: Make
Step 3: With B as the centre, draw an arc of . Name that point as A.
Step 4: With C as the centre, draw an arc of . Name that point as D.
Step 5: Join A and D.
Then, $ABCD$ is the required quadrilateral.

#### Question 10:

Steps of construction:
Step 1: Draw BC$5cm$
Step 2: Make
Step 3: With B as the centre, draw an arc of . Name that point as A.
Step 4: With C as the centre, draw an arc of . Name that point as D.
Step 5: Join A and D.
Then, $ABCD$ is the required quadrilateral.

Steps of construction:
Step 1: Draw QR
Step 2: Make
Step 3: With Q as the centre, draw an arc of . Name that point as P.
Step 4: With R as the centre, draw an arc of $2.7cm$. Name that point as S.
Step 6: Join P and S.
Then, $PQRS$ is the required quadrilateral.

#### Question 11:

Steps of construction:
Step 1: Draw QR
Step 2: Make
Step 3: With Q as the centre, draw an arc of . Name that point as P.
Step 4: With R as the centre, draw an arc of $2.7cm$. Name that point as S.
Step 6: Join P and S.
Then, $PQRS$ is the required quadrilateral.

Steps of construction:
Step 1: Draw AB
Step 2: Make
Step 3: With B as the centre, draw an arc of $4cm$.
Step 3: Sum of all the angles of the quadrilateral is ${360}^{\circ }$.
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }\phantom{\rule{0ex}{0ex}}{50}^{\circ }+{105}^{\circ }+\angle C+{80}^{\circ }={360}^{\circ }\phantom{\rule{0ex}{0ex}}{235}^{\circ }+\angle C={360}^{\circ }\phantom{\rule{0ex}{0ex}}\angle C={360}^{\circ }-{235}^{\circ }\phantom{\rule{0ex}{0ex}}\angle C={125}^{\circ }$
Step 5: With C as the centre, make .
Step 6: Join C and D.
Step 7: Measure $\angle D={80}^{\circ }$
Then, $ABCD$ is the required quadrilateral.

#### Question 12:

Steps of construction:
Step 1: Draw AB
Step 2: Make
Step 3: With B as the centre, draw an arc of $4cm$.
Step 3: Sum of all the angles of the quadrilateral is ${360}^{\circ }$.
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }\phantom{\rule{0ex}{0ex}}{50}^{\circ }+{105}^{\circ }+\angle C+{80}^{\circ }={360}^{\circ }\phantom{\rule{0ex}{0ex}}{235}^{\circ }+\angle C={360}^{\circ }\phantom{\rule{0ex}{0ex}}\angle C={360}^{\circ }-{235}^{\circ }\phantom{\rule{0ex}{0ex}}\angle C={125}^{\circ }$
Step 5: With C as the centre, make .
Step 6: Join C and D.
Step 7: Measure $\angle D={80}^{\circ }$
Then, $ABCD$ is the required quadrilateral.

Steps of construction:
Step 1: Draw PQ$5cm$
Step 2:
$\angle P+\angle Q+\angle R+\angle S={360}^{\circ }\phantom{\rule{0ex}{0ex}}{100}^{\circ }+\angle Q+{100}^{\circ }+{75}^{\circ }={360}^{\circ }\phantom{\rule{0ex}{0ex}}{275}^{\circ }+\angle Q={360}^{\circ }\phantom{\rule{0ex}{0ex}}\angle Q={360}^{\circ }-{275}^{\circ }\phantom{\rule{0ex}{0ex}}\angle Q={85}^{\circ }$
Step 3: Make
Step 3: With Q as the centre, draw an arc of .
Step 4: Make $\angle R={100}^{\circ }$
Step 6: Join R and S.
Step 7: Measure $\angle S={75}^{\circ }$
Then, $PQRS$ is the required quadrilateral.

#### Question 13:

Steps of construction:
Step 1: Draw PQ$5cm$
Step 2:
$\angle P+\angle Q+\angle R+\angle S={360}^{\circ }\phantom{\rule{0ex}{0ex}}{100}^{\circ }+\angle Q+{100}^{\circ }+{75}^{\circ }={360}^{\circ }\phantom{\rule{0ex}{0ex}}{275}^{\circ }+\angle Q={360}^{\circ }\phantom{\rule{0ex}{0ex}}\angle Q={360}^{\circ }-{275}^{\circ }\phantom{\rule{0ex}{0ex}}\angle Q={85}^{\circ }$
Step 3: Make
Step 3: With Q as the centre, draw an arc of .
Step 4: Make $\angle R={100}^{\circ }$
Step 6: Join R and S.
Step 7: Measure $\angle S={75}^{\circ }$
Then, $PQRS$ is the required quadrilateral.

Steps of construction:
Step 1: Draw $AB=4cm$
Step 2:
Step 3: $A{C}^{2}=A{B}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}{5}^{2}={4}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}25-16=B{C}^{2}\phantom{\rule{0ex}{0ex}}BC=3cm$
With B as the centre, draw an arc equal to 3 cm.
Step 4: Make $\angle C={90}^{\circ }$
Step 5: With A as the centre and radius equal to , draw an arc and name that point as D.
Then, $ABCD$ is the required quadrilateral.

#### Question 1:

Steps of construction:
Step 1: Draw $AB=4cm$
Step 2:
Step 3: $A{C}^{2}=A{B}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}{5}^{2}={4}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}25-16=B{C}^{2}\phantom{\rule{0ex}{0ex}}BC=3cm$
With B as the centre, draw an arc equal to 3 cm.
Step 4: Make $\angle C={90}^{\circ }$
Step 5: With A as the centre and radius equal to , draw an arc and name that point as D.
Then, $ABCD$ is the required quadrilateral.

Steps of construction:
Step 1: Draw AB $5.2cm$
Step 2: With B as the centre, draw an arc of .
Step 3: With A as the centre, draw another arc of , 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 $5.2cm$.
Step 6: With A as the centre, draw another arc of , cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

#### Question 2:

Steps of construction:
Step 1: Draw AB $5.2cm$
Step 2: With B as the centre, draw an arc of .
Step 3: With A as the centre, draw another arc of , 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 $5.2cm$.
Step 6: With A as the centre, draw another arc of , cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

Steps of construction:
Step 1: Draw AB= $4.3cm$
Step 2: With B as the centre, draw an arc of .
Step 3: With A as the centre, draw another arc of $4cm$, 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 $4.3cm$.
Step 6: With B as the centre, draw another arc of , cutting the previous arc at C.
Step 7: Join CD and BC.
â€‹then, ABCD is the required parallelogram.

#### Question 3:

Steps of construction:
Step 1: Draw AB= $4.3cm$
Step 2: With B as the centre, draw an arc of .
Step 3: With A as the centre, draw another arc of $4cm$, 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 $4.3cm$.
Step 6: With B as the centre, draw another arc of , cutting the previous arc at C.
Step 7: Join CD and BC.
â€‹then, ABCD is the required parallelogram.

Steps of construction:
Step 1: Draw PQ= 4 cm
Step 2: Make $\angle PQR={60}^{\circ }$
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.

#### Question 4:

Steps of construction:
Step 1: Draw PQ= 4 cm
Step 2: Make $\angle PQR={60}^{\circ }$
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.

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

#### Question 5:

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

We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw AB$4.4cm$
Step 2: With A as the centre and radius $2.8cm$, draw an arc.
Step 3: With B as the centre and radius $3.5cm$, 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.

#### Question 6:

We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw AB$4.4cm$
Step 2: With A as the centre and radius $2.8cm$, draw an arc.
Step 3: With B as the centre and radius $3.5cm$, 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.

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.

#### Question 7:

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.

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

Steps of construction:
Step 1: Draw AC$3.8\mathrm{cm}$
Step 2: Bisect AC at O.
Step 3: Make $\angle COX={60}^{\circ }$
Produce XO to Y.
Step 4:

Step 5: Join AB, BC, CD and AD.
â€‹Thus, ABCD is the required parallelogram.

#### Question 8:

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

Steps of construction:
Step 1: Draw AC$3.8\mathrm{cm}$
Step 2: Bisect AC at O.
Step 3: Make $\angle COX={60}^{\circ }$
Produce XO to Y.
Step 4:

Step 5: Join AB, BC, CD and AD.
â€‹Thus, ABCD is the required parallelogram.

Steps of construction:
Step 1: Draw AB = $11cm$
Step 2: Make
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.

#### Question 9:

Steps of construction:
Step 1: Draw AB = $11cm$
Step 2: Make
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.

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB $6.4cm$
Step 2: Make
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.
â€‹Thus, ABCD is a required square.

#### Question 10:

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB $6.4cm$
Step 2: Make
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.
â€‹Thus, ABCD is a required square.

We know that the diagonals of a square bisect each other at right angles.
Steps of construction:
Step 1: Draw AC
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:

Step 4: Join AB, BC, CD and DA.
ABCD is the required square.

#### Question 11:

We know that the diagonals of a square bisect each other at right angles.
Steps of construction:
Step 1: Draw AC
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:

Step 4: Join AB, BC, CD and DA.
ABCD is the required square.

Steps of construction:
Step 1: Draw QR$3.6cm$
Step 2: Make  $\angle Q={90}^{\circ }\phantom{\rule{0ex}{0ex}}\angle R={90}^{\circ }$
Step 3:

Step 3: Draw an arc of length 4.8 cm from point Q and name that point as P.
â€‹Step 4: Draw an arc of length 6 cm from point R, cutting the previous arc at P.
â€‹Step 5: Join PQ
Step 6: Draw an arc of length 4.8 cm from point R.
F
rom 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.

#### Question 12:

Steps of construction:
Step 1: Draw QR$3.6cm$
Step 2: Make  $\angle Q={90}^{\circ }\phantom{\rule{0ex}{0ex}}\angle R={90}^{\circ }$
Step 3:

Step 3: Draw an arc of length 4.8 cm from point Q and name that point as P.
â€‹Step 4: Draw an arc of length 6 cm from point R, cutting the previous arc at P.
â€‹Step 5: Join PQ
Step 6: Draw an arc of length 4.8 cm from point R.
F
rom 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.

We know that the diagonals of a rhombus bisect each other.
.Steps of construction:
Step 1: Draw AC= $6cm$
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:

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.
â€‹Thus, ABCD is the required rhombus, as shown in the figure.

#### Question 13:

We know that the diagonals of a rhombus bisect each other.
.Steps of construction:
Step 1: Draw AC= $6cm$
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:

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.
â€‹Thus, ABCD is the required rhombus, as shown in the figure.

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

#### Question 14:

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

Steps of construction:
Step1: Draw AB
Step2: Draw
Sum of the adjacent angles is 180°.
$\angle BAX+\angle ABY=180°\phantom{\rule{0ex}{0ex}}=>\angle BAX=180°-60°=120°$
Step 3:

Step 4: Join C and D.
Then, ABCD is the required rhombus.

#### Question 15:

Steps of construction:
Step1: Draw AB
Step2: Draw
Sum of the adjacent angles is 180°.
$\angle BAX+\angle ABY=180°\phantom{\rule{0ex}{0ex}}=>\angle BAX=180°-60°=120°$
Step 3:

Step 4: Join C and D.
Then, ABCD is the required rhombus.

Steps of construction:
Step 1: Draw AB=
Step 2: Make  $\angle ABX={75}^{\circ }$
Step 3: With B as the centre, draw an arc at $4cm$. Name that point as C.
Step 4:  $AB\parallel CD$

Make  $\angle BCY=105°$
At C, draw an arc of length .
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

#### Question 16:

Steps of construction:
Step 1: Draw AB=
Step 2: Make  $\angle ABX={75}^{\circ }$
Step 3: With B as the centre, draw an arc at $4cm$. Name that point as C.
Step 4:  $AB\parallel CD$

Make  $\angle BCY=105°$
At C, draw an arc of length .
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle,
Step3: With B as the centre, draw an arc of . Name that point as C. Join B and C.
Step4:

Draw an angle,

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

Step5: Join A and D.

â€‹Thus, ABCD is the required trapezium.

#### Question 1:

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle,
Step3: With B as the centre, draw an arc of . Name that point as C. Join B and C.
Step4:

Draw an angle,

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

Step5: Join A and D.

â€‹Thus, ABCD is the required trapezium.

( 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.

#### Question 2:

( 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.

Let the angles be
Sum of the angles of a quadrilateral is ${360}^{\circ }$.
$x+2x+3x+4x=360\phantom{\rule{0ex}{0ex}}10x=360\phantom{\rule{0ex}{0ex}}x=\frac{360}{10}\phantom{\rule{0ex}{0ex}}x=36$

$\left(2x\right)°=\left(2×36{\right)}^{\circ }={72}^{\circ }\phantom{\rule{0ex}{0ex}}\left(3x\right)°=\left(3×36{\right)}^{\circ }={108}^{\circ }\phantom{\rule{0ex}{0ex}}\left(4x\right)°=\left(4×36{\right)}^{\circ }={144}^{\circ }$

The angles of the quadrilateral are

#### Question 3:

Let the angles be
Sum of the angles of a quadrilateral is ${360}^{\circ }$.
$x+2x+3x+4x=360\phantom{\rule{0ex}{0ex}}10x=360\phantom{\rule{0ex}{0ex}}x=\frac{360}{10}\phantom{\rule{0ex}{0ex}}x=36$

$\left(2x\right)°=\left(2×36{\right)}^{\circ }={72}^{\circ }\phantom{\rule{0ex}{0ex}}\left(3x\right)°=\left(3×36{\right)}^{\circ }={108}^{\circ }\phantom{\rule{0ex}{0ex}}\left(4x\right)°=\left(4×36{\right)}^{\circ }={144}^{\circ }$

The angles of the quadrilateral are

Sum of any two adjacent angles of a parallelogram is ${180}^{\circ }$.

$\left(2x\right)°=\left(2×36\right)°={72}^{\circ }\phantom{\rule{0ex}{0ex}}\left(3x\right)°=\left(3×36\right)°={108}^{\circ }$

Measures of the angles are .

#### Question 4:

Sum of any two adjacent angles of a parallelogram is ${180}^{\circ }$.

$\left(2x\right)°=\left(2×36\right)°={72}^{\circ }\phantom{\rule{0ex}{0ex}}\left(3x\right)°=\left(3×36\right)°={108}^{\circ }$

Measures of the angles are .

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

$2\left(l+b\right)=180\phantom{\rule{0ex}{0ex}}⇒2\left(4x+5x\right)=180\phantom{\rule{0ex}{0ex}}⇒2\left(9x\right)=180\phantom{\rule{0ex}{0ex}}⇒18x=180\phantom{\rule{0ex}{0ex}}⇒x=10$

#### Question 5:

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

$2\left(l+b\right)=180\phantom{\rule{0ex}{0ex}}⇒2\left(4x+5x\right)=180\phantom{\rule{0ex}{0ex}}⇒2\left(9x\right)=180\phantom{\rule{0ex}{0ex}}⇒18x=180\phantom{\rule{0ex}{0ex}}⇒x=10$

Rhombus is a parallelogram.

Consider:

Therefore, the diagonals bisects at O.

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

Consider :

(corresponding parts of congruent triangles)

Further,

$\angle COD=\angle COB=90°$

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

#### Question 6:

Rhombus is a parallelogram.

Consider:

Therefore, the diagonals bisects at O.

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

Consider :

(corresponding parts of congruent triangles)

Further,

$\angle COD=\angle COB=90°$

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

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 $△AOB$:

Therefore, the length of each side of the rhombus is 10 cm.

#### Question 7:

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 $△AOB$:

Therefore, the length of each side of the rhombus is 10 cm.

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

Opposite angles of a parallelogram are equal.

Therefore, the first and the second angles are:
${\left(3x-2\right)}^{\circ }={\left(2×13-2\right)}^{\circ }={37}^{\circ }\phantom{\rule{0ex}{0ex}}{\left(50-x\right)}^{\circ }={\left(50-13\right)}^{\circ }={37}^{\circ }\phantom{\rule{0ex}{0ex}}$
Sum of adjacent angles in a parallelogram is ${180}^{\circ }$.
Adjacent angles = ${180}^{\circ }-{37}^{\circ }={143}^{\circ }$

#### Question 8:

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

Opposite angles of a parallelogram are equal.

Therefore, the first and the second angles are:
${\left(3x-2\right)}^{\circ }={\left(2×13-2\right)}^{\circ }={37}^{\circ }\phantom{\rule{0ex}{0ex}}{\left(50-x\right)}^{\circ }={\left(50-13\right)}^{\circ }={37}^{\circ }\phantom{\rule{0ex}{0ex}}$
Sum of adjacent angles in a parallelogram is ${180}^{\circ }$.
Adjacent angles = ${180}^{\circ }-{37}^{\circ }={143}^{\circ }$

(d) none of the these

Let the angles be .

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

$x+3x+7x+9x=360\phantom{\rule{0ex}{0ex}}20x=360\phantom{\rule{0ex}{0ex}}x=18$

#### Question 9:

(d) none of the these

Let the angles be .

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

$x+3x+7x+9x=360\phantom{\rule{0ex}{0ex}}20x=360\phantom{\rule{0ex}{0ex}}x=18$

(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.

Breadth of the rectangle = 6 cm

#### Question 10:

(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.

Breadth of the rectangle = 6 cm

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

#### Question 11:

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

(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:

Hence, the two bisectors intersect at right angles.

#### Question 12:

(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:

Hence, the two bisectors intersect at right angles.

(c) 9
Hexagon has six sides.

#### Question 13:

(c) 9
Hexagon has six sides.

(b) 8

It has 8 sides.

#### Question 14:

(b) 8

It has 8 sides.

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

(ii) Sum of all interior angles = $\left(n-2\right)×180°\phantom{\rule{0ex}{0ex}}$

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

#### Question 15:

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

(ii) Sum of all interior angles = $\left(n-2\right)×180°\phantom{\rule{0ex}{0ex}}$

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

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

(ii) Sum of all interior angles of a polygon is

#### Question 16:

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

(ii) Sum of all interior angles of a polygon is

(i) Octagon has 8 sides.

(ii) Sum of the interior angles of a regular hexagon = $\left(6-2\right)×{180}^{\circ }={720}^{\circ }$

(iii) Each exterior angle of a regular polygon is ${60}^{\circ }$.

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.

$72n=360\phantom{\rule{0ex}{0ex}}n=\frac{360}{72}\phantom{\rule{0ex}{0ex}}n=5$

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

(v) A pentagon has 5 diagonals.

#### Question 17:

(i) Octagon has 8 sides.

(ii) Sum of the interior angles of a regular hexagon = $\left(6-2\right)×{180}^{\circ }={720}^{\circ }$

(iii) Each exterior angle of a regular polygon is ${60}^{\circ }$.

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.

$72n=360\phantom{\rule{0ex}{0ex}}n=\frac{360}{72}\phantom{\rule{0ex}{0ex}}n=5$

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

(v) A pentagon has 5 diagonals.

(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.

#### Question 18:

(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.