RS Aggarwal solution class 8 chapter 17 Construction of Quadrilaterals Exercise 17B

Exercise 17B

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Q1 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 1:

Construct a parallelogram ABCD in which AB = 5.2 cm, BC = 4.7 cm and AC = 7.6 cm.

Answer 1:

Steps of construction:
Step 1: Draw AB = $5.2cm$
Step 2: With B as the centre, draw an arc of $4.7 cm$.
Step 3: With A as the centre, draw another arc of $7.6 cm$, 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 $4.7 cm$, cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

Q2 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 2:

Construct a parallelogram ABCD in which AB = 4.3 cm, AD = 4 cm and BD = 6.8 cm.

Answer 2:

Steps of construction:
Step 1: Draw AB= $4.3cm$
Step 2: With B as the centre, draw an arc of $6.8 cm$.
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 $4 cm$, cutting the previous arc at C.
Step 7: Join CD and BC.
​then, ABCD is the required parallelogram.

Q3 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 3:

Construct a parallelogram PQRS in which QR = 6 cm, PQ = 4 cm and ∠PQR = 60° cm.

Answer 3:

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.

Q4 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 4:

Construct a parallelogram ABCD in which BC = 5 cm, ∠BCD = 120° and CD = 4.8 cm.

Answer 4:

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 $4.8 \mathrm{cm}$, 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 $4.8 \mathrm{cm}$ cutting the previous arc at A.
Step 5: Join AD and AB
​then, ABCD is a required parallelogram.

Q5 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 5:

Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side.

Answer 5:

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.

Q6 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 6:

Construct a parallelogram ABCD in which AB = 6.5 cm, AC = 3.4 cm and the altitude AL from A is 2.5 cm. Draw the altitude from C and measure it.

Answer 6:

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.

Q7 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 7:

Construct a parallelogram ABCD, in which diagonal AC = 3.8 cm, diagonal BD = 4.6 cm and the angle between AC and BD is 60°.

Answer 7:

 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.
​Thus, ABCD is the required parallelogram.

Q8 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 8:

Construct a rectangle ABCD whose adjacent sides are 11 cm and 8.5 cm.

Answer 8:

Steps of construction:
Step 1: Draw AB = $11cm$ 
Step 2: Make $$\begin{gathered} \angle A={90}^{\circ } \\ \angle B={90}^{\circ } \end{gathered}$$
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.

Q9 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 9:

Construct a square, each of whose sides measures 6.4 cm.

Answer 9:

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB = $6.4cm$ 
Step 2: Make $$\begin{gathered} \angle A={90}^{\circ } \\ \angle B={90}^{\circ } \end{gathered}$$
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.

Q10 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 10:

Construct a square, each of whose diagonals measures 5.8 cm.

Answer 10:

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

Q11 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 11:

Construct a rectangle PQRS in which QR = 3.6 cm and diagonal PR = 6 cm. Measure the other side of the rectangle.

Answer 11:

Steps of construction:
Step 1: Draw QR = $3.6cm$ 
Step 2: Make  $$\begin{gathered} \angle Q={90}^{\circ } \\ \angle R={90}^{\circ } \end{gathered}$$
Step 3:
 

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.

Q12 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 12:

Construct a rhombus the lengths of whose diagonals are 6 cm and 8 cm.

Answer 12:

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.
​Thus, ABCD is the required rhombus, as shown in the figure.

Q13 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 13:

Construct a rhombus ABCD in which AB = 4 cm and diagonal AC is 6.5 cm.

Answer 13:

Steps of construction:
Step 1: Draw AB = $4cm$
Step 2: With B as the centre, draw an arc of $4 \mathrm{cm}$.
Step 3: With A as the centre, draw another arc of $6.5 \mathrm{cm}$, 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 $4 \mathrm{cm}$, cutting the previous arc at D.
Step 7: Join AD and CD.
ABCD is the required rhombus.

Q14 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 14:

Draw a rhombus whose side is 7.2 cm and one angle is 60°.

Answer 14:

Steps of construction:
Step1: Draw AB = 
Step2: Draw
Sum of the adjacent angles is 180°.

Step 3:  

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

Q15 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 15:

Construct a trapezium ABCD in which AB = 6 cm, BC = 4 cm, CD = 3.2 cm, ∠B = 75° and DC||AB.

Answer 15:

Steps of construction:
Step 1: Draw AB=$6 \mathrm{cm}$
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 
At C, draw an arc of length $3.2 \mathrm{cm}$.
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

Q16 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 16:

Draw a trapezium ABCD in which AB||DC, AB = 7 cm, BC = 5 cm, AD = 6.5 cm and ∠B = 60°.

Answer 16:

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle,
Step3: With B as the centre, draw an arc of $5 \mathrm{cm}$. Name that point as C. Join B and C.
Step4:


Draw an angle,  $\angle BCY, \mathrm{equal} \mathrm{to} 120°.$

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

Step5: Join A and D.

​Thus, ABCD is the required trapezium.