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

Exercise 17A

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

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Question 1:

Construct a quadrilateral ABCD in which AB = 4.2 cm, BC = 6 cm, CD = 5.2 cm, DA = 5 cm and AC = 8 cm.

Answer 1:

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

Thus, ABCD is the required quadrilateral.

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

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Question 2:

Construct a quadrilateral PQRS in which PQ = 5.4 cm, QR = 4.6 cm, RS = 4.3 cm, SP = 3.5 cm and diagonal PR = 4 cm.

Answer 2:

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

Thus, PQRS is the required quadrilateral.

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

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Question 3:

Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 3.8 cm, CD = DA = 4.5 cm and diagonal BD = 5.6 cm.

Answer 3:

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

Thus, ABCD is the required quadrilateral.

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

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Question 4:

Construct a quadrilateral ABCD in which AB = 3.6 cm, BC = 3.3 cm, AD = 2.7 cm, diagonal AC = 4.6 cm and diagonal BD = 4 cm.

Answer 4:

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

Thus, ABCD is the required quadrilateral.

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

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Question 5:

Construct a quadrilateral PQRS in which QR = 7.5 cm, PR = PS = 6 cm, RS = 5 cm and QS = 10 cm. Measure the fourth side.

Answer 5:

Steps of construction:
Step 1: Draw $QR=7.5 \mathrm{cm}.$
Step 2: With Q as the centre and radius equal to $10 \mathrm{cm}$, draw an arc.
Step 3: With R as the centre and radius equal to $5 \mathrm{cm}$, 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 $6 \mathrm{cm},$ draw an arc.
Step 6: With R as the centre and radius equal to $6 \mathrm{cm}$, 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.

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

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Question 6:

construct a quadrilateral ABCD in which AB =3.4 cm, CD = 3 cm, DA = 5.7 cm, AC = 8 cm and BD = 4 cm.

Answer 6:

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

Thus, ABCD is the required quadrilateral.

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

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Question 7:

Construct a quadrilateral ABCD in which AB = BC = 3.5 cm, AD = CD = 5.2 cm and ∠ABC = 120°.

Answer 7:

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

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

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Question 8:

Construct a quadrilateral ABCD in which AB = 2.9 cm, BC = 3.2 cm, CD = 2.7 cm, DA = 3.4 cm and ∠A = 70°.

Answer 8:

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.

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

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Question 9:

Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 5 cm, CD = 4.6 cm, ∠B = 125° and ∠C = 60°.

Answer 9:

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

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

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Question 10:

Construct a quadrilateral PQRS in which PQ = 6 cm, QR = 5.6 cm, RS = 2.7 cm, ∠Q = 45° and ∠R = 90°.

Answer 10:

Steps of construction:
Step 1: Draw QR= $5.6 cm$
Step 2: Make 
Step 3: With Q as the centre, draw an arc of $6 cm$. 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.

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

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Question 11:

Construct a quadrilateral ABCD in which AB = 5.6 cm, BC = 4 cm, ∠A = 50°, ∠B = 105° and ∠D = 80°.

Answer 11:

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

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

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Question 12:

Construct a quadrilateral PQRS in which PQ = 5 cm, QR = 6.5 cm, ∠P = ∠R = 100° and ∠S = 75°.

Answer 12:

Steps of construction:
Step 1: Draw PQ= $5cm$
Step 2:
$$\begin{gathered} \angle P+\angle Q+\angle R+\angle S={360}^{\circ } \\ {100}^{\circ }+\angle Q+{100}^{\circ }+{75}^{\circ }={360}^{\circ } \\ {275}^{\circ }+\angle Q={360}^{\circ } \\ \angle Q={360}^{\circ }-{275}^{\circ } \\ \angle Q={85}^{\circ } \end{gathered}$$
Step 3: Make $\angle P={100}^{\circ } and \angle Q={85}^{\circ }$
Step 3: With Q as the centre, draw an arc of $6.5 cm$.
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.

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

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Question 13:

Construct a quadrilateral ABCD in which AB = 4 cm, AC = 5 cm, AD = 5.5 cm and ∠ABC = ∠ACD = 90°.

Answer 13:

Steps of construction:
Step 1: Draw $AB=4cm$
Step 2: $Make \angle B={90}^{\circ }$
Step 3: $$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ {5}^2=4^2+B{C}^2 \\ 25-16=B{C}^2 \\ BC=3cm \end{gathered}$$
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 $5.5 cm$, draw an arc and name that point as D.
Then, $ABCD$ is the required quadrilateral.