# RD Sharma Solutions Class 7 Constructions Exercise 17.5

## RD Sharma Solutions Class 7 Chapter 17 Exercise 17.5

#### Exercise 17.5

Q1. Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.

Steps of construction:

1. Draw a line segment QR = 4 cm.
2. Draw $\angle QRX$ of measure $90^{\circ}$.
3. With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
4. Join PQ to obtain the desired triangle PQR.

PQR is the required triangle.

Q2. Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.

Steps of construction:

1. Draw a line segment QR = 2.5 cm.
2. Draw $\angle QRX$ of measure $90^{\circ}$.
3. With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
4. Join PQ to obtain the desired triangle PQR.

PQR is the required triangle.

Q3. Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure $30^{\circ}$

Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let cC = 30°.

Therefore $\angle A+\angle B+\angle C=180^{\circ}\\ \angle B=180^{\circ}-30^{\circ}-90^{\circ}=60^{\circ}$

Steps of construction:

1. Draw a line segment BC = 5.4 cm.
2. Draw angle CBY=$60^{\circ}$
3. Draw angle BCX of measure $30^{\circ}$ with X on the same side of BC as Y.
4. Let BY and CX intersect at A.

Then ABC is the required triangle.

Q4. Construct a right triangle ABC in which AB= 5.8 cm, BC= 4.5 cm and $\angle C=90^{\circ}$.

Steps of construction:

1. Draw a line segment BC = 4.5 cm.
2. Draw $\angle BCX$ of measure $90^{\circ}$..
3. With centre B and radius AB =5.8 cm, draw an arc of the circle to intersect ray BX at A.
4. Join AB to obtain the desired triangle ABC.

ABC is the required triangle.

Q5. Construct a right triangle, right angled at C in which AB = 5.2 cm and BC= 4.6 cm.

Steps of construction:

1. Draw a line segment BC = 4.6 cm.
2. Draw $\angle BCX$ of measure $90^{\circ}$
3. With centre B and radius AB =5.2cm, draw an arc of the circle to intersect ray CX at A.
4. Join AB to obtain the desired triangle ABC.

ABC is the required triangle.