RD Sharma Solutions Class 7 Constructions Exercise 17.5

RD Sharma Class 7 Solutions Chapter 17 Ex 17.5 PDF Free Download

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.

1

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.

2

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 \(30^{\circ}\) 

3 

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 \(\angle C=90^{\circ}\).

4

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.

5

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.

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