RD Sharma Solutions For Class 7 Maths Exercise 17.5 Chapter 17 Constructions

In this exercise, we shall discuss RHS triangle construction and steps involved in them. The primary objective is to help students improve their problem-solving abilities, which in turn, helps in boosting their confidence level to achieve high marks in their exams. Students can refer and download RD Sharma Solutions for Class 7 Maths Exercise 17.5 Chapter 17 Constructions. These solutions are formulated by BYJU’S expert team in Maths.

Download the PDF of RD Sharma Solutions For Class 7 Chapter 17 – Constructions Exercise 17.5

 

rd sharma class 7 maths solution ch 17 ex 5
rd sharma class 7 maths solution ch 17 ex 5
rd sharma class 7 maths solution ch 17 ex 5
rd sharma class 7 maths solution ch 17 ex 5

 

Access answers to Maths RD Sharma Solutions For Class 7 Chapter 17 – Constructions Exercise 17.5

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

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 24

Steps of construction:

1. Draw a line segment QR = 4 cm.

2. Draw ∠QRX of measure 90o.

3. With center 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.

5. PQR is the required triangle.

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

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 25

Steps of construction:

1. Draw a line segment QR = 2.5 cm.

2. Draw ∠QRX of measure 90o.

3. With center 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.

5. PQR is the required triangle.

3. Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30o

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 26

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

Therefore ∠A + ∠B + ∠C = 180o

∠B = 180o − 30o− 90o = 60o

Steps of construction:

1. Draw a line segment BC = 5.4 cm.

2. Draw angle CBY = 60o

3. Draw angle BCX of measure 30o with X on the same side of BC as Y.

4. Let BY and CX intersect at A.

5. Then ABC is the required triangle.

4. Construct a right triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and ∠C = 90o.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 27

Steps of construction:

1. Draw a line segment BC = 4.5 cm.

2. Draw ∠BCX of measure 90o

3. With center 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.

5. ABC is the required triangle.

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

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 28

Steps of construction:

1. Draw a line segment BC = 4.6 cm.

2. Draw ∠BCX of measure 90o

3. With center B and radius AB = 5.2 cm, draw an arc of the circle to intersect ray CX at A.

4. Join AB to obtain the desired triangle ABC.

5. ABC is the required triangle.


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