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

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

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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