Mathematics – Class 7 – Chapter 10 – Practical Geometry – Exercise 10.5 – NCERT Exercise Solution

1. Construct the right angled ΔPQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm.

Solution:

Steps:

Step 1. Draw a line segment QR = 8 cm.

Step 2. Draw a ray QY, at point Q to make an angle of 90^o i.e. ∠YQR = 90^o.

Step 3. Now, R as a center with a radius of 10 cm, draw an arc that cuts the ray QY at P.

Step 4. Join PR.

Hence, ΔPQR is the required right-angled triangle.

2. Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Solution:

Let the ΔABC is a right-angled triangle at ∠B = 90^o

Given,

AC is hypotenuse = 6 cm

BC = 4 cm

Now,

According to the question we have to construct the right-angled triangle.

Steps:

Step 1. Draw a line segment BC = 4 cm.

Step 2. Draw a ray BX, at point B to make an angle of 90^o i.e. ∠XBC = 90^o.

Step 3. Now, C as a center with a radius of 6 cm, draw an arc that cuts the ray BX at A.

Step 4. Join AC.

Hence, ΔABC is the required right-angled triangle.

3. Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.

Solution:

Steps:

Step 1. Draw a line segment BC = 6 cm.

Step 2. Draw a ray CX, at point C to make an angle of 90^o i.e. ∠XCB = 90^o.

Step 3. Now, C as a center with a radius of 6 cm, draw an arc that cuts the ray CX at A.

Step 4. Join AB.

Hence, ΔABC is the required right-angled triangle.