Mathematics – Class 7 – Chapter 10 – Practical Geometry – Exercise 10.2 – NCERT Exercise Solution

1. Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm

Solution:

Steps of construction:

Step 1. Draw a line segment YZ = 5 cm.

Step 2. With Z as a center and radius 6 cm, draw an arc outside.

Step 3. With Y as a center and radius 4.5 cm, draw another arc, cutting the previous arc at X.

Step 4. Join XY and XZ.

Now, ΔXYZ is the required triangle.

Figure:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.2 - NCERT Exercise Solution

2. Construct an equilateral triangle of side 5.5 cm.

Solution:

Steps of construction:

Step 1. Draw a line segment AB = 5.5 cm.

Step 2. With A as a center and radius 5.5 cm, draw an arc outside.

Step 3. With B as a center and radius 5.5 cm, draw another arc, cutting the previous arc at C.

Step 4. Join CA and CB.

Now, ΔABC is the required equilateral triangle.

Figure:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.2 - NCERT Exercise Solution

3. Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

Solution:

Steps of construction:

Step 1. Draw a line segment QR = 3.5 cm.

Step 2. With Q as a center and radius 4 cm, draw an arc outside.

Step 3. With R as a center and radius 4 cm, draw another arc, cutting the previous arc at P.

Step 4. Join PQ and PR.

Now, ΔPQR is the required isosceles triangle.

Figure:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.2 - NCERT Exercise Solution

4. Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

Solution:

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

Step 2. With B as a center and radius 2.5 cm, draw an arc outside.

Step 3. With C as a center and radius 6.5 cm, draw another arc, cutting the previous arc at A.

Step 4. Join AB and AC.

Now, ΔABC is the required triangle.

Figure:

Mathematics - Class 7 - Chapter 10 - Practical Geometry - Exercise 10.2 - NCERT Exercise Solution

Now, when we measure the angle B by the protector then we find that angle B is 90o

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