Table of Contents

## Practical Geometry Class 7 Extra Questions Maths Chapter 10

**Extra Questions for Class 7 Maths Chapter 10 Practical Geometry**

### Practical Geometry Class 7 Extra Questions Very Short Answer Type

Question 1.

State whether the triangle is possible to construct if

(a) In ΔABC, m∠A = 80°, m∠B = 60°, AB = 5.5 cm

(b) In ΔPQR, PQ = 5 cm, QR = 3 cm, PR = 8.8 cm

Solution:

(a) m∠A = 80°, m∠B = 60°

m∠A + m∠B = 80° + 60° = 140° < 180°

So, ΔABC can be possible to construct.

(b) PQ = 5 cm, QR = 3 cm, PR = 8.8 cm

PQ + QR = 5 cm + 3 cm = 8 cm < 8.8 cm

or PQ + QR < PR

So, the ΔPQR can not be constructed.

Question 2.

Draw an equilateral triangle whose each side is 4.5 cm.

Solution:

Steps of construction:

(i) Draw AB = 4.5 cm.

(ii) Draw two arcs with centres A and B and same radius of 4.5 cm to meet each other at C.

(iii) Join CA and CB.

(iv) ΔCAB is the required triangle.

Question 3.

Draw a ΔPQR, in which QR = 3.5 cm, m∠Q = 40°, m∠R = 60°.

Solution:

Steps of construction:

(i) Draw QR = 3.5 cm.

(ii) Draw ∠Q = 40°, ∠R = 60° which meet each other at P.

(iii) ΔPQR is the required triangle.

Question 4.

There are four options, out of which one is correct. Choose the correct one:

(i) A triangle can be constructed with the given measurement.

(a) 1.5 cm, 3.5 cm, 4.5 cm

(b) 6.5 cm, 7.5 cm, 15 cm

(c) 3.2 cm, 2.3 cm, 5.5 cm

(d) 2 cm, 3 cm, 6 cm

(ii) (a) m∠P = 40°, m∠Q = 60°, AQ = 4 cm

(b) m∠B = 90°, m∠C = 120° , AC = 6.5 cm

(c) m∠L = 150°, m∠N = 70°, MN = 3.5 cm

(d) m∠P = 105°, m∠Q = 80°, PQ = 3 cm

Solution:

(i) Option (a) is possible to construct.

1.5 cm + 3.5 cm > 4.5 cm

(ii) Option (a) is correct.

m∠P + m∠Q = 40° + 60° = 100° < 180°

Question 5.

What will be the other angles of a right-angled isosceles triangle?

Solution:

In right angled isosceles triangle ABC, ∠B = 90°

∠A + ∠C = 180° – 90° = 90°

But ∠A = ∠B

∠A = ∠C = \(\frac { 90 }{ 2 }\) = 45°

Hence the required angles are ∠A = ∠C = 45°

Question 6.

What is the measure of an exterior angle of an equilateral triangle?

Solution:

We know that the measure of each interior angle = 60°

Exterior angle = 180° – 60° = 120°

Question 7.

In ΔABC, ∠A = ∠B = 50°. Name the pair of sides which are equal.

Solution:

∠A = ∠B = 50°

AC = BC [∵ Sides opposite to equal angles are equal]
Hence, the required sides are AC and BC.

Question 8.

If one of the other angles of a right-angled triangle is obtuse, whether the triangle is possible to construct.

Solution:

We know that the angles other than right angle of a right-angled triangle are acute angles.

So, such a triangle is not possible to construct.

Question 9.

State whether the given pair of triangles are congruent.

Solution:

Here, AB = PQ = 3.5 cm

AC = PR = 5.2 cm

∠BAC = ∠QPR = 70°

ΔABC = ΔPQR [By SAS rule]

### Practical Geometry Class 7 Extra Questions Short Answer Type

Question 10.

Draw a ΔABC in which BC = 5 cm, AB = 4 cm and m∠B = 50°.

Solution:

Steps of construction:

(i) Draw BC = 5 cm.

(ii) Draw ∠B = 50° and cut AB = 4 cm.

(iii) Join AC.

(iv) ΔABC is the required triangle.

Question 11.

Draw ΔPQR in which QR = 5.4 cm, ∠Q = 40° and PR = 6.2 cm.

Solution:

Steps of construction:

(i) Draw QR = 5.4 cm.

(ii) Draw ∠Q = 40°.

(iii) Take R as the centre and with radius 6.2 cm, draw an arc to meet the former angle line at P.

(iv) Join PR.

(v) ΔPQR is the required triangle.

Question 12.

Construct a ΔPQR in which m∠P = 60° and m∠Q = 30°, QR = 4.8 cm.

Solution:

m∠Q = 30°, m∠P = 60°

m∠Q + m∠P + m∠R = 180° (Angle sum property of triangle)

30° + 60° + m∠R = 180°

90° + m∠R = 180°

m∠R = 180° – 90°

m∠R = 90°

Steps of construction:

(i) Draw QR = 4.8 cm.

(ii) Draw ∠Q = 30°.

(iii) Draw ∠R = 90° which meets the former angle line at P.

(iv) ∠P = 180° – (30° + 90°) = 60°

(v) ΔPQR is the required triangle.

### Practical Geometry Class 7 Extra Questions Higher Order Thinking Skills [HOTS] Type

Question 13.

Draw an isosceles right-angled triangle whose hypotenuse is 5.8 cm.

Solution:

Right angled triangle is an isosceles triangle

Each of its acute angles = \(\frac { 90 }{ 2 }\) = 45°

Steps of construction:

(i) Draw AB = 5.8 cm.

(ii) Construct ∠A = 45° and ∠B = 45° to meet each other at C.

(iii) ∠C = 180° – (45° + 45°) = 90°

(iv) ΔACB is the required isosceles right angle triangle.

Question 14.

Construct a ΔABC such that AB = 6.5 cm, AC = 5 cm and the altitude AP to BC is 4 cm.

Solution:

Steps of construction:

(i) Draw a line l and take any point P on it.

(ii) Construct a perpendicular to l at P.

(iii) Cut AP = 4 cm.

(iv) Draw two arcs with centre A and radii 6.5 cm and 5 cm to cut the line l at B and C respectively.

(v) Join AB and AC.

(vi) ΔABC is the required triangle.

Question 15.

Construct an equilateral triangle whose altitude is 4.5 cm.

Solution:

Steps of construction:

(i) Draw any line l and take a point D on it.

(ii) Construct a perpendicular to l at D and cut AD = 4.5 cm.

(iii) Draw the angle of 30° at on either side of AD to meet the line l at B and C.

(iv) ΔABC is the required equilateral triangle.