## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 15 Triangles

**Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 15 Triangles**

### Triangles Exercise 15A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.

Stale, if the triangles are possible with the following angles :

(i) 20°, 70° and 90°

(ii) 40°, 130° and 20°

(iii) 60°, 60° and 50°

(iv) 125°, 40° and 15°

Solution:

Question 2.

If the angles of a triangle are equal, find its angles.

Solution:

Question 3.

In a triangle ABC, ∠A = 45° and ∠B = 75°, find ∠C.

Solution:

Question 4.

In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.

Solution:

Question 5.

Calculate the unknown marked angles in each figure :

Solution:

Question 6.

Find the value of each angle in the given figures:

Solution:

Question 7.

Find the unknown marked angles in the given figure:

Solution:

Question 8.

In the given figure, show that: ∠a = ∠b + ∠c

(i) If ∠b = 60° and ∠c = 50° ; find ∠a.

(ii) If ∠a = 100° and ∠b = 55° : find ∠c.

(iii) If ∠a = 108° and ∠c = 48° ; find ∠b.

Solution:

Question 9.

Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6.

Solution:

Question 10.

One angle of a triangle is 60°. The, other two angles are in the ratio of 5 : 7. Find the two angles.

Solution:

Question 11.

One angle of a triangle is 61° and the other two angles are in the ratio 1 : 1 . Find these angles.

Solution:

Question 12.

Find the unknown marked angles in the given figures :

Solution:

### Triangles Exercise 15B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.

Find the unknown angles in the given figures:

Solution:

Question 2.

Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures :

Solution:

Question 3.

The angle of vertex of an isosceles triangle is 100°. Find its base angles.

Solution:

Question 4.

One of the base angles of an isosceles triangle is 52°. Find its angle of vertex.

Solution:

Question 5.

In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.

Solution:

Question 6.

The vertical angle of an isosceles triangle is 15° more than each of its base angles. Find each angle of the triangle.

Solution:

Question 7.

The base angle of an isosceles triangle is 15° more than its vertical angle. Find its each angle.

Solution:

Question 8.

The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.

Solution:

Question 9.

The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.

Solution:

Question 10.

In the given figure, BI is the bisector of∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.

Solution:

Question 11.

In the given figure, express a in terms of b.

Solution:

Question 12.

(a) In Figure (i) BP bisects ∠ABC and AB = AC. Find x.

(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects ∠ABC and∠ADB = 70°.

Solution:

Question 13.

In each figure, given below, ABCD is a square and ∆ BEC is an equilateral triangle.

Find, in each case : (i) ∠ABE(ii) ∠BAE

Solution:

Question 14.

In ∆ ABC, BA and BC are produced. Find the angles a and h. if AB = BC.

Solution:

### Triangles Exercise 15C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.

Construct a ∆ABC such that:

(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm

(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm

(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm

Solution:

Question 2.

Construct a A ABC such that:

(i) AB = 7 cm, BC = 5 cm and ∠ABC = 60°

(ii) BC = 6 cm, AC = 5.7 cm and ∠ACB = 75°

(iii) AB = 6.5 cm, AC = 5.8 cm and ∠A = 45°

Solution:

Question 3.

Construct a ∆ PQR such that :

(i) PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R.

(ii) QR = 4.4 cm, ∠R = 30° and ∠Q = 75°. Measure PQ and PR.

(iii) PR = 5.8 cm, ∠P = 60° and ∠R = 45°.

Measure ∠Q and verify it by calculations

Solution:

Question 4.

Construct an isosceles A ABC such that:

(i) base BC = 4 cm and base angle = 30°

(ii) base AB = 6-2 cm and base angle = 45°

(iii) base AC = 5 cm and base angle = 75°.

Measure the other two sides of the triangle.

Solution:

Question 5.

Construct an isosceles ∆ABC such that:

(i) AB = AC = 6.5 cm and ∠A = 60°

(ii) One of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles.

(iii) BC = AB = 5-8 cm and ZB = 30°. Measure ∠A and ∠C.

Solution:

Question 6.

Construct an equilateral A ABC such that:

(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.

(ii) Each side is 6 cm.

Solution:

Question 7.

(i) Construct a ∆ ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.

(ii) Construct an isosceles ∆PQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using ruler and compasses only construct a circumcircle to this triangle.

(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.

Construct a circumcircle to this triangle.

Solution:

Question 8.

(i) Construct a ∆ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.

(ii) Construct an isosceles ∆ MNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct an incircle to this triangle and measure its radius.

(iii) Construct an equilateral ∆DEF whose one side is 5.5 cm. Construct an incircle to this triangle.

(iv) Construct a ∆ PQR such that PQ = 6 cm, ∠QPR = 45° and angle PQR = 60°. Locate its incentre and then draw its incircle.

Solution: