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

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

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**POINTS TO REMEMBER**

**1. Definition of a triangle :** A closed figure, having 3 sides, is called a triangle and is usually denoted by the Greek letter ∆ (delta).

The figure, given alongside, shows a triangle ABC (∆ABC) bounded by three sides AB, BC and CA.

**Hence it has six elements** : 3 angles and 3 sides.

**2. Vertex :** The point, where any two sides of a triangle meet, is called a vertex.

Clearly, the given triangle has three vertices; namely : A, B and C. [Vertices is the plural of vertex]

**3. Interior angles :** In ∆ABC (given above), the angles BAC, ABC and ACB are called its interior angles as they lie inside the ∆ ABC. The sum of interior angles of a triangle is always 180°.

**4. Exterior angles :** When any side of a triangle is produced the angle so formed, outside the triangle and at its vertex, is called its** exterior angle.**

e.g. if side BC is produced to the point D; then ∠ACD is its exterior angle. And, if side AC is produced to the point E, then the exterior angle would be ∠BCE.

Thus. at every vertex, two exterior angles can be formed and that these two angles being vertically opposite angles, are always equal.

Make the following figures clear :

5. **Interior opposite angles :** When any side of a triangle is produced; an exterior angle is formed. The two interior angles of this triangle, that are opposite to the exterior angle formed; are called its **interior opposite angles.**

In the given figure, side BC of ∆ABC is produced to the point D, so that the exterior ∠ACD is formed. Then the two interior opposite angles are ∠B AC and ∠ABC.

6. **Relation between exterior angle and interior opposite ****angles :**

Exterior angle of a triangle is always equal to the

sum of its two interior opposite angles.

In ∆ABC,

Ext. ∠ACD = ∠A + ∠B

**7. CLASSIFICATION OF TRIANGLES**

**(A) With regard to their angles :**

**1. Acute angled triangle :** It is a triangle, whose each angle is acute i.c. each angle is less than 90°.

**2. Right angled triangle :** It is a triangle, whose one angle is a right angle i.e. equal to 90”.

The figure, given alongside, shows a right angled triangle XYZ as ∠XYZ = 90°

**Note :** (i) One angle of a right triangle is 90° and the other two angles of it are acute; such that their sum is always 90”.

In ∆XYZ, given above, ∠Y = 90° and each of ∠X and ∠Z is acute such that ∠X + ∠Z = 90°. .

(ii)In a right triangle, the side opposite to the right angle is largest of all its sides and is called the** hypotenuse**. In given right angled ∆ XYZ side XZ is its hypotenuse

**3.Obtuse angled triangle :** If one angle of a triangle is 1

obtuse, it is called an obtuse angled triangle.

**Note :** In case of an obtuse angled triangle, each of the other two angles is always acute and their sum is less than 90”.

**(B) With regard to their sides :**

**(1) Scalene triangle:** If all the sides of a triangle are unequal, it is called a** scalene triangle.**

In a scalene triangle; all its angles are also unequal.

**(2) Isosceles triangle :** If atleast two sides of a triangle are equal, it is called an** isosceles triangle.**

In ∆ ABC, shown alongside, side AB = side AC.

∴∆ ABC is an isosceles triangle.

**Note** : (i) The angle contained by equal sides i.e. ∠BAC is called the **vertical angle** or the **angle of vertex.**

(ii) The third side (i.e. the unequal side) is called the **base** of the isosceles triangle.

(iii) The two other angles (i.e. other than the angle of vertex) are called the **base angles** of the triangle.

**IMPORTANT PROPERTIES OF AN ISOSCELES TRIANGLE**

The base angles i.e. the angles opposite to equal sides of an isosceles triangle are always equal.

In given triangle ABC,

(i) If side AB = side BC; then angle opposite to AB = angle opposite to BC i.e. ∠C = ∠A.

(ii) If side BC = side AC; then angle opposite to BC = angle opposite to AC i.e. ∠A = ∠B and so on.

**Conversely :** If any two angles of a triangle are equal; the sides opposite to these angles are also equal i.e. the triangle is isosceles.

Thus in ∆ ABC,

(i) If ∠B = ∠C => side opposite to ∠B = side opposite to ∠C i.e. side AC = side AB.

(ii) If ∠A = ∠B => side BC = side AC and so on.

**(3) Equilateral triangle :**

If all the sides of a triangle are equal, it is called an equilateral triangle.

In the given figure, A ABC is equilateral, because AB = BC = CA.

Also, all the angles of an equilateral triangle are equal to each other and so each angle = 60°. [∵60° + 60° + 60° = 180°]

Since, all the angles of an equilateral triangle are equal, it is also known as equiangular triangle. Note : An equilateral triangle is always an isosceles triangle, but its converse is not always true.

**(4) Isosceles right angled triangle :** If one angle of an isosceles triangle is 90°, it is called an isosceles right angled triangle.

In the given figure, ∆ ABC is an isosceles right angled triangle, because : ∠ ACB = 90° and AC = BC.

Here, the base is AB, the vertex is C and the base angles are ∠BAC and ∠ABC, which are equal.

Since, the sura of the angles of a triangle = 180″

∴∠ABC = ∠BAC = 45 [∵45° + 45° + 90° = 180°]

### 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\(\frac { 1 }{ 2 }\) : 1 \(\frac { 1 }{ 3 }\). 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:**

**(i) Steps of Construction :**

(i) Draw a line segment BC = 4 cm.

(ii) With centre B and radius 6 cm draw an arc.

(iii) With centre C and radius 5.5 cm, draw another arc intersecting the First are at A.

(iv) Join AB and AC. ∆ABC is the required triangle.

**(ii) Steps of Construction :**

(i) Draw a line segment CB = 6 5 cm

(ii) With centre C and radius 4.2 cm draw an arc.

(iii) With centre B and radius 5.1 cm draw another arc intersecting the first arc at A.

(iv) Join AC and AB.

∆ ABC is the required triangle.

**(iii) Steps of Construction :**

(i) Draw a line segment BC = 4 cm.

(ii) With centre B and radius 3.5 cm, draw an arc

(iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.

(iv) Join AB and AC.

∆ ABC is the required triangle.

**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:**

**(i) Steps of Construction :**

(i) Draw a line segment AB = 7 cm.

(ii) At B, draw a ray making an angle of 60° and cut off BC = 5 cm

(iii) Join AC,

∆ABC is the required triangle.

**(ii) Steps of Construction :**

(i) Draw a line segment BC = 6 cm.

(ii) At C, draw a ray making an angle of 75° and cut off CA = 5.7 cm.

(iii) JoinAB

∆ ABC is the required triangle.

**(iii) Steps of Construction :**

(i) Draw a line segment AB = 6.5 cm

(ii) At A, draw a ray making an angle of 45° and cut off AC = 5.8 cm

(iii) JoinCB.

∆ ABC is the required triangle.

**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:**

**(i) Steps of Construction:**

(i) Draw a line segment PQ = 6 cm.

(ii) At P, draw a ray making an angle of 45°

(iii) At Q, draw another ray making an angle of 60° which intersects the first ray at R.

∆ PQR is the required triangle.

On measuring ∠R, it is 75°.

**(ii) Steps of Construction :**

(i) Draw a line segment QR = 44 cm.

(ii) At Q, draw a ray making an angle of 75°

(iii) At R, draw another arc making an angle of 30° ; which intersects the first ray at R

∆ PQR is the required triangle.

On measuring the lengths of PQ and PR, PQ = 2.1 cm and PR = 4. 4 cm.

**(iii) Steps of Construction :**

(i) Draw a line segment PR = 5.8 cm

(ii) At P, construct an angle of 60°

(iii) At R, draw another angle of 45° meeting each other at Q.

∆ PQR is the required triangle. On measuring ∠Q, it is 75°

Verification : We know that sum of angles of a triangle is 180°

∴∠P + ∠Q + ∠R = 180°

⇒ 60° + ∠Q + 45° = 180°

⇒ ∠Q + 105° = 180°

⇒ ∠Q = 180° – 105° = 75°.

**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:**

**(i) Steps of Construction :**

We know that in an isosceles triangle base angles are equal.

(i) Draw a line segment BC = 4 cm.

(ii) At B and C, draw rays making an angle of 30° each intersecting each other at A.

∆ ABC is the required triangle.

On measuring the equal sides each is 2.5 cm (approx.) in length.

**(ii) Steps of Construction :**

We know that in an isosceles triangle, base angles are equal.

(i) Draw a line segment AB = 6.2 cm

(ii) At A and B, draw rays making an angle of 45° each which intersect each other at C.

∆ABC is the required triangle.

On measuring the equal sides, each is 4.3 cm (approx.) in length.

**(iii) Steps of Construction :**

We know that base angles of an isosceles triangles are equal.

(i) Draw a line segment AC = 5cm.

(ii) At A and C, draw rays making an angle of 75° each which intersect each other at B.

∆ ABC is the required triangle.

On measuring the equal sides, each is 9.3 cm in length.

**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:**

**(i) Steps of Construction :**

(i) Draw a line segment AB = 6.5 cm.

(ii) At A, draw a ray making an angle of 60°.

(iii) Cut off AC = 6.5 cm

(iv) JoinBC.

∆ABC is the required triangle.

**(ii) Steps of Construction :**

(i) Draw a line segment AB = 6 cm

(ii) At A, construct an angle equal to 45°

(iii) Cut off AC = 6 cm

(iv) JoinBC.

∆ ABC is the required triangle.

On measuring, ∠B and ∠C, each is equal 1° to, 67\(\frac { 1 }{ 2 }\)°

**(iii) Steps of Construction :**

(i) Draw a line segment BC = 5.8 cm

(ii) At B, draw a ray making an angle of 30°.

(iii) Cut off BA = 5.8 cm

(iv) Join AC.

∆ ABC is the required triangle On measuring ∠C and ∠A, each is equal to 75°.

**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:**

**(i) Steps of Construction :**

(i) Draw a line segment AB = 5 cm.

(ii) With centres A and B and radius 5 cm each, draw two arcs intersecting each other at C.

(iii) Join AC and BC ∆ABC is the required triangle.

(iv) Draw the perpendicular bisectors of sides AC and BC which intersect each other at P-

(v) Join PA, PB and PC.

On measuring, each is 2.8 cm.

**(ii) Steps of Construction :**

(i) Draw a line segment AB = 6 cm.

(ii) At A and B as centre and 6 cm as radius draw two arcs intersecting each other at C.

(iii) Join AC and BC.

∆ABC is the required triangle.

**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:**