# Triangles ICSE Class-6th Concise Selina Maths Solutions Ch-26

## Chapter / Lesson / Exercise -26 (Including Types, Properties and Constructions)

**Triangles ICSE Class-6th Concise** Selina Mathematics Solutions Chapter-26 (Including Types, Properties and Constructions).We provide step by step Solutions of Exercise / lesson-26 **Triangles** (Including Types, Properties and Constructions) for **ICSE Class-6 Concise** Selina Mathematics.

Our Solutions contain all type Questions of Exe-26 A, Exe-26 B and Revision Exercise to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6.

**Triangles ICSE Class-6th Concise** Selina Mathematics Solutions Chapter-26 (Including Types, Properties and Constructions)

–: Select Topics :–

### Basic Concept of **Triangles for ICSE Class-6th **Mathematics

**1. Collinear Points:** Three or more points which lie on the same straight line, are called collinear points.

**2. Non-Collinear Points:** Three or more points which do not lie on the same line, are called non- col linear points.

**3. Triangle:** By joining the three non-collinear points, a triangle is formed or A triangle is a figure which is enclosed by three lines segments. In the figure, ABC is a triangle.

**4. Parts of triangle:** A triangle has six parts, three sides and three angles which are on the vertices of the triangle.

**5. Sum of angles of a triangle:** The sum of the three angles of a triangle is 180°.

**6. Exterior angle of a triangle:** If one side of a triangle is produced then the exterior angle is formed. Exterior angle of a triangle is equal to sum of its interior opposite angles. In other words, we can say that exterior angle of a triangle is greater than each of its interior opposite angles. In the figure.

∠ACE is exterior angle and ∠ACE = ∠A + ∠B and also ∠ACE > ∠A and ∠ACF > ∠B.

**7. Classification of triangles :**

**(A) According to their sides.**

**(i)** **Equilateral Triangle:** If three sides of a triangle are equal, it is called an equilateral triangle.

**(ii)** **Isosceles Triangle:** If any two sides of a triangle are equal, then it is called an isosceles triangle.

**(iii)** **Scalene Triangle:** If no two sides of the triangle are equal. Then it is called a scalene triangle

**(B) According to Angles :**

**(i)** **Acute-angled Triangle:** A triangle whose each angle is acute, is called an acute angled triangle.

**(ii)** **Right-angled Triangle:** A triangle whose one angle is a right angle i.e. 90°, is called a right angled triangle.

**(iii) Obtused-angled Triangle:** A triangle whose one angle is an obtused angle, is called an obtused-angled triangle

**8. Some special properties of a triangle:**

**(i)** If one angle of a triangle is equal to the sum of other two cycles, the angle is a right angled.

**(ii)** If the acute angles of a right angled triangle are equal, then the triangle is a right angled isosceles triangle and its each acute angle will be of 45°.

**(iii)** Sum of two sides of a triangle is greater than the third side.

**(iv)** There can be only one right angle in a triangle.

**(v)** There can be only one obtuse angle in a triangle.

**(vi)** Side opposite to greater angle is greater.

**(vii)** Angle opposite to shorter side is shorter.

### Exercise – 26 A **Triangles **(Including Types, Properties and Constructions) for **ICSE Class-6th Concise** Selina Mathematics Solutions

**Question -1.**

In each of the following, find the marked unknown angles :

**Answer-1**

**(i)** Since, sum of all angles of triangle = 180°

Hence, 70° + 72° + z = 180°

⇒ 142°+ z = 180° ”

⇒ z= 180°-142°

z = 38°

**(ii)** Since, sum of all angles of a triangle = 180°

1st Triangle 50° + 80° + b = 180°

⇒ 130°+ &= 180°

⇒b= 180° – 130°

b = 50°

2nd Triangle 40° + 45° + a = 180°

⇒ 85° + a = 180°

⇒ a = 180° -85

a = 95°

**(iii)** 60° + 45° + 20° + x = 180°

⇒ 125° + x = 180°

⇒ x = 180° – 125° => x = 55°

**Question- 2.**

Can a triangle together have the following angles ?

(i) 55°, 55° and 80°

(ii) 33°, 74° and 73°

(iii) 85°, 95° and 22°.

**Answer-2**

**(i)** Sum of all angles of a triangle = 180° Here, 55° + 55° + 80° = 180°

190° ≠ 180°

No.

**(ii)** 33°+ 74°+ 73°= 180°

180°= 180°

Yes.

**(iii)** 85° + 95° + 22° = 180°

202° ≠ 180°

No.

**Question -3.**

Find x, if the angles of a triangle are:

(i) x°, x°, x°

(ii) x°, 2x°, 2x°

(iii) 2x°, 4x°, 6x°

**Answer-3**

**(i)** Since, sum of all the angles of a triangle =180

x° + x° + x° = 180

⇒ 3x° = 180

⇒ x° = ^{180}⁄_{3}

x = 60

**(ii)** x° + 2x° + 2x° = 180

5x° = 180

x° = ^{180}⁄_{5}

x° = 36

**(iii)** 2x° + 4x° + 6x° =180

12x° = 180

x° = ^{180}⁄_{12}

x° = 15

**Question -4.**

One angle of a right-angled triangle is 70°. Find the other acute angle.

**Answer-4**

We know that, sum of angles of a triangle = 180°.

Let, the acute angle be ‘x’

∴ x + 90° + 70° = 180°

⇒ x+ 160° = 180°

⇒ x= 180°-160°

⇒ x = 20°

∴The acute angle is 20°.

**Question -5. Triangles ICSE Class-6th Concise**

In ∆ABC, ∠A = ∠B = 62° ; find ∠C.

**Answer-5**

∠A + ∠B + ∠C= 180°

⇒ 62° + 62° + ∠C = 180°

⇒ 124° + ∠C = 180°

⇒ ∠C = 180° – 124°

⇒∠C = 56°

**Question- 6.**

In ∆ABC, C = 56°C = 56° ∠B = ∠C and ∠A = 100° ; find ∠B.

**Answer-6**

∠A + ∠B + ∠C = 180°

⇒ 100° + ∠B + ∠B = 180°

⇒ 2∠B = 180° 100°

∠B = 802°

∠B = 40°

∠C = ∠B = 40°

**Question -7.**

Find, giving reasons, the unknown marked angles, in each triangle drawn below:

**Answer-7**

We know that,

Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)

**(i)** ∴ 110° = x + 30° (by property)

⇒x=110°-30° x = 80°

**(ii)** x+115° = 180°

(linear property of angles)

⇒x = 180°- 115° ⇒x = 65°

∴115° = x + y

⇒ 115° = 65° + _y ⇒ y= 115° – 65° =50°

y = 50°

**(iii)** 110° = 2x + 3x

5x – 110°

x = 1105°

x = 22°

∴2x = 2 x 22 = 44°

3x = 3 x 22 = 66°

**Question -8.**

Classify the following triangles according to angle :

**Answer-8**

**(i)** Since, it has an obtuse angle of 120° Hence, it is obtuse angled triangle.

**(ii)** Since, all the angle of triangle is less than 90°.

Hence, it is an acute angled triangle.

**(iii)** Since ∠MNL = 90°, and sum of two acute angle ∠M + ∠N = 30° + 60° = 90°.

Hence, it is a right angled triangle

**Question -9.**

Classify the following triangles according to sides :

**Answer-9**

(**i)** Since, two sides-are equal.

Hence, Isosceles triangle.

**(ii)** Since, all the three sides are unequal.

Hence, Scalene, triangle.

**(iii)** Since, all the three sides are unequal Hence, Scalene triangle.

**(iv)** All the three sides are equal.

Hence, equilateral triangle

**Triangles **(Including Types, Properties and Constructions) Exercise-26 B for **ICSE Class-6th Concise** Selina Mathematics Solutions

Construct triangle ABC, when :

**Question- 1.**

AB = 6 cm, BC = 8 cm and AC = 4 cm.

**Answer-1**

**Steps of Construction:**

(1) Draw a line AB = 6 cm.

(2) compasses and taking B as centre, draw an arc of 8 cm radius.

(3) With A as centre, draw an arc of 4 cm radius, which cuts the previous arc at C.

(4) Join AC and BC.

Triangle ABC, obtained, is the required triangle.

**Question -2.**

AB = 3.5 cm, AC = 4.8 cm and BC = 5.2 cm.

**Answer-2**

**Steps of Construction :**

(1) Draw a line AB = 3.5 cm.

(2) Using compasses and taking B as centre, draw an arc of 5.2 cm radius.

(3) With A as centre, draw an arc of 4.8

(4) Join AC and BC

**Question- 3.**

AB = BC = 5 cm and AC = 3 cm. Measure angles A and C. Is ∠A = ∠C?

**Answer-3**

**Steps of Construction :**

(1) Draw a line AB = 5 cm.

(2) Using compasses and taking B as centre, draw an arc of 5 cm radius.

(3) With A as centre, draw an arc of 3 cm radius, which cuts the previous arc at C.

(4) Join AC and BC.

**Question- 4.**

AB = BC = CA = 4.5 cm. Measure all the angles of the triangle. Are they equal ?

**Answer-4**

**Steps of Construction :**

(1) Draw a a line AB =4.5

(2) Using compasses and taking BC as centre, draw an arc of 4.5 cm radius.

(3) With AC as centre, draw an arc of 4-5 cm radius, which cuts the previous arc at C.

(4) Join AC and BC.

(5) Measurement, ∠A = ∠B = ∠C = 60°.

**Question -5. Triangles ICSE Class-6th Concise**

AB = 3 cm, BC = 7 cm and ∠B = 90°.

**Answer-5**

**Steps of Construction :**

(1) Draw a line segment AB = 3 cm.

(2) With the help of compasses, construct ∠ABC = 90°.

(3) With B as centre, draw an arc of 7 cm length which cuts BP at C.

(4) Join A and C.

(5) Triangle ABC, so obtained, is the required triangle.

**Question -6.**

AC = 4.5 cm, BC = 6 cm and ∠C = 60°.

**Answer-6**

**Steps of Construction :**

(1) Draw a line AC = 4.5 cm.

(2) With the help of compasses, construct ∠ACB = 60°.

(3) With C as centre, draw an arc of 6 cm radius, which cuts CB at C.

(4) Join B and A.

**Question -7.**

AC = 6 cm, ∠A = 60“ and ∠C = 45°. Measure AB and BC.

**Answer-7**

**Steps of Construction :**

(1) Draw a line segment AC = 6 cm.

(2) At A construct an angle ∠A = 60°.

(3) At C construct an angle ∠C = 45°.

(4) AD and CE intersect each other at B.

(5) ∴ ∆ABC is the required triangle.

(6) On measuring AB = 4-4 cm, BC = 5.4 cm.

**Question -8.**

AB = 5.4 cm, ∠A = 30° and ∠B = 90°. Measure ∠C and side BC.

**Answer-8**

**Steps of Construction :**

(1) Draw a line segment AB = 5.4 cm.

(2) At A construct an angle ∠A = 30°.

(3) At B construct an angle ∠B = 90°.

(4) AD and BE intersect each other at C.

(5) ∴ ∆ABC is the required triangle.

(6) On measuring ∠C = 60° side BC = 31 cm.

**Question -9.**

AB = 7 cm, ∠B = 120° and ∠A = 30°. Measure AC and BC.

**Answer-9**

**Steps of Construction :**

(1) Draw a line segment AB = 7 cm

(2) At A construct an angle ∠A = 30°.

(3) At C construct an angle ∠C = 45°.

(4) AE and BD intersect each other at C.

(5) ∴ ∆ABC is the required triangle.

(6) On measuring length of AC = 12cm and BC = 7 cm respectively.

**Question -10.**

BC = 3 cm, AC = 4 cm and AB = 5 cm. Measure angle ACB. Give a special name to this triangle.

**Answer-10**

**Steps of Construction :**

(1) Draw a line segment AB = 5 cm

(2) From A, with the help of compass cut arc AC = 4cm

(3) From point B cut an arc BC = 3 cm.

(4) Join AC and BC.

(5) AABC is required right-angled triangle.

Measuring ∠ACB = 90°

### Revision Exercise **Triangles **(Including Types, Properties and Constructions) for **ICSE Class-6th Concise** Selina Mathematics Solutions

**Question -1.**

If each of the two equal angles of an isosceles triangle is 68°, find the third angle.

**Answer-1**

Let Δ ABC is an isosceles triangle

∴ In Δ ABC, AB = AC

and ∠B = ∠C = 68°

But ∠A + ∠B + ∠C = 108° …(Sum of angles of a triangle)

⇒ ∠A + 68° + 68° = 180°

⇒ ∠A + 136° = 180°

⇒ ∠A = 180° – 136°

∴ ∠A = 44°

Hence third angle = 44°

**Question -2.**

One of the angles of a triangle is 110°, the two other angles are equal. Find their value.

**Answer-2**

Let in Δ ABC,

∠A = 110° and ∠B = ∠C

But ∠A + ∠B + ∠C = 180°

⇒ 110° + ∠B + ∠B = 180°

⇒ 2 ∠B = 180° – 110°

⇒ 2 ∠B = 70°

⇒ ∠B = ^{70}⁄_{2}=35∘

∴ ∠C = ∠B = 35°

Hence each of two equal angles is 35°

**Question- 3.**

The angles of a triangle are in the ratio 3:5: 7. Find each angle.

**Answer-3**

Ratio in angles of a triangle is 3: 5: 7

But sum of angles of a triangle = 180°

Sum of ratios = 3 + 5 + 7 = 15

∴ First angle = ** ^{180}⁄_{15 }× 3** = 36∘

Second angle = ** ^{180}⁄_{15 }× 5** = 60∘

and third angle = ** ^{180}⁄_{15 }× 7**=84∘

∴ Angles of the triangle are 36°, 60° and 84°.

**Question- 4.**

The angles of a triangle are (2x – 30°),(3x – 40°) and (x + 10°) Find the value of x .

**Answer-4**

∵ The sum of angles of a triangle = 180°

∴ (2x – 30°) + (3x – 40°) + ** ^{5x}⁄_{2 }**+10°)=180°

⇒ 2x – 30° + 3x – 40° + ** ^{5x}⁄_{2}** + 10° = 180°

⇒ 2x + 3x + ** ^{5x}⁄_{2}** – 70° + 10° = 180°

⇒ ** ^{4x+6x+5x}⁄_{2 }**-60°=180°

⇒ ** ^{15x}⁄_{2}** = 180° + 60° = 240°

x = ** ^{240×2}⁄_{15}**=32

∴ x = 32°

**Question- 5.**

In each of the following figures, triangle ABC is equilateral and triangle PBC is isosceles. If PBA = 20°; find in each case:

(a) angle PBC.

(b) angle BPC.

**Answer-5**

**(a) angle PBC.**

In above figure,

Δ ABC is an equilateral triangle and Δ PBC is an isosceles triangle and ∠PBA = 20°.

∵ Δ ABC is an equilateral triangle

∴ ∠ABC = 60°

⇒ ∠PBA + ∠PBC = 60°

⇒ 20° + ∠PBC = 60°

⇒ ∠PBC = 60° – 20° = 40°

∵ ∠PBC is an isosceles triangle,

∴ ∠PBC = ∠PCB = 40°

Now in ΔBPC,

∠PBC + ∠PCB + ∠BPC = 180° (Sum of angles of a triangle)

⇒ 40° + 40° + ∠BPC = 180°

⇒ 80° + ∠BPC = 180°

⇒ ∠BPC = 180° – 80° = 100°

**(b) angle BPC.**

In the above figure,

∠PBA = ∠ABC + ∠PBA = 60° + 20° = 80°

But ΔPBC is an isosceles triangle

∴ ∠PCB = ∠PBC = 80°

But ∠PBC + ∠PCB + ∠BPC = 180° …(Sum of angles of a triangle)

⇒ 80° + 80° + ∠BPC = 180°

⇒ 160° + ∠BPC = 180°

∠BPC = 180° – 160° = 20°

**Question -6.**

Construct a triangle ABC given AB = 6 cm, BC = 5 cm and CA = 5.6 cm. From vertex A draw a perpendicular on to side BC. Measure the length of this perpendicular.

**Answer-6**

**Steps of Construction :**

**(1)** Draw a line AB = 6 cm.

**(2)** Using compass, taking A and B as centre draw arcs of 5 cm and 5.6 cm respectively, which cut each other at C.

**(3)** Join AC and BC.

**(4)** Now, from vertex A draw a bisector AD towards BC.

On measuring length AD = 5 cm.

**Question- 7.**

Construct a triangle PQR, given PQ = 6 cm, ∠P = 60° and ∠Q = 30°. Measure angle R and the length of PR.

**Answer-7**

**Steps of Construction :**

**(1)** Draw a line PQ = 6 cm.

**(2)** Using compass taking P as centre draw an angle ∠P = 60°.

**(3)** Using compass taking Q as a centre draw an angle ∠Q = 30°.

**(4)** PS and QT intersect each other R.

**(5)** ∆RPQ is the required triangle.

On measuring; ∠R = 90°, length of PR = 3 cm

**Question- 8. Triangles ICSE Class-6th Concise**

Construct a triangle ABC given BC = 5 cm, AC = 6 cm and ∠C = 75°. Draw the bisector of the interior angle at A. Let this bisector meet BC at P ; measure BP.

**Answer-8**

**Steps of Construction :**

**(1)** Draw BC = 5 cm.

**(2)** With the help of compass from centre C. Draw an angle ∠C = 75°.

**(3)** From CD, cut an arc AC = 6 cm.

**(4)** Join AB.

**(5)** From A draw an bisector AP.

**(6)** On measuring BP = 2.6 cm.

**Question -9.**

Using ruler and a pair compass only, construct a triangle XYZ given YZ = 7 cm, ∠XYZ = 60° and ∠XZY = 45°. Draw the bisectors of angles X and Y.

**Answer-9**

**Steps of Construction :**

**(1)** Draw a line YZ = 7 cm.

**(2)** Y as a centre draw an ∠XYZ = 60°.

**(3)** Z as a centre draw an ∠XZY = 45°.

**(4)** From X and Y as centre draw bisector of ∠X and ∠Y, which meet at point O.

**Question- 10.**

Using ruler and a pair compass only, construct a triangle PQR, given PQ = 5.5 cm, QR = 7.5 cm and RP = 6 cm. Draw the bisectors of the interior angles at P, Q and R. Do these bisectors meet at the same point ?

**Answer-10**

**Steps of Construction :**

**(1)** Draw a line PQ = 5.5 cm.

**(2)** From Q as a centre draw an arc QR = 7.5 cm.

**(3)** From P as a centre draw an arc PR = 6 m, which intersects previous arc at R.

**(4)** Join PR and QR.

**(5)** Now, draw interior bisectors of ∠P, ∠QR ∠R which meets each other at point S.

**Question -11.**

One angle of a triangle is 80° and the other two are in the ratio 3 : 2. Find the unknown angles of the triangle.

**Answer-11**

Let angle ∠A of a triangle ABC = 80°

But sum of three angles of a triangle = 180°

Sum of remaining two angles = 180° – 80° = 100°

Ratio in their two angles = 3:2

Let second angle = 3x

and third angle = 2x

3x + 2x = 100°

5x = 100 x = ^{100}⁄_{5} = 20°

second angle ∠B = 3x = 3 x 20° = 60°

and third angle ∠C = 2x = 2 x 20° = 40°

Hence other two angles are 60° and 40°.

**Question -12.**

Find the value of x if ∠A = 32°, ∠B = 55° and obtuse angle AED = 115°.

**Answer-12**

In the figure, ∠A = 32, ∠B = 55°

∠AED =115°

In ∆ABC

Exterior ∠ACD = ∠A + ∠B = 32° + 55° = 87°

Similarly in ∆CDE

Ext. ∠AED = ∠D + ∠ACD

⇒ 115° = x + 87° ⇒ x = 115°- 87° = 28°

Hence x = 28°

–: End of** Triangles ICSE Class-6th Concise** Solutions :–

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