**Telanagana SCERT Solution Class VII (7) Math Chapter 5 Triangles and Its Properties.**

**Exercise – 1**

**(1.) Is it possible to have a triangle with the following sides?**

(i) 3 cm, 4 cm and 5 cm. (ii) 6 cm, 6 cm and 6 cm.

(iii) 4 cm, 4 cm and 8 cm. (iv) 3 cm, 5 cm and 7 cm

Solution: (i) Possible

(ii) Possible

(iii) Not Possible

(iv) Possible

(v) Not Possible.

**Exercise – 2**

**Exercise – 3**

**(1) (i) <A + <B + <C + = 180 (angles sum property)**

=> x^{o} + 50^{o} + 60^{o} = 180^{o}

=> x = 180^{o} – 110^{o}

=> x = 70^{o}

**(ii) Here <P = 90 ^{o}**

∴ <P + <Q + <R = 180^{o} (Angles sum property)

=> 90^{o} + 30^{o} + x = 180^{o}

=> x = 180^{o} – 120^{o} -= 60^{o}

**(iii) <x + <y + <z = 180 ^{o} (angles sum property)**

=> 30^{o} + 110^{o} + x = 180^{o}

=> x = 180^{o} – 140^{o} = 40^{o}

**(2) (i) <R = y = 120 ^{o} (linear pair)**

∴ <P + <Q + <R = 180^{0} (angles sum property)

=> x^{o} + 50^{o} + 120^{o} = 180^{o}

=> x = 180^{o} – 170^{o} = 10^{o}

**(ii) <R = x = 80 ^{o} (Opposite angles)**

∴ <R + <S + <T = 180^{o} (angles sum property)

=> x + 50^{o} + y = 180^{o}

=> 80^{o} + 50^{o} + y = 180^{o}

=> y = 180^{o} – 130^{o} = 50^{o}

**(iii) <M + <N + <A = 180 ^{o} (angles sum property)**

=> 50^{o} + y + 60^{o} = 180^{o}

=> y = 180^{o} – 110^{o} = 70^{o}

∴ x = y = 70^{o} (linear pair)

**(iv) <C = x ^{o} = 60^{o}**

∴ <A + <B + <C = 180^{o} (angles sum property)

=> y + 30^{o} + 60^{o} = 180^{o}

=> y = 180^{o} – 90^{o} = 90^{o}

**(v) <E = y ^{o} = 90^{o} (Opposite angles)**

∴ <E + <F + <G = 180^{o} (angles sum property)

=> x + y + x = 180^{o}

=> 2x + 90^{o} = 180^{o}

=> 2x = 180^{o} – 90^{o}

=> 2x = 90^{o}

=> x = 90^{o}/2

=> x = 45^{o}

∴ <E = 45^{o} = <G

**(4.) In a right-angled triangle, one acute angle is 30 ^{o}. Find the other acute angle.**

Solution:

**(5.) State true or false for each of the following statements.**

**(i) A triangle can have two right angles.**

**Ans. **False

**(ii) A triangle can have two acute angles.**

**Ans. **True

**(iii) A triangle can have two obtuse angles.**

**Ans. **False

**(iv) Each angle of a triangle can be less than 60 ^{o}**

**Ans. **False

**(6.) The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.**

**Solution: **let the angles be 1x, 2x and 3x

By angle sum property.

1x + 2x + 3x = 180^{o}

= 6x = 180^{o}

= x = 180^{o}/6 = 30

Hence, the angles are 30, 2 x 30^{o} = 60 and 3 x 30^{o} = 90

**(8.) In the figure, ****∠****ABD = 3 ****∠****DAB and ****∠****CDB = 96 ^{o} . Find **

**∠**

**ABD**

**Solution: **let <DAB = x^{o} and <ABD = 3x^{o}

Sum of interior opposite angles = exterior angle at D.

=> <ABD + <DAB = <BDS

=> 3x^{o} + x^{o} = 96

=> 4x^{o} = 96

=> x^{o} = 96^{o}/4 = 24^{o}

Therefore,

<ABD = 3x ^{o}= 3 x 24^{o} = 72^{o}

**(9.) In ΔPQR ****∠****P= 2 ****∠****Q and 2 ****∠****R = 3 ****∠****Q , calculate the angles of ΔPQR.**

**Solution: **let the <Q = x^{o}

then, <p = 2<Q = 2x^{o} and <R = 3 <Q = 3x^{o}

By angle sum property

<P + <Q + <R = 180^{o}

=> 2x^{o} + x^{o} + 3x^{o} = 180^{o}

=> 6x^{o} = 180^{o}

=> x = 180^{o}/6 = 30^{o}

Therefore, <P = 2x = 2 x 30^{o} = 60^{o}

<Q = x = 30^{o} and <R = 2 x x = 3 x 30^{o} = 90^{o}

**(10) If the angles of a triangle are in the ratio 1 : 4 : 5, find the angles.**

Solution: let the angles be 1x, 4x and 5x.

By angle sum property,

x + 4x + 5x = 180^{o}

=> 10x = 180^{o}

=> x = 180^{o}/10 = 18^{o}

Therefore, the angles are x = 18^{o} and 4x = 4 x 18^{o} = 72^{o} and 5x = 5 x 18^{o} = 90^{o}

**(11.) The acute angles of a right triangle are in the ratio 2 : 3. Find the angles of the triangle**

**Solution: **let the acute angles be 2x and 3x

right angle of right triangle is 90^{o}

By angle sum property,

2x + 3x + 90 = 180^{o}

5x = 180^{o} – 90^{o} = 90^{o}

x = 90^{o}/5 = 18^{o}

Therefore acute angles are (2 x 18^{o}) = 36^{o} and (3 x 18^{o}) = 54^{o}

**Exercise – 4**

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