Telangana SCERT Class 7 Math Solution Chapter 5 Triangles and Its Properties

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 = 180⁰ (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