# 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)

=> xo + 50o + 60o = 180o

=> x = 180o – 110o

=> x = 70o

(ii) Here <P = 90o

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

=> 90o + 30o + x = 180o

=> x = 180o – 120o -= 60o

(iii) <x + <y + <z = 180o (angles sum property)

=> 30o + 110o + x = 180o

=> x = 180o – 140o = 40o

(2) (i) <R = y = 120o (linear pair)

∴ <P + <Q + <R = 1800 (angles sum property)

=> xo + 50o + 120o = 180o

=> x = 180o – 170o = 10o

(ii) <R = x = 80o (Opposite angles)

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

=> x + 50o + y = 180o

=> 80o + 50o + y = 180o

=> y = 180o – 130o = 50o

(iii) <M + <N + <A = 180o (angles sum property)

=> 50o + y + 60o = 180o

=> y = 180o – 110o = 70o

∴ x = y = 70o (linear pair)

(iv) <C = xo = 60o

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

=> y + 30o + 60o = 180o

=> y = 180o – 90o = 90o

(v) <E = yo = 90o (Opposite angles)

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

=> x + y + x = 180o

=> 2x + 90o = 180o

=> 2x = 180o – 90o

=> 2x = 90o

=> x = 90o/2

=> x = 45o

∴ <E = 45o = <G (4.) In a right-angled triangle, one acute angle is 30o. 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 60o

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 = 180o

= 6x = 180o

= x = 180o/6 = 30

Hence, the angles are 30, 2 x 30o = 60 and 3 x 30o = 90 (8.) In the figure, ABD = 3 DAB and CDB = 96o . Find ABD

Solution: let <DAB = xo and <ABD = 3xo

Sum of interior opposite angles = exterior angle at D.

=> <ABD + <DAB = <BDS

=> 3xo + xo = 96

=> 4xo = 96

=> xo = 96o/4 = 24o

Therefore,

<ABD = 3x o= 3 x 24o = 72o

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

Solution: let the <Q = xo

then, <p = 2<Q = 2xo and <R = 3 <Q = 3xo

By angle sum property

<P + <Q + <R = 180o

=> 2xo + xo + 3xo = 180o

=> 6xo = 180o

=> x = 180o/6 = 30o

Therefore, <P = 2x = 2 x 30o = 60o

<Q = x = 30o and <R = 2 x x = 3 x 30o = 90o

(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 = 180o

=> 10x = 180o

=> x = 180o/10 = 18o

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

(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 90o

By angle sum property,

2x + 3x + 90 = 180o

5x = 180o – 90o = 90o

x = 90o/5 = 18o

Therefore acute angles are (2 x 18o) = 36o and (3 x 18o) = 54o ### Exercise – 4       Updated: October 17, 2020 — 3:19 pm

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