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
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