RS Aggarwal Class 7 Math Fifteenth Chapter Properties of Triangles Exercise 15A Solution
EXERCISE 15A
(1) In a ∆ABC, if ∠A = 72o and ∠B = 63o, find ∠C.
Solution: We know that the sum of the angles of a triangle is 180o.
∠A + ∠B + ∠C = 180o
or, 72o + 63o + ∠C = 180o
or, ∠C = 180o – 135o
or, ∠C = 45o
(2) In a ∆DEF, if ∠E = 105o and ∠F = 40o, find ∠D.
Solution: We know that the sum of the angles of a triangle is 180o.
∠D + ∠E + ∠F = 180o
or, ∠D + 105o + 40o = 180o
or, ∠D = 180o – 145o
or, ∠D = 35o
(3) In a ∆XYZ, if ∠X = 90o and ∠Z = 48o, find ∠Y.
Solution: We know that the sum of the angles of a triangle is 180o.
∠X + ∠Y + ∠Z = 180o
or, 90o + ∠Y + 48o = 180o
or, ∠Y = 180o – 138o
or, ∠Y = 42o
(4) Find the angles of a triangle which are in the ratio 4 : 3 : 2.
Solution: Let the measures of the given angles of the triangle be (4x)o, (3x)o and (2x)o. Then,
4x + 3x + 2x = 180
or, 9x = 180
or, x = 180/9
or, x = 20
Hence, the angles of the triangle are 80o, 60o and 40o.
(5) One of the acute angles of a right triangle is 36o, find the other.
Solution: Let the measure of unknown angle be xo.
We know that the sum of the angles of a triangle is 180o.
∴ x + 90 + 36 = 180
or, x = 180 – 126
or, x = 54
Hence, each unknown angle is 54o.
(6) The acute angles of a right triangle are in the ratio 2 : 1. Find each of these angles.
Solution: Let the measure of each angles be (2x)o and xo.
We know that the sum of the angles of a triangle is 180o.
∴ 2x + x + 90 = 180
or, 3x = 180 – 90
or, 3x = 90
or, x = 30.
Hence the required angles are (2 × 30)o = 60o and 30o.
(7) One of the angles of a triangle is 100o and the other two angles are equal. Find each of the equal angles.
Solution: Let the measure of each unknown angle be xo.
We know that the sum of the angles of a triangle is 180o.
∴ x + x + 100 = 180
or, 2x = 180 – 100
or, 2x = 80
or, x = 40
Hence, each unknown angle is 40o.
(8) Each of the two equal angles of an isosceles triangle is twice the third angle. Find the angles of the triangle.
Solution: Let the third angle be xo.
∴ 2x + 2x + x = 180
or, 5x = 180
or, x = 36
Hence, the third angle is 36o and each others are (36 × 2) = 72o.
(9) If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
Solution: Given, ∠A = ∠B + ∠C
∴ ∠A + ∠B + ∠C = 180
or, ∠A + ∠A = 180
or, 2∠A = 180
or, ∠A = 90
Hence, prove that the triangle is right-angled.
(10) In a ∆ABC, if 2∠A = 3∠B = 6∠C, calculate ∠A, ∠B and ∠C.
(11) What is the measure of each angle of an equilateral triangle?
Ans: Each angle of an equilateral triangle measure 60o.
(12) In the given figure, DE ∥ BC. If ∠A = 65o and ∠B = 55o, find (i) ∠ADE; (ii) ∠AED; ∠C.
(i) Since, DE ∥ BC and ADB is the transversal, so
∠ADE = ∠ABC = 55o (corresponding angles)
(iii) In ∆ABC, ∠A = 65o, ∠B = 55o
∴ ∠A + ∠B + ∠C = 180
or, 65 + 55 + ∠C = 180
or, ∠C = 180 – 120
or, ∠C = 60
(ii) Since, DE ∥ BC and ADB is the transversal, so
∠C = ∠AED = 60o (corresponding angles)
(13) Can a triangle have
(i) Two right angles? = NO
(ii) Two obtuse angles? = No
(iii) Two acute angles? = Yes
(iv) All angles more than 60o? = No
(v) All angles less than 60o? = No
(vi) All angles equal to 60o? = Yes
(14) Answer the following in “Yes” or “No”.
(i) Can a isosceles triangle be a right triangle? = Yes
(ii) Can a right triangle be a scalene triangle? = Yes
(iii) Can a right triangle be an equilateral triangle? = No
(iv) Can an obtuse triangle be an isosceles triangle? = Yes
(15) Fill in the blanks:
(i) A right triangle cannot have an obtuse angle.
(ii) The acute angles of a right triangle are complementary.
(iii) Each acute angle of an isosceles right triangle measures 45o.
(iv) Each angle of an equilateral triangle measures 60o.
(v) The side opposite the right triangle is called hypotenuse.
(vi) The sum of the lengths of the side of a triangle is called its perimeter.
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