Selina Concise Class 6 Math Chapter 25 Quadrilateral Exercise 25A Solution

Selina Concise Class 6 Math Chapter 25 Quadrilateral Exercise 25A Solution

EXERCISE 25A

(1) Let the other equal angles measure be xo and xo.

We know that, sum of four angle of a quadrilateral = 360o.

Then, x + x + 89 + 113 = 360

⇒ 2x + 202 = 360

⇒ 2x = 360 – 202 = 158

⇒ x = 79

Therefore the measure of each equal angle is 79o.

(2) Let the measure of required two angles are 5x and 7x.

Then, 68 + 76 + 5x + 7x = 360

⇒ 12x + 144 = 360

⇒ 12x = 360 – 144 = 216

⇒ x = 18

Therefore the measure of the angle is (5 × 18) = 90o and (7 × 18) = 126o.

(3) (i) the value of x

∴ 4x + {5(x + 2)} + (7x – 20) + {6(x + 3)} = 360

⇒ 4x + 5x + 10 + 7x – 20 + 6x + 18 = 360

⇒ (4x + 5x + 7x + 6x) + 10 – 20 + 18 = 360

⇒ 22x – 10 + 18 = 360

⇒ 22x + 8 = 360

⇒ 22x = 360 – 8 = 352

⇒ x = 16

(ii) Each angle of the quadrilateral.

(4x)o = (4 × 16)o = 64o

{5 (x + 2)}o = {(5 × 16) + 10}o = 90o

(7x – 20)o = (7 × 16) – 20 = 92o

{6 (x + 3)}o = (6 × 16) + 3 = 99o

(4) Use the information given in the following figure to find:

∠A = 90o (given)

∴ 3x – 5 + 8x – 15 + 2x + 4 + 90 = 360

⇒ (3x + 8x + 2x) – 5 – 15 + 4 +90 = 360

⇒ 13x – 20 + 94 = 360

⇒ 13x = 360 – 74 = 286

⇒ x = 22

(i) Value of x = 22

(ii) ∠B = 2x + 4 = (2 × 22) + 4 = 48o

∠C = 3x – 5 = (3 × 22) – 5 = 61o

(5) Let the measure of ∠A = x

∠D = 2x

∠C = 4x

∠B = 5x

Then, x + 5x + 4x + 2x = 360

⇒ 12x = 360

⇒ x = 30

(i) ∠A = 30o

∠B = (5 × 30) = 150o

∠C = (4 × 30) = 120o

∠D = (2 × 30) = 60o

(ii) Trapezium

(6) (i) From

⇒ 4x + 48 + 3x + 4x + x = 360

⇒ 12x = 360 – 48 = 312

⇒ x = 26

(ii) ∠ABC = 4 × 26 = 104o

(iii) ∠ACD = 180 – 4x – 48

⇒ ∠ACD = 132 – 104 = 28o

(7) ∠A + ∠B + ∠C + ∠D = 360

⇒ {5 (a + 2)} + {2 (2a + 7)} + 64 + (64 – 8) = 360

⇒ 5a + 10 + 4a + 14 + 64 + 56 = 360

⇒ 9a + 144 = 360

⇒ 9a = 360 – 144 = 216

⇒ a = 24

Therefore, ∠A = 5 (a + 2) = 5 (24 + 2) = 5 × 26 = 130

(8) ∠a + ∠b + ∠c + 70 = 360

⇒ a + 2a + 15 + 3a + 5 + 70 = 360

⇒ 6a + 90 = 360

⇒ 6a = 360 – 90

⇒ 6a = 270

⇒ a = 45

Value of b = 2a + 15 = (2 × 45) + 15 = 90 + 15 = 105o

Value of c = 3a + 5 = (3 × 45) + 5 = 135 + 5 = 140o

(9) Let the each equal angle measure be xo.

Then, x + x + x + 69 = 360

⇒ 3x = 360 – 69

⇒ 3x = 291

⇒ x = 97

Hence, the measure of each angle is 97o.

(10) Let the required angles measure be 3x, 4x, 6x and 7x.

Then, 3x + 4x + 6x + 7x = 360

⇒ 20x = 360

⇒ x = 18

Therefore, ∠P = 3 × 18 = 54o

∠Q = 4 × 18 = 72o

∠R = 6 × 18 = 108o

∠S = 7 × 18 = 126o

Hence, PQ ∥ RS

As ∠P + ∠Q = 54o + 72o = 126o

Which is not equal to180o.

∴ Ps and QR are not parallel.

(12) Let, ∠A = 4x and ∠D = 5x

Since, ∠A + ∠D = 180o [∵ AB ∥ DC]

∴ 4x + 5x = 180o

⇒ 9x = 180o

⇒ x = 20o

Then, ∠A + ∠B + ∠C + ∠D = 360

⇒ (4 × 20) + 3x – 15 + 4x + 20 + (5 × 20) = 360

⇒ 80 + 3x – 15 + 4x + 20 + 100 = 360

⇒ 7x + 200 – 15 = 360

⇒ 7x + 185 = 360

⇒ 7x = 360 – 185 = 175

⇒ x = 25

∴ ∠A = 80o

∠B = (3x – 15)o = [(3 × 25) – 15]o = (75 – 15)o = 60o

∠C = (4x + 20)o = [(4 × 25) + 20]o = (100 + 20)o = 120o

∠D = 100o


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