Geometry - Study Mode

[#161] ABC is a triangle and the sides AB, BC and CA are produced to E, F and G respectively. If ∠CBE = ∠ACF = 130°, then the value of ∠GAB is:
Correct Answer

(A) 100°

Explanation

Solution: We know that ⇒ Add of total exterior angle of a triangle (polygon) = 360° ⇒ So, 130° + 130° + x° = 360° ⇒ x° = 100°

[#162] A triangle ABC is inscribed in a circle with centre O. AO is produced to meet the circle at K and AD ⊥ BC. If ∠B = 80° and ∠C = 64°, then the measure of ∠DAK is:
Correct Answer

(B) 16°

Explanation

Solution: ∠ACK = 90° (In semicircle) ∠B = ∠K = 80° In ΔAKC, ∠A + ∠K + ∠C = 180° ∠A + 80° + 90° = 180° ∠A = 10° In ΔABD, ∠A = 90° - 80° = 10° ∠CAD = 90° - 64° = 26° ∠DAK = 26° - 10° = 16°

[#163] In the figure, chords AB and CD of a circle intersect externally at P. If AB = 4 cm, CD = 11 cm and PD = 15 cm, then the length of PB is:
Correct Answer

(A) 10 cm

Explanation

Solution: Let, PA = x PA × PB = PC × PD x × (x + 4) = 4 × 15 x × (x + 4) = 60 x 2 + 4x - 60 = 0 x 2 + 10x - 6x - 60 = 0 x(x + 10) - 6(x + 10) = 0 x = 6 PB = x + 4 = 6 + 4 = 10 cm

[#164] In the given figure, if AC, DE are parallel and ∠CAB = 38°, then the value of ∠ABC + 5∠CBD is:
Correct Answer

(C) 218°

Explanation

Solution: If AC || BD, 2a + b + a = 180° . . . . . . (1) 2a = 38° a = 19° From equation (1) 2 × 19° + b + 19° = 180° b = 123° Now, ∠ABC + 5∠CBD = 123° + 5 × 19° = 218°

[#165] A, B and C are three points on the circle. If AB = AC = 7√2 cm and ∠BAC = 90° men the radius is equal to:
Correct Answer

(B) 7 cm

Explanation

Solution: $${ ext{R}} = frac{{{ ext{Hypotenuse}}}}{2} = frac{{14}}{2} = 7$$