Triangles - Study Mode

[#46] In ΔABC, AD ⊥ BC and AD 2 = BD × DC. The measure of ∠BAC is :
Correct Answer

(B) 90°

Explanation

Solution: AD 2 = BD.DC ΔADC ∼ ΔCAB (Property of a right angle Δ) ∠BAC = ∠ADC = 90°

[#47] In ΔABC, AB = BC = K, AC = $$sqrt 2 $$ k, then ΔABC is a :
Correct Answer

(A) Right isosceles triangle

Explanation

Solution: ∵ AB = BC = K ⇒ AC = $$sqrt 2 $$ K ⇒ (AC) 2 = (AB) 2 + (BC) 2 ⇒ ($$sqrt 2 $$ K) 2 = K 2 + K 2 ⇒ 2K 2 = 2K 2 ⇒ Therefore ΔABC will be a right isosceles triangle.

[#48] In ΔABC and ΔPQR, ∠B = ∠Q, ∠C = ∠R. M is the midpoint on QR, If AB : PQ = 7 : 4, then $$frac{{{ ext{area}},left( {vartriangle ABC}
ight)}}{{{ ext{area}},left( {vartriangle PMR}
ight)}}$$ xa0 is :
Correct Answer

(D) $$frac{{49}}{8}$$

Explanation

Solution: $$eqalign{
& frac{{{ ext{area}}{mkern 1mu} left( {vartriangle ABC}
ight)}}{{{ ext{area}}{mkern 1mu} left( {vartriangle PMR}
ight)}} cr
& = frac{{{{left( 7
ight)}^2}}}{{frac{1}{2} imes {{left( 4
ight)}^2}}} cr
& = frac{{49}}{8} cr} $$

[#49] In ΔABC, ∠B = 70° and ∠C = 30°, AD and AE are respectively the perpendicular on side BC and bisector of ∠A. The measure of ∠DAE is:
Correct Answer

(D) 20°

Explanation

Solution: ∠A = 180° - (∠B + ∠C) ∠A = 180° - 100° ∠A = 80° ∠BAE = ∠EAC = $$frac{1}{2}$$ ∠A = 40° In ΔBAD ∠BAD = 90° - ∠B ∠BAD = 90° - 70° ∠BAD = 20° ∠DAE = ∠BAE - ∠BAD ∠DAE = 40° - 20° ∠DAE = 20°

[#50] In ΔABC and ΔDEF, if ∠A = 50°, ∠B = 70°, ∠C = 60°, ∠D = 60°, ∠E = 70° and ∠F = 50°, then
Correct Answer

(A) ΔABC ∼ ΔFED

Explanation

Solution: From figure it is clear = ΔABC ∼ ΔFED