Math Olympiad Test: Triangles - 1 - Class 10 MCQ

In ΔABD and ΔACD

By Angle Bisector Theorem
it is given,
and ∠BAC = 180 - 70 - 50
= 180 - 120 = 60°
so, ∠BAD = 

∵ P and Q are mid-points of ∠M and ∠N respectively
∠ PQ || MN (according to midpoint theorem) and, also,

where G and H are feets of  perpendicular dropped from L on PQ and MN respectively.

∠ CAB = ∠ACD
{ Alternate opposite int erior angles}
∠BDC = ∠DBA
∴ ΔAOB ~ ΔCOD (SS similarity)

CB = CD - BD = 11m - 6m = 5m
ED = AB = 12 m

∠A = 180° - (∠B + ∠C)
= 180° - (55° + 35°) = 90°
∴ ∠A is 90° or the triangle is right angled at A
∴ AB² + AC² = BC² (pythagoras’ theorem)

In ΔABC
∠BAC = 180° - (100° + 40°) = 40°
∠DAE = 180° - (50° + 40°) = 90° [linear pair]

∵ Vertical angles of isosceles Ds are identical
∴ Other two equal angles of both the Ds will be identical.
∴ By using S.S.S. similarity, we can say, that the two triangles are similar.
∴ Ratio of altitudes = 

Given:

ΔABC is right-angle at C

AC = √3BC

Formula used:

tan θ = Perpendicular/Base

tan 30° = 1/√3

Calculation: