Triangles(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

TRIANGLES MCQs

Total Questions : 83 | Page 4 of 9 pages

Question 31. ABC is a triangle, PQ is line segment intersecting AB is P and AC in Q and PQ || BC. The ratio of AP : BP = 3 : 5 and length of PQ is 18 cm. The length of BC is:

28 cm
48 cm
84 cm
42 cm

Discuss Question

Answer: Option B. -> 48 cm
∵ PQ || BC
So ∠AQP = ∠ACB = α
and
∠APQ = ∠ABC = β
So, ΔABC and ΔAPQ
$$\eqalign{
& \frac{{AP}}{{AB}} = \frac{{PQ}}{{BC}} \cr
& \frac{3}{8} = \frac{{PQ}}{{BC}} \cr
& \frac{3}{8} = \frac{{18}}{{BC}} \cr
& BC = 48\,{\text{cm}} \cr} $$

Question 32. Which of the following is a true statement

Two similar triangles are always congruent
Two similar triangles have equal areas
Two triangles are similar if their corresponding sides are proportional
Two polygons are similar if their corresponding sides are proportional

Discuss Question

Answer: Option C. -> Two triangles are similar if their corresponding sides are proportional
Two triangles are similar if their corresponding sides are proportional

Question 33. ΔABC is similar to ΔDEF is area of ΔABC is 9 sq. cm. and area of ΔDEF is 16 sq. cm. and BC = 21 cm. Then the length of EF will be:

5.6 cm
2.8 cm
3.7 cm
1.4 cm

Discuss Question

Answer: Option B. -> 2.8 cm
∵ ΔABC ≅ ΔDEF
$$\eqalign{
& \therefore \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{\sqrt 9 }}{{\sqrt {16} }} \cr
& = \frac{{2.1}}{{EF}} = \frac{3}{4} \cr
& EF = 2.8\,{\text{cm}} \cr} $$

Question 34. In ΔABC and ΔPQR, ∠B = ∠Q, ∠C = ∠R. M is the midpoint on QR, If AB : PQ = 7 : 4, then $$\frac{{{\text{area}}\,\left( {\vartriangle ABC} \right)}}{{{\text{area}}\,\left( {\vartriangle PMR} \right)}}$$ is :

$$\frac{{35}}{8}$$
$$\frac{{35}}{{16}}$$
$$\frac{{49}}{{16}}$$
$$\frac{{49}}{8}$$

Discuss Question

Answer: Option D. -> $$\frac{{49}}{8}$$
$$\eqalign{
& \frac{{{\text{area}}{\mkern 1mu} \left( {\vartriangle ABC} \right)}}{{{\text{area}}{\mkern 1mu} \left( {\vartriangle PMR} \right)}} \cr
& = \frac{{{{\left( 7 \right)}^2}}}{{\frac{1}{2} \times {{\left( 4 \right)}^2}}} \cr
& = \frac{{49}}{8} \cr} $$

Question 35. 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:

24°
10°
15°
20°

Discuss Question

Answer: Option D. -> 20°
∠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°

Question 36. In ΔABC, the line parallel to BC intersect AB & AC at P & Q respectively. If AB : AP = 5 : 3, then AQ : QC is:

3 : 2
1 : 2
3 : 5
2 : 3

Discuss Question

Answer: Option A. -> 3 : 2
$$\eqalign{
& \frac{{AP}}{{PB}} = \frac{{AQ}}{{QC}} \cr
& \frac{{AQ}}{{QC}} = \frac{3}{2} \cr
& AQ:QC = 3:2 \cr} $$

Question 37. In ΔABC and ΔDEF, if ∠A = 50°, ∠B = 70°, ∠C = 60°, ∠D = 60°, ∠E = 70° and ∠F = 50°, then

ΔABC ∼ ΔFED
ΔABC ∼ ΔDFE
ΔABC ∼ ΔEDF
ΔABC ∼ ΔDEF

Discuss Question

Answer: Option A. -> ΔABC ∼ ΔFED
From figure it is clear
= ΔABC ∼ ΔFED

Question 38. In ΔABC, ∠B = 60° and ∠C = 40°; AD and AE are respectively the bisector of ∠A and perpendicular on BC. The measure of ∠EAD is:

9°
11°
10°
12°

Discuss Question

Answer: Option C. -> 10°
As we know
$$\eqalign{
& \angle EAD = \frac{{\angle B - \angle C}}{2} \cr
& \angle EAD = \frac{{60 - 40}}{2} \cr
& \angle EAD = {10^ \circ } \cr} $$

Question 39. The orthocenter of a triangle is the point where?