Welcome to Part 1 of our new 8-part series on Chapter 6, Triangles. This post contains the top 25 Multiple Choice Questions (MCQs) to help you master concepts like BPT, Similarity, and Area Ratios.
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Top 25 MCQs - Triangles
Question 1: BPT Application
In triangle ABC, DE || BC. If AD = 1.5 cm, DB = 3 cm, and AE = 1 cm, then EC is:
1.5 / 3 = 1 / EC
1 / 2 = 1 / EC => EC = 2 cm.
Question 2: Similar Triangles Areas
If ΔABC ~ ΔPQR and Area(ΔABC)/Area(ΔPQR) = 9/16, then BC/QR is:
Area(ABC)/Area(PQR) = (BC/QR)²
9/16 = (BC/QR)²
BC/QR = √(9/16) = 3/4.
Question 3: Similar Triangles Sides
If ΔABC ~ ΔDEF such that 2AB = DE and BC = 8 cm, then EF is:
Given 2AB = DE => AB/DE = 1/2.
1/2 = 8/EF => EF = 16 cm.
Question 4: Right Triangle Check
The lengths of the sides of a triangle are 7 cm, 24 cm, and 25 cm. Is this a right-angled triangle?
25² = 625.
Since 7² + 24² = 25², it is a right-angled triangle.
Question 5: BPT Converse
In ΔABC, D and E are points on AB and AC such that AD/DB = AE/EC. Then:
Question 6: Corresponding Angles
If ΔABC ~ ΔPQR, ∠A = 50° and ∠B = 70°, then ∠R is:
Since ΔABC ~ ΔPQR, corresponding angles are equal.
So, ∠R = ∠C = 60°.
Question 7: Congruency vs Similarity
Which of the following statements is false?
Question 8: Area Ratio Median
The ratio of the areas of two similar triangles is equal to the square of the ratio of their:
Question 9: Isosceles Right Triangle
ABC is an isosceles triangle right-angled at C. Therefore:
By Pythagoras theorem: AB² = AC² + BC².
AB² = AC² + AC² = 2AC².
Question 10: Ladder Problem
A ladder 10 m long reaches a window 8 m above the ground. The distance of the foot of the ladder from the base of the wall is:
By Pythagoras theorem: 10² = 8² + x²
100 = 64 + x²
x² = 36 => x = 6 m.
Question 11: Equilateral Triangle Altitude
The altitude of an equilateral triangle of side 2a is:
Using Pythagoras: (2a)² = h² + a²
4a² = h² + a² => h² = 3a² => h = a√3.
Question 12: Rhombus Diagonals
The diagonals of a rhombus are 16 cm and 12 cm. The length of the side of the rhombus is:
Half diagonals = 8 cm and 6 cm.
Side² = 8² + 6² = 64 + 36 = 100.
Side = √100 = 10 cm.
Question 13: BPT Application 2
In ΔABC, DE || BC such that AD/DB = 3/5. If AC = 4.8 cm, then AE is:
Let AE = 3x, EC = 5x. AC = 8x = 4.8.
x = 0.6.
AE = 3x = 3(0.6) = 1.8 cm.
Question 14: Similarity Ratio
If ΔABC ~ ΔPQR with AB/PQ = 1/3, then Area(ΔABC)/Area(ΔPQR) is:
Question 15: Midpoint Theorem
In ΔABC, D and E are midpoints of AB and AC respectively. If DE = 4 cm, then BC is:
4 = (1/2)BC => BC = 8 cm.
Question 16: Similar Triangles
ΔABC ~ ΔDEF. If AB=4 cm, BC=3.5 cm, CA=2.5 cm and DF=7.5 cm, the perimeter of ΔDEF is:
Since similar, Ratio of Perimeters = Ratio of Sides.
CA/DF = 2.5/7.5 = 1/3.
Perimeter(ABC)/Perimeter(DEF) = 1/3
10 / P(DEF) = 1/3 => P(DEF) = 30 cm.
Question 17: Trapezium Diagonals
In trapezium ABCD, AB || DC. Diagonals intersect at O. If AO/OC = 1/2 and AB = 5 cm, then DC is:
AO/OC = AB/DC
1/2 = 5/DC => DC = 10 cm.
Question 18: Pythagorean Triplet
Which of the following sets of sides represents a right-angled triangle?
Question 19: Isosceles Angle
In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:
AB² + BC² = 108 + 36 = 144 = AC².
It satisfies Pythagoras theorem, so angle opposite to longest side (AC) is 90°. Angle B = 90°.
Question 20: Shadow Problem
A vertical pole of length 6 m casts a shadow 4 m long on the ground. At the same time, a tower casts a shadow 28 m long. The height of the tower is:
Height/Shadow = Height/Shadow
6/4 = h/28
h = (6 × 28) / 4 = 6 × 7 = 42 m.
Question 21: Congruent & Similar
Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are:
Question 22: Area Ratio
If ΔABC ~ ΔQRP, Area(ABC)/Area(QRP) = 9/4, AB = 18 cm and BC = 15 cm, then PR is:
9/4 = (BC/RP)²
3/2 = 15/PR
3PR = 30 => PR = 10 cm.
(Note order: ABC ~ QRP, so BC corresponds to RP).
Question 23: Equilateral Area
ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC. Ratio of areas of ΔABC and ΔBDE is:
Since both are equilateral, they are similar.
Area(ABC)/Area(BDE) = (BC/BD)² = (BC / (BC/2))² = (2)² = 4/1.
Question 24: Corresponding Altitudes
If two similar triangles have a scale factor of a:b, then the ratio of their corresponding altitudes is:
Question 25: Square Diagonals
If the diagonal of a square is 10√2 cm, then its perimeter is:
10√2 = Side × √2 => Side = 10 cm.
Perimeter = 4 × Side = 4 × 10 = 40 cm.
