Triangles, Similarity, Medians and Altitudes Previous Year Questions (PYQs) for CAT: 7+ Solved Questions with Step-by-Step Solutions

Solve 7+ Triangles, Similarity, Medians and Altitudes previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Triangles, Similarity, Medians and Altitudes

Geometry Chapter Journey

Triangles, Similarity, Medians and Altitudes

12 CAT PYQs

Step 1 · Selected Topic · 5 PYQs · Highest weight inside this chapter

📐 Altitudes, Areas and Right Triangles

Master how height creates area, how one triangle gives many altitudes, and how right triangles appear inside CAT geometry.

Step 2 · 4 PYQs · Medium-high weight

📍 Medians, Centroids and Section Ratios

Later, you will learn how medians split area and how centroid ratios simplify triangle division problems.

Step 3 · 3 PYQs · Conceptual finish

🔎 Similarity, Isosceles and Angle Chasing

Finally, you will combine equal sides, equal angles, and proportional lengths to solve compact but tricky CAT questions.

By the end of this chapter: you should be able to see a triangle not as a drawing, but as a system of areas, heights, ratios, and hidden right triangles.

Topic Hero: Height Is the Secret Handle of a Triangle

Selected Topic

Altitudes, Areas and Right Triangles

This topic teaches one exam weapon: convert triangle information into area and right-triangle information.

5 direct CAT PYQs Area ratios Pythagoras

Area appears when base meets perpendicular height.

Triangles, Similarity, Medians and Altitudes: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

In a right-angled triangle $△ A B C$ , the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of $△ A B P$ , $△ A B Q$ and $△ A B C$ are in arithmetic progression. If the area of $△ A B C$ is 1.5 times the area of $△ A B P$ , the length of PQ, in cm, is

Correct Answer:

Step-by-Step Solution

This is an areas-in-arithmetic-progression-on-a-base-line question. You know this because several triangles all share the same vertex and the same height, with their bases lying along one straight line, and their areas are said to form an AP.

Why this method applies: since angle B = 90° (AB is the altitude and BC is the base), AB is perpendicular to BC. That means for ANY point X on line BC, the height from A down to BC is always AB = 5. So triangle ABX's area only depends on the base length BX.

Step 1 — Write area as a function of base length.

$[A B X] = \frac{1}{2} \times B X \times A B = \frac{1}{2} \times B X \times 5 = 2.5 B X$

Step 2 — Find the area of the full triangle ABC.

$[A B C] = 2.5 \times 12 = 30$

Step 3 — Use the given ratio to find area of ABP.

$[A B C] = 1.5 \times [A B P] \Rightarrow [A B P] = \frac{30}{1.5} = 20$

So $2.5 B P = 20 \Rightarrow B P = 8$ .

Step 4 — Use the AP condition to find area of ABQ.

The three areas $[A B P], [A B Q], [A B C]$ = $20, [A B Q], 30$ are in AP, so the middle term is the average of the other two:

$2 [A B Q] = 20 + 30 = 50 \Rightarrow [A B Q] = 25$

So $2.5 B Q = 25 \Rightarrow B Q = 10$ .

Step 5 — Find PQ.

$P Q = ∣ B Q - B P ∣ = ∣10 - 8∣ = 2 cm$

So $P Q = 2$ cm.

Common trap: assuming the altitude is somewhere other than AB, or forgetting that because angle B is the right angle, AB itself acts as the constant height for every triangle ABX with X on BC — this is what lets you convert "area in AP" directly into "base length in AP".

Question 2 · Quantitative Ability MCQ

In $△ A B C$ , $A B = A C = 12$ cm and $D$ is a point on side $B C$ such that $A D = 8$ cm. If $A D$ is extended to a point $E$ such that $∠ A C B = ∠ A E B$ , then the length, in cm, of $A E$ is

A.
20
B.
16
C.
18
D.
14

Correct Answer:

Step-by-Step Solution

Identify the hidden similarity. Since $A B = A C$ , $△ A B C$ is isosceles, meaning the base angles are equal: $∠ A B C = ∠ A C B$ . The problem states $∠ A C B = ∠ A E B$ . By transitivity, $∠ A B C = ∠ A E B$ . Now, compare $△ A B D$ and $△ A E B$ . They share the same angle at $A$ ( $∠ B A D = ∠ E A B$ ). They also have $∠ A B D = ∠ A E B$ (since $∠ A B D$ is just $∠ A B C$ ). By AA similarity, $△ A B D \sim △ A E B$ . The ratio of corresponding sides must be equal: $\frac{A B}{A E} = \frac{A D}{A B}$ . Cross-multiplying gives $A B^{2} = A E \times A D$ . Substitute the given lengths: $1 2^{2} = A E \times 8 ⟹ 144 = 8 \times A E ⟹ A E = 18$ .

Question 3 · Quantitative Ability NAT

A triangle ABC is formed with $A B = A C = 50$ cm and $B C = 80$ cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is

Correct Answer:

378

Step-by-Step Solution

This is a digit-formation / counting-up-to-a-limit question. Recognise it because it asks for numbers "up to" a specific value with a digit restriction (non-repeating digits) — the trigger word "up to" means we must split by number of digits and handle the boundary carefully.

Step 1: Split by number of digits, since 1-digit, 2-digit and 3-digit numbers are filled differently.

1-digit numbers (1 to 9): every single digit is automatically non-repeating.

Count = 9

2-digit numbers (10 to 99): first digit has 9 choices (1–9, not 0), second digit has 9 remaining choices (0–9 except the first digit).

Count = $9 \times 9 = 81$

3-digit numbers (100 to 499, since we must stay "up to 500"): the hundreds digit can only be 1, 2, 3 or 4 — NOT 5, because any 3-digit number starting with 5 in this range is 500 itself, and 500 has repeated 0's, so it fails the non-repeating condition anyway.

Hundreds digit: 4 choices

Tens digit: 9 choices (any digit except the hundreds digit)

Units digit: 8 choices (any digit except the two already used)

Count = $4 \times 9 \times 8 = 288$

Step 2: Check the boundary value 500 separately — its digits are 5, 0, 0, which repeat, so it is correctly excluded by not including "5" as a hundreds-digit option.

Step 3: Add all cases.

Total = $9 + 81 + 288 = 378$

Trap check: a common mistake is to allow the hundreds digit 5 choices (1–5) assuming any 3-digit number starting with 5 up to 500 counts — but only 500 itself starts with 5 in this range, and it fails the non-repetition rule, so 5 should not be an option at all.

Final answer: 378

Question 4 · Quantitative Ability MCQ

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is

A.
$8$
B.
$6$
C.
$7$
D.
$5$

Correct Answer:

Step-by-Step Solution

This is the "Point on Base of an Equilateral Triangle" pattern: a point divides one side into two parts, and an area-ratio condition tells you exactly where that point is, after which the Law of Cosines gives the required length.

Step 1 — Recognize the trigger.

D lies on BC, splitting it into BD and DC. The condition "area(ADC) is half area(ABD)" is a ratio between the two smaller triangles sharing the same apex A and the same height (the perpendicular distance from A to line BC).

Step 2 — Convert the area ratio into a length ratio.

Triangles ABD and ADC share the same height from A, so their areas are proportional to their bases:

$\frac{Area ( A B D )}{Area ( A D C )} = \frac{B D}{D C}$

Given Area(ADC) = (1/2)·Area(ABD), this means Area(ABD)/Area(ADC) = 2, so:

$\frac{B D}{D C} = 2$

Step 3 — Solve for the actual lengths.

Let $D C = x$ , so $B D = 2 x$ . Since $B D + D C = B C = 3$ :

$2 x + x = 3 \Rightarrow x = 1$

So $D C = 1$ and $B D = 2$ .

Step 4 — Use the Law of Cosines in triangle ABD.

In triangle ABD, $A B = 3$ , $B D = 2$ , and $∠ A B D = ∠ A B C = 6 0^{\circ}$ (equilateral triangle):

$A D^{2} = A B^{2} + B D^{2} - 2 \cdot A B \cdot B D \cdot cos 6 0^{\circ} = 9 + 4 - 2 (3) (2) (\frac{1}{2}) = 13 - 6 = 7$

Step 5 — Take the square root.

$A D = 7$

Trap avoided: it's tempting to assume the right angle exists at D and use Pythagoras directly (e.g. $A D^{2} = A B^{2} - B D^{2}$ ), but D is not the foot of a perpendicular here — the 60° angle must be used via the Law of Cosines, not a right-angle shortcut.

Answer: $A D = 7$ cm.

Question 5 · Quantitative Ability MCQ

If a triangle ABC, $∠ B C A = 5 0^{\circ}$ . D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then $∠ F D E$ , in degrees, is equal to

A.
72
B.
80
C.
100
D.
96

Correct Answer:

Step-by-Step Solution

This is a pure angle-chasing question built from two isosceles triangles hidden inside a bigger triangle. You can recognise it because two equal-length conditions are given (AD = DE and BD = DF) with no numeric side lengths — only angles matter.

Step 1: Let $∠ B A C = α$ and $∠ A B C = β$ . Since the angles of triangle ABC sum to $18 0^{\circ}$ and $∠ B C A = 5 0^{\circ}$ :

$α + β = 18 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}$

Step 2: Look at triangle ADE. Since D lies on AB and E lies on AC, the angle at vertex A inside this small triangle is the same as $∠ B A C = α$ . We are told $A D = D E$ , so triangle ADE is isosceles with the two equal sides being AD and DE. Equal sides sit opposite equal angles, so the angle opposite DE (which is $∠ D A E = α$ ) equals the angle opposite AD (which is $∠ D E A$ ). So:

$∠ D E A = α \Rightarrow ∠ A D E = 18 0^{\circ} - 2 α$

Step 3: Look at triangle BDF. Since D lies on AB and F lies on BC, the angle at vertex B inside this small triangle is $∠ D B F = β$ . We are told $B D = D F$ , so by the same isosceles logic, the angle opposite BD (which is $∠ D F B$ ) equals the angle opposite DF (which is $∠ D B F = β$ ). So:

$∠ D F B = β \Rightarrow ∠ B D F = 18 0^{\circ} - 2 β$

Step 4: Since A, D, B are collinear (D is a point on segment AB), the three angles at D on one side of line AB — namely $∠ A D E$ , $∠ E D F$ , and $∠ F D B$ — must add up to a straight angle:

$∠ A D E + ∠ E D F + ∠ F D B = 18 0^{\circ}$

Step 5: Substitute the known expressions:

$(18 0^{\circ} - 2 α) + ∠ E D F + (18 0^{\circ} - 2 β) = 18 0^{\circ}$

$∠ E D F = 18 0^{\circ} - (18 0^{\circ} - 2 α) - (18 0^{\circ} - 2 β) = 2 α + 2 β - 18 0^{\circ}$

Step 6: Use $α + β = 13 0^{\circ}$ from Step 1:

$∠ E D F = 2 (13 0^{\circ}) - 18 0^{\circ} = 26 0^{\circ} - 18 0^{\circ} = 8 0^{\circ}$

Answer: $∠ F D E = 8 0^{\circ}$ , option B.

Common trap: students often try to find $α$ and $β$ individually, which is impossible since only their sum is fixed — the beauty of this question is that the final angle depends only on $α + β$ , not on the individual values.

Triangles, Similarity, Medians and Altitudes Previous Year Questions (PYQs) for CAT: 7+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Triangles, Similarity, Medians and Altitudes

Triangles, Similarity, Medians and Altitudes

📐 Altitudes, Areas and Right Triangles

📍 Medians, Centroids and Section Ratios

🔎 Similarity, Isosceles and Angle Chasing

Topic Hero: Height Is the Secret Handle of a Triangle

Altitudes, Areas and Right Triangles

Triangles, Similarity, Medians and Altitudes: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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