Triangles, Similarity, Medians and Altitudes Practice Questions for CAT: 114+ Solved Questions with Step-by-Step Solutions

Solve 114+ Triangles, Similarity, Medians and Altitudes practice 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 MCQ

A line passing through the centroid $G$ of $△ A B C$ intersects the sides $A B$ and $A C$ at points $P$ and $Q$ respectively. If $A P : P B = 3 : 2$ , what is the ratio $A Q : QC$ ?

A.
1:2
B.
2:1
C.
3:1
D.
1:3

Correct Answer:

Step-by-Step Solution

Key idea: This is a centroid transversal question. The trigger is a line passing exactly through the centroid and cutting the two adjacent sides, asking for the relationship between the two section ratios.

Step 1: Recall the centroid transversal property. For any line passing through the centroid $G$ that intersects sides $A B$ and $A C$ at $P$ and $Q$ , the sum of the reciprocals of the side fractions is always exactly 3.

Mathematically: $\frac{A B}{A P} + \frac{A C}{A Q} = 3$ .

Step 2: Convert the given ratio $A P : P B$ into the fraction $\frac{A P}{A B}$ .

$A P : P B = 3 : 2$ means $A B$ is split into $3 + 2 = 5$ parts.

So, $\frac{A P}{A B} = \frac{3}{5}$ .

Step 3: Substitute this into the centroid property formula.

The reciprocal is $\frac{A B}{A P} = \frac{5}{3}$ .

$\frac{5}{3} + \frac{A C}{A Q} = 3$ .

Step 4: Solve for $\frac{A C}{A Q}$ .

$\frac{A C}{A Q} = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$ .

Step 5: Convert back to the section ratio $A Q : QC$ .

If $\frac{A C}{A Q} = \frac{4}{3}$ , then $\frac{A Q}{A C} = \frac{3}{4}$ .

This means $A Q$ is 3 parts and $A C$ is 4 parts, leaving $QC = 4 - 3 = 1$ part.

Therefore, $A Q : QC = 3 : 1$ .

Answer: 3:1

Question 2 · Quantitative Ability MCQ

In an isosceles triangle where sides $A B$ and $A C$ are equal in length, what is the relationship between the altitude drawn to side $A B$ and the altitude drawn to side $A C$ ?

A.
They are equal in length.
B.
The altitude to $A B$ is strictly longer.
C.
The altitude to $A C$ is strictly longer.
D.
They are always perpendicular to each other.

Correct Answer:

Step-by-Step Solution

Key idea: This is a symmetry property question. Recognise it because it asks about the altitudes drawn to the equal sides of an isosceles triangle.

Step 1: Recall the area formula: Area = $\frac{1}{2} \times base \times height$ .

Step 2: The area of the triangle is constant. If we choose $A B$ as the base, the height is the altitude to $A B$ ( $h_{c}$ ). If we choose $A C$ as the base, the height is the altitude to $A C$ ( $h_{b}$ ).

Step 3: Since Area = $\frac{1}{2} \times A B \times h_{c} = \frac{1}{2} \times A C \times h_{b}$ , and we are given that $A B = A C$ , it strictly follows that $h_{c} = h_{b}$ .

Answer: They are equal in length.

Question 3 · Quantitative Ability NAT

Let $D, E$ , and $F$ be the midpoints of the sides $B C, C A$ , and $A B$ of $△ A B C$ , respectively. The medians $A D, B E$ , and $C F$ intersect the line segments $E F, D F$ , and $D E$ at points $X, Y$ , and $Z$ , respectively. If the area of $△ A B C$ is $320$ sq cm, what is the area, in sq cm, of $△ X Y Z$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the interaction between the medians of a triangle and its medial triangle. The trigger is the medians intersecting the sides of the midpoint triangle, forming a new inner triangle.

Step 1: Understand the medial triangle $△ D E F$ . Connecting the midpoints of the sides of $△ A B C$ creates the medial triangle $△ D E F$ . By the midpoint theorem, $△ D E F$ is similar to $△ A B C$ with a side ratio of $\frac{1}{2}$ .

Therefore, Area( $△ D E F$ ) = $(\frac{1}{2})^{2} \times$ Area( $△ A B C$ ) = $\frac{1}{4} \times 320 = 80$ sq cm.

Step 2: Locate $X, Y$ , and $Z$ . Consider the median $A D$ . It is a known property that a median of the main triangle bisects any segment parallel to its base that is bounded by the other two sides.

Since $E F$ is the midsegment parallel to $B C$ , the median $A D$ must intersect $E F$ exactly at its midpoint. Thus, $X$ is the midpoint of $E F$ .

By identical logic, $Y$ is the midpoint of $D F$ , and $Z$ is the midpoint of $D E$ .

Step 3: Analyze $△ X Y Z$ . Since $X, Y$ , and $Z$ are the midpoints of the sides of $△ D E F$ , $△ X Y Z$ is simply the medial triangle of $△ D E F$ !

Step 4: Calculate the final area. The area of a medial triangle is always $\frac{1}{4}$ of its parent triangle.

Area( $△ X Y Z$ ) = $\frac{1}{4} \times$ Area( $△ D E F$ ) = $\frac{1}{4} \times 80 = 20$ sq cm.

Answer: 20

Question 4 · Quantitative Ability MCQ

In a right-angled triangle $A B C$ with the right angle at $B$ , the altitude $B D$ is drawn to the hypotenuse $A C$ . The incircles of triangles $A B D$ and $B D C$ have radii $r_{1}$ and $r_{2}$ respectively. If $r_{1} = 3$ cm and $r_{2} = 4$ cm, which of the following is the area of triangle $A B C$ (in sq cm)?

A.
120
B.
150
C.
180
D.
200

Correct Answer:

Step-by-Step Solution

Key idea: This is a "similar right triangles and inradius" question. The altitude to the hypotenuse divides a right triangle into two smaller right triangles that are similar to each other and to the original triangle.

Step 1: Because $△ A B D \sim △ C B D$ , the ratio of their corresponding linear dimensions (including their inradii) is equal to the ratio of their hypotenuses.

Ratio of similarity $k = \frac{r _{1}}{r _{2}} = \frac{3}{4}$ .

Step 2: This means the sides of $△ A B D$ and $△ C B D$ are in the ratio $3 : 4$ . Consequently, the main triangle $A B C$ must be a $3 : 4 : 5$ right triangle. Let its sides be $A B = 3 x$ , $B C = 4 x$ , and $A C = 5 x$ .

Step 3: Find the segments of the hypotenuse.

The altitude $B D = \frac{A B \times B C}{A C} = \frac{3 x \times 4 x}{5 x} = 2.4 x$ .

The segment $A D = \frac{A B ^{2}}{A C} = \frac{9 x ^{2}}{5 x} = 1.8 x$ .

Step 4: Use the inradius formula for a right triangle: $r = \frac{leg _{1} + leg _{2} - hypotenuse}{2}$ .

For $△ A B D$ , the legs are $A D$ and $B D$ , and the hypotenuse is $A B$ .

$r_{1} = \frac{1.8 x + 2.4 x - 3 x}{2} = \frac{1.2 x}{2} = 0.6 x$ .

Step 5: Solve for $x$ .

We are given $r_{1} = 3$ , so $0.6 x = 3 ⟹ x = 5$ .

Step 6: Calculate the area of $△ A B C$ .

The legs of $△ A B C$ are $A B = 3 (5) = 15$ and $B C = 4 (5) = 20$ .

Area $= \frac{1}{2} \times 15 \times 20 = 150$ sq cm.

Answer: 150

Question 5 · Quantitative Ability NAT

In $△ A B C$ , the measure of $∠ B$ is exactly twice the measure of $∠ C$ . The internal angle bisector of $∠ A$ meets the side $B C$ at point $D$ . If $A B = 14$ cm and $A C = 21$ cm, what is the length, in cm, of the side $B C$ ?

Correct Answer:

17.5

Step-by-Step Solution

Key idea: This question tests a classic geometric construction involving the double-angle property ( $∠ B = 2∠ C$ ) and an angle bisector. The trigger is the specific $2 : 1$ angle ratio combined with an angle bisector.

Step 1: Construct an auxiliary point. Extend the side $A B$ past $B$ to a point $E$ such that $B E = B D$ .

Step 2: Analyze $△ B D E$ . Since $B E = B D$ , $△ B D E$ is isosceles, meaning $∠ E = ∠ B D E$ .

Step 3: Use the Exterior Angle Theorem on $△ B D E$ . The exterior angle at $B$ is $∠ A B C$ .

$∠ A B C = ∠ E + ∠ B D E = 2∠ E$ .

Step 4: Link to $∠ C$ . We are given $∠ A B C = 2∠ C$ . Therefore, $2∠ E = 2∠ C ⟹ ∠ E = ∠ C$ .

Step 5: Analyze $△ A E C$ . Since $∠ E = ∠ C$ , $△ A E C$ is isosceles with $A E = A C$ .

We are given $A C = 21$ , so $A E = 21$ .

Step 6: Find $B D$ . Since $A E = A B + B E$ and we defined $B E = B D$ , we have:

$21 = 14 + B D ⟹ B D = 7$ cm.

Step 7: Use the Angle Bisector Theorem in $△ A B C$ . The bisector $A D$ divides $B C$ in the ratio of the adjacent sides:

$\frac{A B}{A C} = \frac{B D}{D C} ⟹ \frac{14}{21} = \frac{7}{D C} ⟹ \frac{2}{3} = \frac{7}{D C}$ .

Solving for $D C$ : $D C = \frac{21}{2} = 10.5$ cm.

Step 8: Calculate $B C$ .

$B C = B D + D C = 7 + 10.5 = 17.5$ cm.

Answer: 17.5

Triangles, Similarity, Medians and Altitudes Practice Questions for CAT: 114+ 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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