Triangles, Similarity, Medians and Altitudes Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Triangles, Similarity, Medians and Altitudes notes for CAT: 52 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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.

Altitude Means Perpendicular Height, Not Just a Side

Altitude = perpendicular height

In a triangle, an altitude is drawn from one vertex to the opposite side at exactly $9 0^{\circ}$ .

Correct idea

Height must be perpendicular to the chosen base.

Common wrong idea

Students often use a slant side as height. That is valid only if it is perpendicular.

Exam language: altitude, height, perpendicular distance, distance from vertex to side — all point toward the same core idea.

The Area Formula Is the Central Engine

Triangle area

Δ = \frac{1}{2} \times base \times height

If you know	You can find
Area and base	Height: $h = \frac{2Δ}{b}$
Area and height	Base: $b = \frac{2Δ}{h}$
Base and height	Area directly

Triangles, Similarity, Medians and Altitudes: Solved Questions with Step-by-Step Explanations (2 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.

More notes in this unit

chapter

Triangles, Similarity, Medians and Altitudes Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Triangles, Similarity, Medians and Altitudes notes for CAT: 52 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questio

A question from this chapter

Question 1

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 ?

Question 2

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 ?

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