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

Triangles, Similarity, Medians and Altitudes short notes for CAT: 41 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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

Same Height or Same Base Gives Instant Area Ratios

Area ratio shortcut

Same height

$\frac{Δ _{1}}{Δ _{2}} = \frac{b _{1}}{b _{2}}$

Use when two triangles stand on the same line.

Same base

$\frac{Δ _{1}}{Δ _{2}} = \frac{h _{1}}{h _{2}}$

Use when triangles have the same base but different top vertices.

CAT cue: point on a side, areas given in ratio, or areas in arithmetic progression.

One Area Gives All Three Altitudes

Altitude from area

For a triangle with area $Δ$ and side lengths $a, b, c$ :

to side

a

h_{a} = \frac{2Δ}{a}

to side

b

h_{b} = \frac{2Δ}{b}

to side

c

h_{c} = \frac{2Δ}{c}

Memory hook: Longer side means smaller altitude, because the same area is being spread over a longer base.

PYQ Pattern: Sum of All Altitudes in an Isosceles Triangle

PYQ model: $A B = A C = 50$ , $B C = 80$

Isosceles altitude splits base: $40$ and $40$ .
Height from A: $h = 5 0^{2} - 4 0^{2} = 900 = 30$ .
Area: $Δ = \frac{1}{2} \cdot 80 \cdot 30 = 1200$ .
Other altitudes: $\frac{2Δ}{50} = \frac{2400}{50} = 48$ each.

Sum of all altitudes

= 30 + 48 + 48 = 126

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.

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

Triangles, Similarity, Medians and Altitudes short notes for CAT: 41 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice q

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

The Area Formula Is the Central Engine

Triangle area

Same Height or Same Base Gives Instant Area Ratios

Area ratio shortcut

Same height

Same base

One Area Gives All Three Altitudes

Altitude from area

PYQ Pattern: Sum of All Altitudes in an Isosceles Triangle

PYQ model: AB=AC=50, BC=80

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

Step-by-Step Solution

Step-by-Step Solution

More short notes in this unit

A question from this chapter

Login to view the complete short notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

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Questions calibrated to the real exam

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Honest progress

Unlock the whole course

PYQ model: $A B = A C = 50$ , $B C = 80$