Areas of Similar Triangles Theorem and Corollaries for CBSE Class 10

When two geometric figures are similar, their linear dimensions (sides, perimeters, heights) grow or shrink in direct proportion. But what happens to two-dimensional measures like surface area? If you double the length of every side of a triangle, does its area double?

In CBSE Class 10 Mathematics, the Areas of Similar Triangles Theorem answers this question with mathematical precision: the area scales not linearly, but with the square of the side ratio! Doubling the sides actually quadruples the area ($2^2 = 4$). This theorem is a frequent source of 3-mark and 4-mark questions in board examinations.

What You Will Learn

• Statement of the Areas of Similar Triangles Theorem
• Complete step-by-step geometric proof using altitudes and similarity
• High-yield corollaries: ratio of areas in terms of altitudes, medians, and angle bisectors
• The congruence condition: what happens when similar triangles have equal areas?
• Solved CBSE board examination problems and midpoint triangle relations
• Common algebraic traps and error-prevention tips

1. Statement of the Theorem

Theorem Statement (CBSE Theorem 6.6)

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\text{If } \Delta ABC \sim \Delta PQR, \quad \text{then:}$ $\mathbf{\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2}$

2. Step-by-Step Geometric Proof

                     A                                        P
                    /|\                                      /|                   / | \                                    / |                   /  |  \                                  /  |                   /   |   \                                /   |                   B----M----C                              Q----N----R

Given:

Two similar triangles $\Delta ABC \sim \Delta PQR$.

To Prove:

$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$

Construction:

Draw altitude $AM \perp BC$ and altitude $PN \perp QR$.

Proof:

1. Express areas of both triangles: $\text{Area}(\Delta ABC) = \frac{1}{2} \times BC \times AM$ $\text{Area}(\Delta PQR) = \frac{1}{2} \times QR \times PN$
2. Find the ratio of their areas: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times QR \times PN} = \frac{BC}{QR} \times \frac{AM}{PN} \quad \text{--- (1)}$
3. Relate altitudes using small triangles: In $\Delta ABM$ and $\Delta PQN$:
   • $\angle B = \angle Q$ (Corresponding angles of similar triangles $\Delta ABC \sim \Delta PQR$).
   • $\angle AMB = \angle PNQ = 90^\circ$ (By construction). Therefore, by the AA Similarity Criterion: $\Delta ABM \sim \Delta PQN$
4. Equate ratios of corresponding sides: $\frac{AM}{PN} = \frac{AB}{PQ} \quad \text{--- (2)}$
5. Use similarity of the original triangles: Since $\Delta ABC \sim \Delta PQR$, all corresponding sides are in the same ratio: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \quad \text{--- (3)}$
6. Substitute (2) and (3) into Equation (1): $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \frac{BC}{QR} \times \frac{AM}{PN} = \frac{AB}{PQ} \times \frac{AB}{PQ} = \left(\frac{AB}{PQ}\right)^2$
7. By transitive property from Equation (3): $\mathbf{\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2}$ Hence, proved.

3. Important Corollaries

If two triangles are similar, the ratio of their areas equals:

1. Square of the ratio of corresponding altitudes: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{\text{Altitude } AM}{\text{Altitude } PN}\right)^2$
2. Square of the ratio of corresponding medians: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{\text{Median } AD}{\text{Median } PM}\right)^2$
3. Square of the ratio of corresponding perimeters: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{\text{Perimeter}(\Delta ABC)}{\text{Perimeter}(\Delta PQR)}\right)^2$

The Congruence Theorem:

If the areas of two similar triangles are equal, then the triangles are congruent.

• Proof: $\frac{\text{Area}_1}{\text{Area}_2} = 1 \implies \left(\frac{AB}{PQ}\right)^2 = 1 \implies AB = PQ$. Similarly, $BC = QR$ and $CA = RP$. By SSS Congruence, $\Delta ABC \cong \Delta PQR$.

4. Solved CBSE Board Examination Problems

Solved Example 1: Finding Missing Area

Problem: Let $\Delta ABC \sim \Delta DEF$ and their areas be respectively $64\text{ cm}^2$ and $121\text{ cm}^2$. If $EF = 15.4\text{ cm}$, find $BC$.

Solution:

1. Since $\Delta ABC \sim \Delta DEF$, by the Areas Theorem: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2$
2. Substitute the given values: $\frac{64}{121} = \left(\frac{BC}{15.4}\right)^2$
3. Take square roots on both sides: $\sqrt{\frac{64}{121}} = \frac{BC}{15.4} \implies \frac{8}{11} = \frac{BC}{15.4}$
4. Cross-multiply: $BC = \frac{8 \times 15.4}{11} = 8 \times 1.4 = 11.2\text{ cm}$
5. Therefore, <u>$BC = 11.2\text{ cm}$</u>.

Solved Example 2: Midpoints Triangle Area Ratio (NCERT Classic)

Problem: $D, E,$ and $F$ are respectively the mid-points of sides $AB, BC,$ and $CA$ of $\Delta ABC$. Find the ratio of the areas of $\Delta DEF$ and $\Delta ABC$.

Solution:

1. In $\Delta ABC$, $D$ is the midpoint of $AB$ and $F$ is the midpoint of $AC$.
2. By the Mid-Point Theorem (Class 9): $DF \parallel BC \quad \text{and} \quad DF = \frac{1}{2}BC$ Similarly, $DE = \frac{1}{2}AC$ and $EF = \frac{1}{2}AB$.
3. Thus, the sides of $\Delta DEF$ and $\Delta ABC$ are proportional: $\frac{DF}{BC} = \frac{DE}{AC} = \frac{EF}{AB} = \frac{1}{2}$
4. By SSS Similarity, $\Delta DEF \sim \Delta CAB$.
5. By the Areas Theorem: $\frac{\text{Area}(\Delta DEF)}{\text{Area}(\Delta ABC)} = \left(\frac{DF}{BC}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
6. Therefore, <u>the ratio of their areas is $1 : 4$</u>.

5. Summary and Examination Tips

Exam Tip: In 1-mark board MCQs: If the sides of two similar triangles are in the ratio $4:9$, the ratio of their areas is $4^2 : 9^2 = 16:81$. If the question gives areas as $16:81$ and asks for sides, the answer is $\sqrt{16} : \sqrt{81} = 4:9$!

Common Mistake: Forgetting to take the square root when finding sides from given areas. Students often write $\frac{BC}{15.4} = \frac{64}{121}$ directly, forgetting that the theorem equates area ratio to the square of the side ratio!