NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

The Areas of Similar Triangles Theorem

The core mathematical tool for this exercise is the Theorem on Areas of Similar Triangles. It states:

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

If we have two triangles, $\Delta ABC$ and $\Delta PQR$, such that $\Delta ABC \sim \Delta PQR$, then:

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

Why does this happen? Let's visualize this using geometry. When two triangles are similar, all their linear dimensions (sides, altitudes, medians, and perimeters) scale by the same constant ratio, let's call it $k$. Since the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$, and both the base and height scale by $k$, the area must scale by $k \times k = k^2$. This elegant scaling property is what you will apply to solve every question in Exercise 6.4.

How to Solve Area Ratio Problems Step-by-Step

1. Step 1: Check or Establish Similarity — Before applying the area theorem, verify if the two triangles are similar. If it is not given directly in the problem, use AA, SAS, or SSS similarity criteria to prove $\Delta ABC \sim \Delta PQR$.
2. Step 2: State the Area Theorem Formula — Write down the theorem formula clearly: $\frac{\text{Area}(\Delta 1)}{\text{Area}(\Delta 2)} = \left(\frac{\text{Side}_1}{\text{Side}_2}\right)^2$. Ensure that the corresponding sides are matched correctly based on the similarity ordering.
3. Step 3: Substitute the Known Values — Plug in the given values of areas or sides. For instance, if the ratio of sides is given as $3:5$, substitute this on the right-hand side and square it to get the ratio of areas as $9:25$.
4. Step 4: Solve for the Unknown Variable — Perform algebraic manipulation. If you need to find a side length when areas are given, take the square root of both sides of the equation first to make the calculations simpler.

Common Board Exam Mistakes & Tips

Here are a few critical points to keep in mind to secure full marks in your CBSE Class 10 board exams:

1. Forgetting to Square: The most common mistake students make is equating the ratio of areas directly to the ratio of sides (i.e., writing $\frac{\text{Area 1}}{\text{Area 2}} = \frac{\text{Side 1}}{\text{Side 2}}$). Always remember to square the side ratio!
2. Incorrect Side Matching: Ensure you match corresponding sides accurately. If $\Delta ABC \sim \Delta DEF$, then $AB$ corresponds to $DE$, $BC$ to $EF$, and $AC$ to $DF$. Writing $\frac{AB}{EF}$ will lead to an incorrect answer.
3. Medians and Altitudes: Note that the ratio of the areas is also equal to the square of the ratio of their corresponding altitudes, medians, or angle bisectors. This property is frequently used in proving complex questions.

Practice Questions with Solutions

• Q: Let $\Delta ABC \sim \Delta DEF$ and their areas be $64 \text{ cm}^2$ and $121 \text{ cm}^2$ respectively. If $EF = 15.4 \text{ cm}$, find $BC$. A: Step 1: Write down the given values and similarity relation. Given: $\Delta ABC \sim \Delta DEF$, $\text{Area}(\Delta ABC) = 64 \text{ cm}^2$, $\text{Area}(\Delta DEF) = 121 \text{ cm}^2$, and $EF = 15.4 \text{ cm}$. Step 2: Apply the Area of Similar Triangles Theorem. $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2$ Step 3: Substitute the known values into the equation. $\frac{64}{121} = \left(\frac{BC}{15.4}\right)^2$ Step 4: Take the square root on both sides to simplify. $\sqrt{\frac{64}{121}} = \frac{BC}{15.4}$ $\frac{8}{11} = \frac{BC}{15.4}$ Step 5: Solve for $BC$. $BC = \frac{8 \times 15.4}{11}$ $BC = 8 \times 1.4 = 11.2 \text{ cm}$ Final answer: The length of $BC$ is $11.2 \text{ cm}$.
• Q: Diagonals of a trapezium $ABCD$ with $AB \parallel CD$ intersect each other at the point $O$. If $AB = 2CD$, find the ratio of the areas of triangles $AOB$ and $COD$. A: Step 1: Identify similar triangles. In $\Delta AOB$ and $\Delta COD$: - $\angle AOB = \angle COD$ (Vertically opposite angles) - $\angle OAB = \angle OCD$ (Alternate interior angles, since $AB \parallel CD$) Therefore, by AA similarity criterion, $\Delta AOB \sim \Delta COD$. Step 2: Apply the Areas of Similar Triangles Theorem. $\frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta COD)} = \left(\frac{AB}{CD}\right)^2$ Step 3: Substitute the relation $AB = 2CD$. $\frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta COD)} = \left(\frac{2CD}{CD}\right)^2$ $\frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta COD)} = (2)^2 = 4$ Final answer: The ratio of the areas of triangles $AOB$ and $COD$ is $4:1$.
• Q: $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Find the ratio of the areas of triangles $ABC$ and $BDE$. A: Step 1: Establish similarity. All equilateral triangles are equiangular (each angle is $60^\circ$). Therefore, any two equilateral triangles are similar by AAA similarity criterion. Thus, $\Delta ABC \sim \Delta BDE$. Step 2: Find the relation between corresponding sides. Since $D$ is the mid-point of $BC$, we have $BD = \frac{1}{2} BC$, which means $BC = 2BD$. Step 3: Apply the Area Theorem. $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta BDE)} = \left(\frac{BC}{BD}\right)^2$ Substitute $BC = 2BD$: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta BDE)} = \left(\frac{2BD}{BD}\right)^2 = (2)^2 = 4$ Final answer: The ratio of the areas of triangles $ABC$ and $BDE$ is $4:1$.
• Q: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. A: Step 1: Set up the triangles. Let $\Delta ABC \sim \Delta PQR$. Let $AD$ be the median of $\Delta ABC$ to side $BC$, and $PM$ be the median of $\Delta PQR$ to side $QR$. Step 2: Prove similarity of sub-triangles $\Delta ABD$ and $\Delta PQM$. Since $\Delta ABC \sim \Delta PQR$, we have: $\frac{AB}{PQ} = \frac{BC}{QR}$ Since $D$ and $M$ are midpoints, $BC = 2BD$ and $QR = 2QM$. Substituting these yields: $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$ Also, $\angle B = \angle Q$ (as $\Delta ABC \sim \Delta PQR$). Thus, by SAS similarity criterion, $\Delta ABD \sim \Delta PQM$. Step 3: Relate medians to sides. From the similarity of sub-triangles, we have: $\frac{AB}{PQ} = \frac{AD}{PM}$ Step 4: Use the main Area Theorem. $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ} ight)^2$ Substituting $\frac{AB}{PQ} = \frac{AD}{PM}$: $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AD}{PM}\right)^2 = \frac{AD^2}{PM^2}$ Final answer: Hence proved, the ratio of the areas is equal to the square of the ratio of their corresponding medians.

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