Mathematics — Standard (041) — CBSE Class 10 board question

Step 1: Find the sides of the triangle.

Let the base = $b$ cm.
Each equal side = $\dfrac{6}{8} \times b = \dfrac{3b}{4}$ cm.

Perimeter = base + 2 × equal side
$$b + 2 \times \frac{3b}{4} = 32$$
$$b + \frac{3b}{2} = 32$$
$$\frac{5b}{2} = 32$$
$$b = \frac{64}{5} = 12.8 \text{ cm}$$

Each equal side $= \dfrac{3}{4} \times 12.8 = 9.6$ cm

Step 2: Find the height of the triangle.

Draw perpendicular from apex to base; it bisects the base (isosceles triangle).
Half base $= \dfrac{12.8}{2} = 6.4$ cm

$$h = \sqrt{(9.6)^2 - (6.4)^2} = \sqrt{92.16 - 40.96} = \sqrt{51.2}$$

$$h = \sqrt{51.2} = \sqrt{\frac{256}{5}} = \frac{16}{\sqrt{5}} = \frac{16\sqrt{5}}{5} \text{ cm}$$

Step 3: Find the area.

$$\text{Area} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 12.8 \times \frac{16\sqrt{5}}{5}$$

$$= \frac{1}{2} \times \frac{64}{5} \times \frac{16\sqrt{5}}{5} = \frac{1}{2} \times \frac{1024\sqrt{5}}{25} = \frac{512\sqrt{5}}{25}$$

$$\boxed{\text{Area} = \frac{512\sqrt{5}}{25} \approx 45.95 \text{ cm}^2}$$

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• The key steps examiners look for: setting up the equation using perimeter, solving for the base, finding the height using Pythagoras (since the altitude of an isosceles triangle bisects the base), then applying Area = ½ × base × height.
• Note: $\frac{6}{8} = \frac{3}{4}$ — simplify early to avoid fraction errors.
• Leave the answer in surd form $\frac{512\sqrt{5}}{25}$ or as a decimal (~45.95 cm²); both are acceptable. Show both for full marks.
• This is a pure mensuration problem; the source passages are from the Similarity chapter but the calculation method is standard Heron's formula or base-height method from earlier classes.