CBSE 2025 · Region 3 · Set 1 · Q32 · 5 marks
The perimeter of an isosceles triangle is $\displaystyle 32$ cm. If each equal side is $\displaystyle \frac{5}{6}$ th of the base, find the area of the triangle.
Marking-scheme solution
Let each equal side of triangle be x and base be y
ATQ, \(\displaystyle x+x+y=32\)
\(\displaystyle 2 \mathrm{x}+\mathrm{y}=32\)
Also, \(\displaystyle \mathrm{x}=\frac{5}{6} \mathrm{y}\)
On solving these equations, we get \(\displaystyle \mathrm{x}=10\) and \(\displaystyle \mathrm{y}=12\)
∴ sides of the traingle are $\displaystyle 10$ cm, $\displaystyle 10$ cm, $\displaystyle 12$ cm
Semi - perimeter of the triangle \(\displaystyle =16 \mathrm{~cm}\)
Area of the triangle \(\displaystyle =\sqrt{16 \times 6 \times 6 \times 4}\)
\(\displaystyle =48 \mathrm{~cm}^{2}\)
Let first term = a and common difference = d
ATQ, \(\displaystyle (\mathrm{a}+2 \mathrm{~d})+(\mathrm{a}+6 \mathrm{~d})=6\)
\(\displaystyle \mathrm{a}+4 \mathrm{~d}=3\)
\(\displaystyle \mathrm{a}=3-4 \mathrm{~d}\)
Also, \(\displaystyle (\mathrm{a}+2 \mathrm{~d})(\mathrm{a}+6 \mathrm{~d})=8\)
\(\displaystyle (3-4 d+2 d)(3-4 d+6 d)=8\)
\(\displaystyle 9-4 \mathrm{~d}^{2}=8\)
\(\displaystyle \mathrm{d}= \pm \frac{1}{2}\)
When \(\displaystyle \mathrm{d}=\frac{1}{2} \Rightarrow \mathrm{a}=1\)
\(\displaystyle \mathrm{S}_{16}=\frac{16}{2}\left[2 \times 1+15 \times \frac{1}{2}\right]\)
\(\displaystyle =76\)
When \(\displaystyle \mathrm{d}=-\frac{1}{2} \Rightarrow \mathrm{a}=5\)
\(\displaystyle \mathrm{S}_{16}=\frac{16}{2}\left[2 \times 5+15 \times\left(-\frac{1}{2}\right)\right]\)
= $\displaystyle 20$
The ages of the participants form the following AP
\(\displaystyle 8,8 \frac{1}{3}, 8 \frac{2}{3}, 9, \cdots\)
where first term \(\displaystyle =8\) and common difference \(\displaystyle =\frac{1}{3}\)
Let the number of participants be n
\(\displaystyle \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \times 8+(\mathrm{n}-1) \frac{1}{3}\right]=168\)
\(\displaystyle \mathrm{n}^{2}+47 \mathrm{n}-1008=0\)
⇒ n = $\displaystyle 16$
∴ the age of the eldest participant \(\displaystyle =8+15 \times \frac{1}{3}=13\) years
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.