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Mathematics · 2025 · 5 marks
CBSE 2025 · Region 4 · Set 1 · Q32
(A) The sides of a right triangle are such that the longest side is 4 m more than the shortest side and the third side is 2 m less than the longest…
The sides of a right triangle are such that the longest side is $\displaystyle 4$ m more than the shortest side and the third side is $\displaystyle 2$ m less than the longest side. Find the length of each side of the triangle. Also, find the difference between the numerical values of the area and the perimeter of the given triangle.
Express the equation $\displaystyle \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} ;(x \neq 3,5)$ as a quadratic equation in standard form. Hence, find the roots of the equation so formed.
Marking-scheme solution
Let the length of shortest side be \(\displaystyle x\) m
\(\displaystyle \therefore\) length of longest side \(\displaystyle =(x+4) \mathrm{m}\)
and length of third side \(\displaystyle =(x+2) \mathrm{m}\)
\[\begin{aligned}
& \text { Now, }(x+4)^{2}=x^{2}+(x+2)^{2} \\
& \Rightarrow x^{2}-4 x-12=0 \\
& \Rightarrow(x-6)(x+2)=0 \\
& \Rightarrow x=6
\end{aligned}
\]
\(\displaystyle \therefore\) sides are $\displaystyle 6$ m, $\displaystyle 8$ m and $\displaystyle 10$ m
Area \(\displaystyle =\frac{1}{2} \times 6 \times 8=24 \mathrm{~m}^{2}\)
Perimeter \(\displaystyle =6+8+10=24 \mathrm{~m}\)
Difference \(\displaystyle =0\)
\[\begin{aligned}
& \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \\
& \Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}
\end{aligned}
\]
Simplifying, we get \(\displaystyle 2 x^{2}-19 x+42=0\)
\[\begin{aligned}
& \Rightarrow(x-6)(2 x-7)=0 \\
& \Rightarrow x=6 \text { or } x=\frac{7}{2}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.