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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 4 · Set 1 · Q31
Rectangle ABCD circumscribes the circle of radius 10 cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.
Rectangle ABCD circumscribes the circle of radius $\displaystyle 10$ cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.
Marking-scheme solution
\[\mathrm{AP}=\mathrm{AS}
\]
\(\displaystyle \mathrm{BP}=\mathrm{BQ}\)
\(\displaystyle \mathrm{CR}=\mathrm{CQ}\)
DR = DS
Adding the above four equations,
\[\begin{aligned}
& \mathrm{AP}+\mathrm{BP}+\mathrm{CR}+\mathrm{DR}=\mathrm{AS}+\mathrm{BQ}+\mathrm{CQ}+\mathrm{DS} \\
& \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{CB} \quad \text{--- (i)}
\end{aligned}
\]
Since ABCD is a rectangle
\(\displaystyle \therefore \mathrm{AB}=\mathrm{CD}\) and \(\displaystyle \mathrm{BC}=\mathrm{AD}\)
⇒ from (i), \(\displaystyle 2 \mathrm{AB}=2 \mathrm{AD}\) or \(\displaystyle \mathrm{AB}=\mathrm{AD}\)
Hence ABCD is a square
Clearly side of square = diameter of circle \(\displaystyle =20 \mathrm{~cm}\)
∴ Perimeter of square \(\displaystyle =4 \times 20 \mathrm{~cm}=80 \mathrm{~cm}\)
Let the length of shortest side be \(\displaystyle x\) m
∴ length of longest side \(\displaystyle =(x+4) \mathrm{m}\) and length of third side \(\displaystyle =(x+2) \mathrm{m}\)
Now, \(\displaystyle (x+4)^{2}=x^{2}+(x+2)^{2}\)
\(\displaystyle \Rightarrow x^{2}-4 x-12=0\)
\(\displaystyle \Rightarrow(x-6)(x+2)=0\)
\(\displaystyle \Rightarrow x=6\)
∴ 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 = $\displaystyle 0$
\(\displaystyle \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}\)
\(\displaystyle \Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}\)
Simplifying, we get \(\displaystyle 2 x^{2}-19 x+42=0\)
\(\displaystyle \Rightarrow(x-6)(2 x-7)=0\)
\(\displaystyle \Rightarrow x=6\) or \(\displaystyle x=\frac{7}{2}\)
As, \(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5}\)
\[\begin{aligned}
& \Rightarrow \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\
& \Rightarrow \angle \mathrm{C}=\angle \mathrm{R}
\end{aligned}
\]
(i) In \(\displaystyle \triangle \mathrm{ADC}\) and \(\displaystyle \triangle \mathrm{PSR}\),
\[\begin{aligned}
& \angle \mathrm{ADC}=\angle \mathrm{PSR} \\
& \text { and } \angle \mathrm{C}=\angle \mathrm{R} \\
& \therefore \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR} \\
& \text { (ii) } \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5} \\
& \Rightarrow \frac{4}{\mathrm{PS}}=\frac{3}{5} \\
& \Rightarrow \mathrm{PS}=\frac{20}{3} \mathrm{~cm}
\end{aligned}
\]
\[\text { (iii) } \begin{aligned}
\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{PQR})} & =\frac{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{AD}}{\dfrac{1}{2} \times \mathrm{QR} \times \mathrm{PS}} \\
& =\frac{3}{5} \times \frac{3}{5}=\frac{9}{25}
\end{aligned}
\]
\[\therefore \operatorname{ar}(\triangle \mathrm{ABC}): \operatorname{ar}(\triangle \mathrm{PQR})=9: 25
\]
Join AF intersecting line \(\displaystyle m\) at G In \(\displaystyle \triangle \mathrm{ACF}, \mathrm{BG} \| \mathrm{CF}\)
\[\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AG}}{\mathrm{GF}}
\]
In \(\displaystyle \Delta \mathrm{FDA}, \mathrm{GE} \| \mathrm{AD}\)
\[\Rightarrow \frac{\mathrm{EF}}{\mathrm{DE}}=\frac{\mathrm{GF}}{\mathrm{AG}} \text { or } \frac{\mathrm{DE}}{\mathrm{EF}}=\frac{\mathrm{AG}}{\mathrm{GF}}
\]
From, (i) and (ii), we get \(\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}\)
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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.