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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 4 · Set 1 · Q22
(A) A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at the speed of 2.5 m/sec. Find the length of his shadow after…
(A)
A $\displaystyle 1.5$ m tall boy is walking away from the base of a lamp post which is $\displaystyle 12$ m high, at the speed of $\displaystyle 2.5$ m/sec. Find the length of his shadow after $\displaystyle 3$ seconds.
OR
(B)
In parallelogram ABCD , side AD is produced to a point E and BE intersects CD at F . Prove that $\displaystyle \triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$
Marking-scheme solution
Let AB be the lamp post and CD be the boy $\displaystyle 1.5$ m tall.
Let the length of shadow be x m
Speed of boy \(\displaystyle =2.5 \mathrm{~m} / \mathrm{sec}\)
\(\displaystyle \therefore\) Distance covered in $\displaystyle 3$ seconds \(\displaystyle =7.5 \mathrm{~m}\)
Now, \(\displaystyle \Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\)
\(\displaystyle \Rightarrow \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BE}}\)
\(\displaystyle \Rightarrow \frac{1.5}{12}=\frac{\mathrm{x}}{7.5+\mathrm{x}}\)
Solving, we get \(\displaystyle \mathrm{x}=\frac{15}{14}\) or $\displaystyle 1.07$ approx.
Hence length of shadow is $\displaystyle 1.07$ m
In \(\displaystyle \Delta \mathrm{ABE}\) and \(\displaystyle \Delta \mathrm{CFB}\),
\(\displaystyle \angle \mathrm{AEB}=\angle \mathrm{CBF}\)
\(\displaystyle \angle \mathrm{A}=\angle \mathrm{C}\)
\(\displaystyle \therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}\)
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