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Ncert exemplar solutions

CHAPTERS

A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Let TW = 15 m be the height of the tower and SW = 24 m be its shadow's length. Also, let PL = x metres be the height of the telephone pole and AL = 16 m be its shadow's length

In ΔTWS and ΔPLA,

∠W=∠L=90°

∠S = ∠A [Since both the figures occur at the same ∴ angles of elevation will be equal]

∴ ΔTWS ~ ΔPLA ( By AA similarity criterion )

\Rightarrow \frac{T W}{P L}=\frac{T S}{P A}=\frac{W S}{L A} \text { ( Corresponding sides of two similar triangles are proportional) }\\ \Rightarrow \frac{15}{x}=\frac{24}{16}\\ \Rightarrow \mathrm{x}=\frac{15 \times 16}{24}=5 \times 2\\ \Rightarrow \mathrm{x}=10 \mathrm{~m}

Hence, the height of the telephone pole is 10 m

Let TW = 15 m be the height of the tower and SW = 24 m be its shadow's length. Also, let PL = x metres be the height of the telephone pole and AL = 16 m be its shadow's length

In ΔTWS and ΔPLA,

∠W=∠L=90°

∠S = ∠A [Since both the figures occur at the same ∴ angles of elevation will be equal]

∴ ΔTWS ~ ΔPLA ( By AA similarity criterion )

\Rightarrow \frac{T W}{P L}=\frac{T S}{P A}=\frac{W S}{L A} \text { ( Corresponding sides of two similar triangles are proportional) }\\ \Rightarrow \frac{15}{x}=\frac{24}{16}\\ \Rightarrow \mathrm{x}=\frac{15 \times 16}{24}=5 \times 2\\ \Rightarrow \mathrm{x}=10 \mathrm{~m}

Hence, the height of the telephone pole is 10 m

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