Question

Abhay walks in a straight line away from an electric lamp post of height 8 metres. When he is at a distance of 8 metres from the foot of the pole, his shadow is 2 metres in length.

1. Find Abhay's height (in cm).
2. Find the length of Abhay's shadow (in metres) when he is at a distance of 12 metres from the foot of the pole.
3. When Abhay is 12 metres away from the foot of the pole, he meets his friend Sachin who then moves towards the pole at the rate of 5 cm/s. If after 2 minutes the length of Sachin's shadow is 1.2 metres, what is his height?
4. Name the similarity criterion used in solving the above problem.

Answer 1

i. Let AB be the pole, CD Abhay and DE his shadow. In triangles ABE and CDE, angle ABE equals angle CDE (both 90 degrees) and angle AEB equals angle CED (common), so triangle ABE is similar to triangle CDE by AA-similarity.

BE = 8 + 2 = 10 m.

\[\frac{AB}{CD}=\frac{BE}{DE} \Rightarrow \frac{8}{CD}=\frac{10}{2}\] 

\[CD=\frac{8\times 2}{10}=1.6\ \text{m}=160\ \text{cm}\]

Abhay's height is 160 cm. 

ii. Let the shadow length be x metres.

BE = 12 + x and CD = 1.6 m. 

\[\frac{AB}{CD}=\frac{BE}{DE} \Rightarrow \frac{8}{1.6}=\frac{12+x}{x}\]

\[5x=12+x \Rightarrow 4x=12 \Rightarrow x=3\]

The shadow is 3 m long. 

iii. Distance covered by Sachin in 2 minutes (120 s): 

\[5\times 120=600\ \text{cm}=6\ \text{m}\]

So his distance from the pole is 12 – 6 = 6 m. Let his height be h metres. Then BQ = 6 m, QR = 1.2 m and BR = 7.2 m. Triangles ABR and PQR are similar by AA-similarity. 

\[\frac{AB}{PQ}=\frac{BR}{QR} \Rightarrow \frac{8}{h}=\frac{7.2}{1.2}=6\]

\[h=\frac{8}{6}=\frac{4}{3}=1\frac{1}{3}\ \text{m}\] 

Sachin's height is `1 1/3` m. 

iv. The criterion used is AA-similarity (two pairs of corresponding angles are equal).