### Question 12

SSC-CGL 2018 June 4 Shift 2

ΔABC is similar to ΔDEF. The area of ∆ABC is 100 sq. cm and the area of ∆DEF is 49 sq. cm. If ΔABC's altitude is 5 cm, then ΔDEF will have corresponding altitude?

- 4.5 cm
- 6 cm
- 7 cm
- 3.5 cm

### Solution in Short

Areas of two triangles have been given as 100 sq. cm and 49 sq. cm, and the altitude of one of them is given as 5 cm.

REMEMBER: Areas of two similar triangles are in proportion to the squares of the corresponding sides and altitudes.Hence the ratio of altitudes of the given triangles should be same as the square root of the ratios of their areas, i.e., 7/10. Altitude of ΔABC has been given as 5 cm, so the altitude of ΔDEF is 5 x (7/10) = 3.5 answer!

### Solution in Detail

Let base and altitude of ∆ABC be b and 5 cm.

Given ∆ABC ~ ΔDEF

$\displaystyle \therefore $ base, altitude of ΔDEF = kb and 5k.

Ratio of areas is given $\displaystyle \frac{49}{100}$

$\displaystyle \therefore \frac{\frac12 kb \times 5k}{\frac12 b \times 5} = \frac{49}{100}$

$\displaystyle \implies k = \frac{7}{10}$

$\displaystyle \therefore \text{altitude of } Δ \text{DEF} = 5k $

$\displaystyle \text{i.e., }= 5 \times \frac{7}{10} = 3.5 \text{ cm }\:\underline{Ans}$

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