(solved)Question 12 SSC-CGL 2018 June 4 Shift 2

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?

1. 4.5 cm
2. 6 cm
3. 7 cm
4. 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.

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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