Question

Here AB $\parallel$ CD. If the altitude of $△$ACD is 6 cm and CD is 8 cm then the area of $△$BCD is:

• A. $14c{m}^2$
• B. $44c{m}^2$
• C. $34c{m}^2$
• D. $24c{m}^2$

Answer 1

The correct option is D $24c{m}^2$

Triangle between the same parallels and sharing same base have equal areas.

So, Area($△$ACD) = Area($△$BCD)

Now, Area($△$ACD) = $\frac{1}{2}\times 8\times 6=24c{m}^2$

$\therefore$ Area($△$BCD) = $24c{m}^2$