Question

Two triangles have an equal area of $24c{m}^2$ and stand on the same base of 6 cm. If the distance PQ (on base) between their altitudes is equal to thrice the length of one of the altitude. Then the area of the quadrilateral formed by the altitudes and the base PQ is equal to:

• A. $134c{m}^2$
• B. $144c{m}^2$
• C. $154c{m}^2$
• D. $192c{m}^2$

Answer 1

The correct option is D

$192c{m}^2$

$△ALM$ and $△BLM$ are the two triangles on base LM. Since they have equal areas and same base, they are between the same parallels. Therefore, the length of their altitudes should also be equal.

Area of triangle = $\frac{1}{2}\times \left(base\right)\times \left(altitude\right)$

$24=\frac{1}{2}\times \left(6\right)\times altitude$

Therefore, altitude = 8 cm

We know that triangle with equal area and with common base have equal altitude. So, AP = BQ.

And PQ = 3AP (given)

So, ABQP is a rectangle with sides 8 cm and 24 cm.

Therefore, area of rectangle = 8 × 24 = $192c{m}^2$