Question

Two triangles have 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 the length of one of the altitude. Then the area of quadrilateral formed by the altitudes and the base PQ is equal to:

• A. $54c{m}^2$
• B. $64c{m}^2$
• C. $34c{m}^2$
• D. $44c{m}^2$

Answer 1

The correct option is B

$64c{m}^2$

$△$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 = $\frac{1}{2}$ × (base) × (altitude)

24 = $\frac{1}{2}$ × (6) × altitude

Therefore, altitude = 8 cm

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

And PQ = AP (given)

So, ABQP is a square with side 8 cm

Therefore, area of square = 8 × 8 = $64c{m}^2$