Question

Two triangles ALM and BLM have same base LM = 6cm. These two triangles have an equal area of $24c{m}^2$. The distance between their altitudes is PQ. If PQ is equal to the length of one of the altitudes then the area of the quadrilateral formed by the altitudes and the base PQ is equal to:

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

Answer 1

The correct option is D $64c{m}^2$

$△$ALM and $△$BLM are the two triangles on base LM.
Since they have equal areas and same base, they are between the same parallels.
i.e, AP || PQ

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

Area of triangle = $\frac{1}{2}$ × (base) × (altitude)

24 = $\frac{1}{2}$ × (6) ×AP

AP = 8 cm

i.e, length of side of the square = 8 cm

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