Question

ABCD is a parallelogram in which CD is produced to P. DC = 6 cm and height BQ = 4 cm. Find the area of $△$ APB

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Answer 1

Theorem used:
Area of triangle is half the area of parallelogram, if both are on same base and between the same parallels.

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From the given figure:
$AB\parallel CD$ ($\because$ ABCD is a parallelogram)
AB = CD = 6 cm (given)
Altitude BQ = 4 cm (given)
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$△ABP$ and parallelogram ABCD are on same base and between the same parallels.

$\therefore \text{Area of}\Delta APB=\frac{1}{2}\times \text{Area of}ABCD$

$\text{Area of}ABCD=\text{Base}\times \text{Altitude}$

$=6\times 4=24c{m}^2$
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$\text{Area of}\Delta APB=\frac{1}{2}\times \text{Area of}ABCD$

$=\frac{1}{2}\times 24$

$\text{Area of}\Delta APB=12c{m}^2$
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