Question : In a trapezium ABCD, AB and DC are parallel sides and $\angle ADC=90^\circ$. If AB = 15 cm, CD = 40 cm and diagonal AC = 41 cm, then the area of the trapezium ABCD is:
Option 1: 245 cm2
Option 2: 240 cm2
Option 3: 247.5 cm2
Option 4: 250 cm2
Correct Answer: 247.5 cm 2
Solution :
In trapezium ABCD, $\angle ADC = 90^\circ$, AB = 15 cm, DC = 40 cm and AC = 41 cm.
In $\Delta ADC$,
AC
2
= AD
2
+ DC
2
⇒ 41
2
= AD
2
+ 40
2
⇒ 1681= AD
2
+ 1600
⇒ AD = 9 cm
The area of the trapezium = $\frac{1}{2}$ × (40 + 15) × (9) = 247.5 cm
2
Hence, the correct answer is 247.5 cm
2
.
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