Question : Parallel sides of a trapezium are $26\;\mathrm{cm}$ and $40\;\mathrm{cm}$ and the area is $792\;\mathrm{cm^2}$. What is the value of the distance (in $\mathrm{cm}$) between parallel sides?
Option 1: $24\;\mathrm{cm}$
Option 2: $48\;\mathrm{cm}$
Option 3: $12\;\mathrm{cm}$
Option 4: $36\;\mathrm{cm}$
Correct Answer: $24\;\mathrm{cm}$
Solution :
Given that the parallel sides are $26\;\mathrm{cm}$ and $40\;\mathrm{cm}$, and the area is $792\;\mathrm{cm^2}$.
Let $d$ and $s$ as the distance between parallel sides and the sum of parallel sides.
The sum of parallel sides $=26+40 =66\;\mathrm{cm}$
The area of a trapezium $=\frac{1}{2} \times s\times d$
⇒ $792 = \frac{1}{2} \times 66 \times d$
⇒ $d = \frac{792 \times 2}{66} = 24\;\mathrm{cm}$
Hence, the correct answer is $24\;\mathrm{cm}$.
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