Question : The base of a right pyramid is a square of side $8 \sqrt{2}$ cm and each of its slant edges is of length 10 cm. What is the volume (in cm$^3$) of the pyramid?
Option 1: $256$
Option 2: $224$
Option 3: $426 \frac{2}{3}$
Option 4: $96 \sqrt{2}$
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Correct Answer: $256$
Solution :
Given: The base of a right pyramid is a square of side = $8\sqrt{2}$ cm and slant edge = 10 cm.
DP = Half of the diagonal of the square
⇒ DP = $\frac{1}{2} \times [\sqrt 2\times 8\sqrt 2] = 8$ cm
Now, In a right-angled triangle OPD
$OP^2= OD^2- DP^2$ [since $\angle OPD = 90^\circ$]
⇒ $OP^2= 10^2- 8^2= 6^2$
⇒ $OP = 6$
So, the height of the pyramid = 6 cm
Area of square base = $a^2=(8\sqrt 2)^2=128$ cm
Now, the volume of the pyramid = $\frac{1}{3} \times \text{area of base}\times 6$
= $\frac{1}{3} \times 128\times 6$
= $256$ cm$^3$
Hence, the correct answer is $256$.
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