Question:
The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is
- $8 \sqrt{3}$
- 12
- $5 \sqrt{5}$
- $10 \sqrt{2}$
Correct Answer: Option: 4
From the diagram, it is obvious that AB is the height of the equilateral triangle and is also the slant height of the pyramid.
Therefore, $AB=\frac{\sqrt{3}}{2}\times side=\frac{\sqrt{3}}{2}\times 20=10\sqrt{3}$
And $AO=\frac{1}{2}\times side=\frac{1}{2}\times 20=10$
Applying Pythagoras theorem in triangle AOB
$O{{B}^{2}}=A{{B}^{2}}-O{{A}^{2}}$
$={{\left( 10\sqrt{3} \right)}^{2}}-{{10}^{2}}$
$=200$
Hence, the height of the pyramid (OB) =$10\sqrt{2}$
CAT Online Course
Also Check: 841+ CAT Quant Questions with Solutions