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# CAT 2019 Quant Question with Solution 62

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

1. $8 \sqrt{3}$
2. 12
3. $5 \sqrt{5}$
4. $10 \sqrt{2}$

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) =$10\sqrt{2}$

Also Check: 841+ CAT Quant Questions with Solutions

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