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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. $10\sqrt{2}$
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We can draw the diagram, using the given information in the question.

The base of a regular pyramid is square. That means $\square\text{WXYZ}$ will be the square.

The four sides of a regular pyramid is an equilateral triangle. That means $\triangle \text{AXW}, \triangle \text{AWZ}, \triangle \text{AXY}$ and $\triangle \text{AYZ}$ are equilateral triangle, and side of each equilateral triangle is $20\; \text{cm}.$



Now, consider the  $\triangle \text{AYZ}.$ 

Let $`h\text{’ cm}$ be the height  of a triangle.



In $\triangle \text{ABY},$ using the Pythagoras’ theorem,

$\boxed{ \text{(AY)}^{2} = \text{(AB)}^{2} + \text{(YB)}^{2}}$

$ \Rightarrow (20)^{2} = (h)^{2} + (10)^{2}$

$ \Rightarrow 400 = h^{2} + 100$

$ \Rightarrow h^{2} = 300$

$ \Rightarrow h = \sqrt{300}$

$ \Rightarrow h = \sqrt{ 3 \times 100}$

$ \Rightarrow h = \text{AB} = 10 \sqrt{3}\; \text{cm}$

Here, $ \text{OB} = \frac{\text{side of the square}}{2} = \frac{\text{XY}}{2} = \frac{20}{2} = 10\; \text{cm}.$

In $ \triangle \text{AOB},$ using the Pythagoras’ theorem,

$ \text{(AB)}^{2} = \text{(OA)}^{2} + \text{(OB)}^{2}$

$ \Rightarrow (10 \sqrt{3})^{2} = \text{(OA)}^{2} + (10)^{2}$

$ \Rightarrow 300-100 = \text{(OA)}^{2}$

$ \Rightarrow \text{(OA)}^{2} = 200$

$ \Rightarrow \text{OA} = \sqrt{200} = \sqrt{2 \times 100} $

$ \Rightarrow \boxed {\text{OA} = 10 \sqrt{2}\; \text{cm}}$

$\therefore$ The vertical height of the pyramid $ = 10 \sqrt{2}\; \text{cm}.$

Correct Answer $: \text{D}$

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