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CAT 2019 Slot 2 — QA Question 18

PyramidEasy

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

Answer & solution

  • 10√2

  • B

    8√3

  • C

    12

  • D

    5√5

Solution

Easy

The apex sits above the centre of the square base. A right triangle is formed by the height, the half-diagonal of the base, and a lateral edge. The lateral edge equals the side of the equilateral faces (2020), so Pythagoras gives the height.

T (apex) O A h 10√2 20
1

Lateral edge and half-diagonal. Each slant face is equilateral with side 2020, so every lateral edge TA=20TA=20. The base is a 20×2020\times20 square; its diagonal is 20220\sqrt{2}, and the centre OO is at half that distance from any corner.

TA=20OA=diagonal2=2022=102\begin{aligned} &TA=20\\ &OA=\frac{\text{diagonal}}{2}=\frac{20\sqrt{2}}{2}=10\sqrt{2} \end{aligned}
2

Pythagoras on △TOA\triangle TOA. The height h=TOh=TO is perpendicular to the base, so TO2+OA2=TA2TO^{2}+OA^{2}=TA^{2}.

h2=TA2−OA2⇒ h2=202−(102)2(from step 1)⇒ h2=400−200=200⇒ h=200=102\begin{aligned} &h^{2}=TA^{2}-OA^{2}\\ &\Rightarrow\ h^{2}=20^{2}-(10\sqrt{2})^{2} \quad\text{(from step 1)}\\ &\Rightarrow\ h^{2}=400-200=200\\ &\Rightarrow\ h=\sqrt{200}=10\sqrt{2} \end{aligned}
h=102 cmh=10\sqrt{2}\ \text{cm}
CAT 2019 Slot 2 QA Q18: The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length — Solution | TheCATExam