CUET UG 2025 — General Test Geometry

2025mcqmedium

The height of an equilateral triangle is $5\sqrt{3}$ cm, then its area is:

Official previous-year question

Held on 22 May 2025 · Verified 13 Jul 2026.

Options

A
$100\sqrt{3}$ $cm^2$
B
$75\sqrt{3}$ $cm^2$
C
$50\sqrt{3}$ $cm^2$
D
$25\sqrt{3}$ $cm^2$

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Solution

For an equilateral triangle with side length $a$ :

Height: $h = \frac{\sqrt{3}}{2} \times a$

Area: $Area = \frac{\sqrt{3}}{4} \times a^2$

Given height $h = 5\sqrt{3}$ cm.

Using the height formula:

$5\sqrt{3} = \frac{\sqrt{3}}{2} \times a$

$a = \frac{5\sqrt{3} \times 2}{\sqrt{3}}$

$a = \frac{10\sqrt{3}}{\sqrt{3}}$

$a = 10$ cm

Using the area formula with $a = 10$ cm:

$Area = \frac{\sqrt{3}}{4} \times a^2$

$Area = \frac{\sqrt{3}}{4} \times (10)^2$

$Area = \frac{\sqrt{3}}{4} \times 100$

$Area = 25\sqrt{3}$ cm²

Therefore, the area of the equilateral triangle is $25\sqrt{3}$ cm².

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Solution

For an equilateral triangle with side length $a$ :

Height: $h = \frac{\sqrt{3}}{2} \times a$

Area: $Area = \frac{\sqrt{3}}{4} \times a^2$

Given height $h = 5\sqrt{3}$ cm.

Using the height formula:

$5\sqrt{3} = \frac{\sqrt{3}}{2} \times a$

$a = \frac{5\sqrt{3} \times 2}{\sqrt{3}}$

$a = \frac{10\sqrt{3}}{\sqrt{3}}$

$a = 10$ cm

Using the area formula with $a = 10$ cm:

$Area = \frac{\sqrt{3}}{4} \times a^2$

$Area = \frac{\sqrt{3}}{4} \times (10)^2$

$Area = \frac{\sqrt{3}}{4} \times 100$

$Area = 25\sqrt{3}$ cm²

Therefore, the area of the equilateral triangle is $25\sqrt{3}$ cm².

CUET UG 2025 — General Test Geometry

Options

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Solution

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CUET UG 2025 — General Test Geometry

Options

Did you get this right?

Solution

Your note

CUET UG Geometry in other years

More CUET UG Geometry questions

Other General Test topics for CUET UG

General Knowledge

Logical Reasoning

Arithmetic

Quantitative Reasoning

Current Affairs

Keep practicing Geometry