Question : The area of an equilateral triangle is $4 \sqrt{3} \mathrm{~cm}^2$. Find the side (in cm) of the triangle.
Option 1: $2$
Option 2: $4$
Option 3: $\sqrt{3}$
Option 4: $2 \sqrt{3}$
Correct Answer: $4$
Solution : It is known that the area of the equilateral triangle = $\frac{\sqrt{3}}{4}a^2$ Where $a$ is the side of the equilateral triangle. It is given that the area of the equilateral triangle is $4\sqrt{3}$ cm$^2$ So, $\frac{\sqrt{3}}{4}a^2$ = $4\sqrt{3}$ ⇒ $a^2 = 16$ ⇒ $a = 4$ cm Hence, the correct answer is 4. .
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Question : Find the area of triangle whose sides are 10 cm, 12 cm, and 18 cm.
Option 1: $22 \sqrt{2} \mathrm{~cm}^2$
Option 2: $30 \sqrt{2} \mathrm{~cm}^2$
Option 3: $28 \sqrt{2} \mathrm{~cm}^2$
Option 4: $40 \sqrt{2} \mathrm{~cm}^2$
Question : Find the area of a triangle whose length of two sides are 4 cm and 5 cm and the angle between them is 45°.
Option 1: $4 \sqrt{2} \mathrm{~cm}^2$
Option 2: $7 \sqrt{2} \mathrm{~cm}^2$
Option 3: $5 \sqrt{2} \mathrm{~cm}^2$
Option 4: $6 \sqrt{2} \mathrm{~cm}^2$
Question : If each side of an equilateral triangle is 12 cm, then its altitude is equal to:
Option 1: $6 \sqrt{3}\ \text{cm}$
Option 2: $3 \sqrt{6}\ \text{cm}$
Option 3: $6 \sqrt{2}\ \text{cm}$
Option 4: $3 \sqrt{2}\ \text{cm}$
Question : The length of each side of an equilateral triangle is $14 \sqrt{3}$ cm. The area of the incircle (in cm2), is:
Option 1: 450
Option 2: 308
Option 3: 154
Option 4: 77
Question : In a triangle ABC, if $\angle B=90^{\circ}, \angle C=45^{\circ}$ and AC = 4 cm, then the value of BC is:
Option 1: $\sqrt{2} \mathrm{~cm}$
Option 2: $4 \mathrm{~cm}$
Option 3: $2 \sqrt{2} \mathrm{~cm}$
Option 4: $4 \sqrt{2} \mathrm{~cm}$
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