Question : The sides of similar triangle $\triangle ABC$ and $\triangle DEF$ are in the ratio of $\frac{\sqrt{3}}{\sqrt{5}}$. If the area of $\triangle ABC$ is $90 \text{ cm}^2$, then the area of $\triangle DFF\left(\right.$ in $\left.\text{cm}^2\right)$ is:
Option 1: 150
Option 2: 152
Option 3: 154
Option 4: 156
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Correct Answer: 150
Solution : Ratio of their sides = $\frac{\sqrt3}{\sqrt5}$ $\therefore$ Ratio of their areas = $(\frac{\sqrt3}{\sqrt5})^2 = \frac{3}{5}$ Given the area of $\triangle ABC$ = $90\text{ cm}^2$ $\therefore$ Area of $\triangle DEF=90\times \frac{5}{3} = 150\text{ cm}^2$ Hence, the correct answer is 150.
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Question : The sides of a triangle are 9 cm, 6 cm, and 5 cm. What is the value of the circumradius of this triangle?
Option 1: $\frac{9 \sqrt{2}}{2} \mathrm{~cm}$
Option 2: $\frac{9 \sqrt{3}}{5} \mathrm{~cm}$
Option 3: $\frac{9 \sqrt{3}}{4} \mathrm{~cm}$
Option 4: $\frac{27 \sqrt{2}}{8} \mathrm{~cm}$
Question : $ABC$ is a right-angled triangle, right-angled at B and $\angle A=60°$ and $AB=20$ cm, then the ratio of sides $BC$ and $CA$ is:
Option 1: $\sqrt{3}:1$
Option 2:
$1:\sqrt{3}$
Option 3:
$\sqrt{3}:\sqrt{2}$
Option 4:
$\sqrt{3}:2$
Question : The sides of a triangle are 20 cm, 21 cm, and 29 cm. The area of the triangle formed by joining the midpoints of the sides of the triangle will be:
Option 1: $67 \frac{2}{3}$ cm2
Option 2: $52 \frac{1}{2}$ cm2
Option 3: $47 \frac{1}{2}$ cm2
Option 4: $58 \frac{1}{3}$ cm2
Question : If $\triangle ABC$~$\triangle PQR$, the ratio of perimeter of $\triangle ABC$ to perimeter of $\triangle PQR$ is 36 : 23 and QR = 3.8 cm, then the length of BC is:
Option 1: $4 \frac{103}{121} cm$
Option 2: $3 \frac{109}{121} cm$
Option 3: $5 \frac{109}{115} cm$
Option 4: $3 \frac{107}{115} cm$
Question : Find the area of an equilateral triangle whose sides are 12 cm.
Option 1: $38 \sqrt{3} \mathrm{~cm}^2$
Option 2: $35 \sqrt{3} \mathrm{~cm}^2$
Option 3: $34 \sqrt{3} \mathrm{~cm}^2$
Option 4: $36 \sqrt{3} \mathrm{~cm}^2$
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