Question : The area of the floor of a cubical room is 64 m2. The length of the longest rod that can be kept in the room is:
Option 1: $16 \sqrt{3} \mathrm{~m}$
Option 2: $4 \sqrt{3} \mathrm{~m}$
Option 3: $12 \sqrt{3} \mathrm{~m}$
Option 4: $8 \sqrt{3} \mathrm{~m}$
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Correct Answer: $8 \sqrt{3} \mathrm{~m}$
Solution : Given: The area of the floor of a cubical room is 64 m2. Let the sides of the room be $a$. So, $a^2 = 64$ ⇒ $a = 8\ \mathrm{m}$ The length of the longest rod = the diagonal of the cubical room = $a\sqrt3$ The length of the longest rod = $8 \times\sqrt3=8\sqrt3\ \mathrm{m}$ Hence, the correct answer is $8\sqrt3\ \mathrm{m}$.
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Question : The breadth ‘b’ of a room is twice its height and half of its length. Find the length of the longest diagonal of the room.
Option 1: $\frac{\sqrt{20} \mathrm{~b}}{2}$
Option 2: $\frac{\mathrm{b}}{2}$
Option 3: $\frac{\sqrt{21} \mathrm{~b}}{2}$
Option 4: $\frac{\sqrt{19} \mathrm{~b}}{2}$
Question : The length of each side of a triangle is 12 cm. What is the length of the circumradius of the triangle?
Option 1: $8 \sqrt{3} \mathrm{~cm}$
Option 2: $2 \sqrt{3} \mathrm{~cm}$
Option 3: $6 \sqrt{3} \mathrm{~cm}$
Option 4: $4 \sqrt{3} \mathrm{~cm}$
Question : Find the area of an equilateral triangle whose sides are 16 cm.
Option 1: $60\sqrt{3}\mathrm{~cm}^2$
Option 2: $62\sqrt{3} \mathrm{~cm}^2$
Option 3: $64\sqrt{3} \mathrm{~cm}^2$
Option 4: $66\sqrt{3} \mathrm{~cm}^2$
Question : If the side of an equilateral triangle is 16 cm, then what is its area?
Option 1: $81 \sqrt{3} \mathrm{~cm}^2$
Option 2: $48 \sqrt{3} \mathrm{~cm}^2$
Option 3: $32 \sqrt{3} \mathrm{~cm}^2$
Option 4: $64 \sqrt{3} \mathrm{~cm}^2$
Question : The perimeter of an equilateral triangle is 48 cm. Find its area (in cm2).
Option 1: $81\sqrt{3}$
Option 2: $8\sqrt{3}$
Option 3: $25\sqrt{3}$
Option 4: $64\sqrt{3}$
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