Question : The lengths of the three medians of a triangle are $9\;\mathrm{cm}$, $12\;\mathrm{cm}$, and $15\;\mathrm{cm}$. The area (in $\mathrm{cm^2}$) of the triangle is:
Option 1: $24$
Option 2: $72$
Option 3: $48$
Option 4: $144$
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Correct Answer: $72$
Solution : The area of a triangle in terms of its medians, Area $=\frac{4}{3} \sqrt{s(s - m_1)(s - m_2)(s - m_3)}$ where $m_1$, $m_2$ and $m_3$ are the lengths of the medians, and $s$ is the semi-perimeter of the medians, which is $s = \frac{m_1 + m_2 + m_3}{2}$ The lengths of the medians are $9\;\mathrm{cm}$, $12\;\mathrm{cm}$ and $15\;\mathrm{cm}$. $s = \frac{9 + 12 + 15}{2} =18\;\mathrm{cm}$ Area $=\frac{4}{3} \sqrt{18(18 - 9)(18 - 12)(18 - 15)}=\frac{4}{3} ×9×6= 72 \text{ cm}^2$ Hence, the correct answer is $72$.
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Question : If $\triangle ABC \sim \triangle QRP, \frac{\operatorname{area}(\triangle A B C)}{\operatorname{area}(\triangle Q R P)}=\frac{9}{4}, A B=18 \mathrm{~cm}, \mathrm{BC}=15 \mathrm{~cm}$, then the length of $\mathrm{PR}$ is:
Question : In the given figure, in $\triangle \operatorname{STU}$, $\mathrm{ST }= 8$ cm, $\mathrm{TU} = 9$ cm and $\mathrm{SU} = 12$ cm. $\mathrm{QU }= 24$ cm, $\mathrm{SR} = 32$ cm and $\mathrm{PT }= 27$ cm. What is the ratio of the area of $\triangle \operatorname{PUQ}$ and the
Question : The height of an equilateral triangle is $9 \sqrt{3} \mathrm{~cm}$. What is the area of this equilateral triangle?
Question : Three medians AD, BE, and CF of $\triangle ABC$ intersect at G. The area of $\triangle ABC$ is $36\text{ cm}^2$. Then the area of $\triangle CGE$ is:
Question : In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle is:
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