Question : AD is the median of $\triangle \mathrm{ABC}$. G is the centroid of $\triangle \mathrm{ABC}$. If AG = 14 cm, then what is the length of AD?
Option 1: 42 cm
Option 2: 28 cm
Option 3: 35 cm
Option 4: 21 cm
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Correct Answer: 21 cm
Solution : AD is the median of triangle ABC. G is the centroid of triangle ABC. If AG = 14 cm So, $\frac{AG}{GD}=\frac{2}{1}$ ⇒ $\frac{14}{GD} = \frac{2}{1}$ ⇒ $GD =7$ cm $\therefore$ AD = AG + GD = 14 + 7 = 21 cm Hence, the correct answer is 21 cm.
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Question : In an equilateral triangle, the circumradius is 14 cm. What is the length of the median in this triangle?
Option 1: $14 \sqrt{3} \mathrm{~cm}$
Option 2: $21 \mathrm{~cm}$
Option 3: $18 \sqrt{3} \mathrm{~cm}$
Option 4: $7 \sqrt{3} \mathrm{~cm}$
Question : $\triangle \mathrm {ABC}$ is similar to $\triangle \mathrm{PQR}$ and $\mathrm{PQ}=10 \mathrm{~cm}$. If the area of $\triangle \mathrm{ABC}$ is $32 \mathrm{~cm}^2$ and the area of $\triangle \mathrm{PQR}$ is $50 \mathrm{~cm}^2$, then the length of $A B$ (in $\mathrm{cm}$ ) is equal to:
Option 1: 10
Option 2: 4
Option 3: 6
Option 4: 8
Question : MX is the median of $\triangle M N O$. Y is the centroid of $\triangle M N O$. If YX = 12 cm, then what is the length of MX?
Option 1: 30 cm
Option 2: 36 cm
Option 3: 28 cm
Option 4: 24 cm
Question : In $\triangle$ABC, D is the midpoint of BC. Length AD is 27 cm. N is a point in AD such that the length of DN is 12 cm. The distance of N from the centroid of ABC is equal to:
Option 1: 3 cm
Option 2: 6 cm
Option 3: 9 cm
Option 4: 15 cm
Question : In the isosceles triangle ABC with BC is the unequal side of the triangle, and line AD is the median drawn from the vertex A to the side BC. If the length AC = 5 cm and the length of the median is 4 cm, then find the length of BC (in (cm).
Option 1: 5
Option 2: 3
Option 3: 4
Option 4: 6
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