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In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.

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#### Solution

\[s = \frac{a + b + c}{2}\]

\[ = \frac{13 + 13 + 10}{2}\]

\[ = \frac{36}{2}\]

\[ = 18 cm\]

\[\text{Area of the triangle} = \sqrt{18\left( 18 - 13 \right)\left( 18 - 13 \right)\left( 18 - 10 \right)}\]

\[ = \sqrt{2 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2}\]

\[ = 60 sq . cm\]

\[\text{Also}, \]

\[\text{Area of the triangle} = \frac{1}{2} \times base \times height\]

\[ \Rightarrow 60 = \frac{1}{2} \times 10 \times \text{height}\]

\[ \Rightarrow \text{height} = \frac{60}{5}\]

\[ \Rightarrow \text{height} = 12 cm\]

The centroid is located two third of the distance from any vertex of the triangle.

Hence, the distance between the vertex opposite the base and the centroid is 8 cm.

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