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Question
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
\[\text{Area of the triangle} = \sqrt{s\left( s - a \right)\left( s - b \right)\left( s - c \right)}\]
\[s = \frac{a + b + c}{2}\]
\[ = \frac{13 + 13 + 10}{2}\]
\[ = \frac{36}{2}\]
= 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
\[\text{Also}, \]
\[\text{Area of the triangle} = \frac{1}{2} \times base \times height\]
\[ \Rightarrow 60 = \frac{1}{2} \times 10 \times \text{height}\]
\[ \Rightarrow \text{height} = \frac{60}{5}\]
= 12 cm
The centroid is located two third of the distance from any vertex of the triangle.
\[\therefore \text{Distance between the vertex and the centroid} = \frac{2}{3} \times 12 = 8 cm\]
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