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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

## $\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}s=\frac{a+b+c}{2}\phantom{\rule{0ex}{0ex}}=\frac{13+13+10}{2}\phantom{\rule{0ex}{0ex}}=\frac{36}{2}\phantom{\rule{0ex}{0ex}}=18\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\sqrt{18\left(18-13\right)\left(18-13\right)\left(18-10\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{2×3×3×5×5×2×2×2}\phantom{\rule{0ex}{0ex}}=60\mathrm{sq}.\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}×\mathrm{base}×\mathrm{height}\phantom{\rule{0ex}{0ex}}⇒60=\frac{1}{2}×10×\mathrm{height}\phantom{\rule{0ex}{0ex}}⇒\mathrm{height}=\frac{60}{5}\phantom{\rule{0ex}{0ex}}⇒\mathrm{height}=12\mathrm{cm}$ The centroid is located two third of the distance from any vertex of the triangle. $\therefore \mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{vertex}\mathrm{and}\mathrm{the}\mathrm{centroid}=\frac{2}{3}×12=8\mathrm{cm}$ Hence, the distance between the vertex opposite the base and the centroid is 8 cm.

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