Question

In an isoscels triangle, length of the congruent sides is 3 cm and its base is 4 cm. Find the distance between the vertex opposite the base and the centroid$\times 1000$.

Answer 1

$Areaoftriangle=\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{a+b+c}{2}$
$Areaoftriangle=\frac{1}{2}\times base\times height$
$Given$
$sidesoftriangle=3,3,4$
$s=\frac{3+3+4}{2}=5$
$Area=\sqrt{5(5-3)(5-3)(5-2)}$
$=\sqrt{5\left(2\right)\left(2\right)\left(3\right)}$
$=2\sqrt{15}$
$Areaoftriangle=\frac{1}{2}\times base\times height$
$2\sqrt{15}=\frac{1}{2}\times 4\times height$
$height=\sqrt{15}$
$The\text{centrod is located 2/3 of the distance from any vertx of the triangle}$
$\text{Hence distance between the vertex opposit the base and the centroid}$
$\frac{2}{3}\times \sqrt{15}=2.582$