The medians of a Delta ABC are 9 , cm, 12 , c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the medians be $m_1 = 9, m_2 = 12, m_3 = 15$. The area of a triangle with medians $m_1, m_2, m_3$ is given by the formula: $	ext{Area} = \fra

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(C) Let the medians be $m_1 = 9, m_2 = 12, m_3 = 15$. The area of a triangle with medians $m_1, m_2, m_3$ is given by the formula: $	ext{Area} = \frac{4}{3} \sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$,where $s_m = \frac{m_1 + m_2 + m_3}{2}$. Here,$s_m = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18$. $	ext{Area} = \frac{4}{3} \sqrt{18(18 - 9)(18 - 12)(18 - 15)}$ $	ext{Area} = \frac{4}{3} \sqrt{18 	imes 9 	imes 6 	imes 3}$ $	ext{Area} = \frac{4}{3} \sqrt{2916} = \frac{4}{3} 	imes 54 = 72 \, cm^2$.

The medians of a $\Delta ABC$ are $9 \, cm, 12 \, cm$ and $15 \, cm$ respectively. Then the area of the triangle is.....$sq \, cm$.

Similar Questions

In $\triangle ABC$,if $A=(1,2)$ and the equations of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively,then the area of $\triangle ABC$ is

For a triangle formed by the vertices $(0,0)$,$(4,0)$,and $(3,4)$,the orthocenter is:

If two vertices of a triangle are $(5, -1)$ and $(-2, 3)$ and its orthocenter is at $(0, 0)$,find the third vertex.

The orthocentre of the triangle formed by the lines $x + y = 1$ and $xy = 0$ is

The points $(11,9), (2,1)$ and $(2,-1)$ are the mid-points of the sides of a triangle. Then,the centroid is

Vedclass Test Series

Exam Paper Generator

Online Exam Module