Question

if A(0,5) B (6,11) and C (10,7) are the vertices of a triangle D and E are the midpoints of AB and AC respectively. Then find the area of triangle ADE
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Answer 1

Given that,

A(0,5) B (6,11) and C (10,7) are the vertices of a triangle. D and E are the midpoints of AB and AC respectively.

To find,

The area of triangle ADE.

Solution,

If D and E are the midpoints of AB and AC. Using mid point formula, coordinates of D are :

$(\dfrac{0+6}{2}, \dfrac{5+11}{2})=(3,8)$

The coordinates of E are :

$(\dfrac{0+10}{2}, \dfrac{5+7}{2})=(5,6)$

The area of ADE is given by :

$A=\dfrac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(x_1-x_2))$

Putting all the values in above formula we get :

$A=\dfrac{1}{2}(0(8-6)+3(6-5)+5(5-8))\\\\A=\dfrac{1}{2}\times (-12)\\\\A=-6\ \text{unit}^2$

Nut area can't be negative.

So,

$A=6\ \text{unit}^2$

So, the area of triangle ADE is 6 sq. units.