Question

IA(0,5), B(6, 11 ) and C[ 10,7) are the vertices of a A ABC, D and E
are the mid-points of AB and AC respectively. Then find the area of A ADE.​

Answer 1

Step-by-step explanation:

hope it will help you

it's very easy

first I just mid point formula that is X1+x2/2

then I applied the formula of the area of triangle

Answer 2

area of triangle  ADE = 6 sq. units

Step-by-step explanation:

if A (0,5) , B(6,11) , C(10,7)  and D and E are the mid points of AB and AC

using mid point formula

$\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$

then coordinates of D

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

coordinates of E

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

then area of triangle ADE

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

area =

$\frac{1}{2} (0(8-6) +3(6-5)+5(5-8))\\\\\frac{1}{2} (0+3-15)\\\\\frac{1}{2} \times -12=-6$

since , the area of triangle can never be negative therefore , neglecting the negative sign we get

area of triangle  ADE = 6 sq. units

#Learn more:

If A ( 0,5) ,B(6,11)and C (4,5) are the vertices of a ABC

https://brainly.in/question/15522017