Question

A(6 1) B(8 2) and C(9 4) are three vertices of a parallelogram ABCD. If E is the mid point of DC. Find the area of the ∆ADE

Answer 1

midpoint of point A and point C is same as the midpoint of point B and point D so I assume a point D and find its coordinates . After D you can find the coordinates of E as E is the midpoint of DC. Now you have got all the three points of the triangle ADE. With the coordinates of these points determine the area of triangle ADE

Answer 2

Answer:

The area of ADE is 0.75 unit².

Step-by-step explanation:

A(6 1) B(8 2) and C(9 4) are three vertices of a parallelogram ABCD.

Area of triangle ABC is

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

$A=\frac{1}{2}(6(2-4)+8(4-1)+9(1-2))=1.5$

ABCD is a parallelogram. E is mid point of DC.

Here AC is the diagonal of ABCD. It means AC divides the area of parallelogram in two equal parts.

$Area(ABC)=Area(ACD)$

E is the mid point of DC. It means AE is median of ACD and divide the area of triangle ACD in two equal parts.

$Area(ADE)=\frac{Area(ACD)}{2}$

$Area(ADE)=\frac{Area(ABC)}{2}$

$Area(ADE)=\frac{1.5}{2}=0.75$

Therefore the area of ADE is 0.75 unit².