Question

Area of the parallelogram ABCD = 640 sq. units.

(i) Find the area of $\Delta$ AFC. Here, E and F are the mid points of AB and AE respectively.

(ii) If Area of $\Delta$ DAE is 46 sq.cm then what is the Area of $\Delta DEF$ ?

Answer 1

(i) Join the vertices D and B

Diagonal BD divides the parallelogram ABCD into two triangles of equal area.

Area of $\Delta ABD=\frac{1}{2}\times$ area of parallelogram ABCD

Area of $\Delta ABD=\frac{1}{2}\times$ 640 = 320 sq. units

Given, E is the midpoint of AB

DE is the median of $\Delta AB$ and it divides the triangle into two triangles of equal area.

Area of $\Delta ADE=\frac{1}{2}\times$ area of $\Delta ABD$

$\frac{1}{2}\times$ 320 = 160 sq. units

Similarly, DF is the median of $\Delta ADE$,

Area of $\Delta AFD=\frac{1}{2}\times$ area of $\Delta ADE$

$\frac{1}{2}\times$ 160 = 80 sq. units

Now, the area of $\Delta AFD$ = area of $\Delta AFC$ [Triangles on same base and between same parallel lines]

Area of $\Delta AFC$ = 80 sq. units

(ii) We know that F is a midpoint of seg. AE

$\therefore A(\Delta DEF=\frac{1}{2}\times$ 46 = 23 sq.cm.