Question

In the figure a given below ABCD is a rectangle with side 24 cm and 7cm. Find the area of angle A B D and angle BDC

Answer 1

area of ABD is equal to area of BDC
since
7*24*1/2=7*12=84

Answer 2

ABCD is a rectangle

AB = DC = 24 cm each

AD = BC = 7 cm each

Area of the rectangle

= lb unit sq.

[ In which l is the length and b is the breadth of the rectangle ]

= 24 × 7

= 168 cm sq.

if we make BD diagonal we will get two triangles

ΔABD and ΔBDC

Bothe the triangles are congruent by SAS congruence condition.

Both are equal,

$ar(\triangle ABD) = ar(\triangle BCD)\\\\= ar(\triangle ABD) + ar(\triangle BCD)=ar(ABCD)\\\\=ar(\triangle ABD) + ar(\triangle ABD) = 168\\\\= 2[ar(\triangle ABD)] =168\\\\=ar(\triangle ABD) = \frac{168}{2}\\\\=ar(\triangle ABD) = 84 cm^{2}$

hence

Area of ΔABD and ΔBDC = 84 cm sq. each