Question

Find the area of the quad ABCD in which AB=7cm,BC=6cm,CD=12cmDA=15cm and AC=9cm

Answer 1

area of quadrilateral ABCD= Area of triangle ABC + Area of triangle ADC
Area of trianlge= sqrt(s(s-a)(s-b)(s-c))
FOR area of triangle ABC

s=(7+6+9)/2=11
area of triangle ABC=sqrt(11(11-7)(11-6)(11-9))
=sqrt(440)
= 20.9761
similarly,
area of triangle ADC=54

therefore area of quadrilateral ABCD=20.9761+ 54
=74.9761

Answer 2

ANSWer:

The diagonal AC divides the quadilateral ABCD into two triangles ABC and ACD.

† $\bf \tiny \: area \: of \: quad. \: abcd \: = area \: of \: \triangle \: abc + area \: of \: \triangle \: acd$

For ∆ABC, we have

$\qquad \tt \: s = \frac{6 + 7 + 9}{2} = 11 \: cm.$

$\therefore \sf \small \: Area \: of \: \triangle \: ABC \: = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \\ \implies \sf \small \: Area \: of \: \triangle \: ABC = \sqrt{11(11 - 6)(11 - 7)(11 - 9)} \\ \\ \\ \implies \sf \small \sqrt{11 \times 5 \times 4 \times 2} = \sqrt{440} \: sq.cm \\ \\ \\ \implies \sf \small \: Area \: of \: \triangle \: ABC = 20.98 \: {cm}^{2}$

_________________________

For ∆ACD, we have

$\qquad \tt \: s = \frac{9 + 12 + 15}{2} = 18 \: cm$

$\therefore \sf \small \: Area \: of \: \triangle \: ACD = \sqrt{18(18 - 9)(18 - 12)(18 - 15)} \\ \\ \\ \implies \sf \small \: Area \: of \: \triangle \: ACD = \sqrt{18 \times 9 \times 6 \times 3} \\ \\ \\ \implies \sf 54 \: sq.unit$

Hence, Area of Quadilateral ABCD = ( 20.98 + 54) cm² = 74.98 cm².