Question

Find area of quadrilateral ABCD AB=7cm,BC=12cm,CD=12cm,DA=9cm,AC=15cm

Answer 1

Step-by-step explanation:

Area of ΔABC = $\sqrt{s(s-a)(s-b)(s-c)}$

where, 2s=$a+b+c$

s=$\frac{a+b+c}{2}$

an putting value

s=$\frac{7+12+15}{2}$

s=17cm

Area of ΔABC = $\sqrt{17(17-7)(17-12)(17-15)}$

=$\sqrt{17(10)(5)(2)}$

=$41.23cm^2$

Now, area of ΔACD = $\sqrt{s(s-a)(s-b)(s-c)}$

s=$\frac{a+b+c}{2}=\frac{15+12+9}{2}$

s=18cm

on putting value:

Area of ΔACD = $\sqrt{18(18-15)(18-12)(18-9)}$

=$\sqrt{2916}=54cm^2$

Area of quadrilateral ABCD = Δ$ABC$+Δ$ABD$

=41.23+54

=$95.23cm^2$

Answer 2

Answer:

Step-by-step explanation:

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

ANOTHER WAY...

let it be 2 triangles rather than 1 big quadrilateral.

ABC = 7/6/9 and ADC = 15/12/9

total area

= √(11 * 4 * 5 * 2) + (12 * 9) / 2

= 2√110 + 54

= 74.976177

=~ 75 cm^2