Question

find the area of quadrilateral ABCD in which angle A = 90°, AB = 30cm, BC = 42 CM, CD = 20 and DA = 16 cm​

Answer 1

Answer:

Sorry bro but there is no formula for such question.

I knew only this . you can solve this question by this method

(side 1 ×side 2 ) ×sin (angle)

Answer 2

The area of the given quadrilateral is $A=576cm^{2}$.

Given,

A quadrilateral ABCD with

∠$A=90$°

Sides,

$AB=30cm\\BC=42cm\\CD=20cm\\AD=16cm$

To find,

Area of the quadrilateral, $A$

Solution,

Start with joining vertices B and D

Now,

In the right-angled triangle formed by the above construction, ABD

$DB=\sqrt[]{(AB)^{2} +(AD)^2}$

⇒$DB=\sqrt{30^{2} +16^{2} }$

⇒$DB=\sqrt{900+256}$

⇒$DB=\sqrt{1156}$

⇒ $DB=34cm$

Area of triangle ABD is,

$A_{abd} =\frac{1}{2} .AD.AB$

⇒$A_{abd}=\frac{1}{2}.(16)(30)$

⇒$A_{abd}=240cm^{2}$

Area of ΔABD=$240cm^2$

To find the area of triangle BCD, we must use Heron's formula

⇒$Area=\sqrt{s(s-a)(s-b)(s-c)}$

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

So,

For ΔBCD,

$s=\frac{BC+CD+BD}{2}$

⇒$s=\frac{42+20+34}{2}$

⇒$s=\frac{96}{2}$

⇒$s=48$

Putting this value in the formula, we get

$A_{bcd}=\sqrt{s(s-BC)(s-CD)(s-DB)}$

⇒$A_{bcd}=\sqrt{48(48-42)(48-20)(48-34)}$

⇒$A_{bcd}=\sqrt{48(6)(28)(14)}$

⇒$A_{bcd}=\sqrt{112896}$

⇒$A_{bcd}=336cm^2$

Area of ΔBCD$=336cm^2$

Finally, the area of quadrilateral ABCD can be obtained by adding both the areas

⇒$A=A_{abd}+A_{bcd}$

⇒$A=240cm^2+336cm^2$

⇒$A=576cm^2$

Therefore, the area of the given quadrilateral is $A=576cm^2$.