Question

find the area of quadrilateral ABCD given in below figure​

Answer 1

Answer:

The area of $quadrilateral$ $ABCD$ is $41.326\ cm^2$.

Step-by-step explanation:

Since $ABCD$ is a $quadrilateral$ in which $AB$ = 7 cm, $BC$ = 15 cm, $CD$ = 12 cm, $AD$ = 6 cm and diagonal $AC$ = 9 cm.

Recall the heron's formula,

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

where $s$ is the semi perimeter and $a,b,c$ are the sides of triangle.

Now,

In Δ$ADC$,

Semi perimeter, $s=\frac{AD+CD+AC}{2}$

⇒ $s=\frac{6+12+9}{2}$

      $=\frac{27}{2}$

      $=13.5$ cm

Using heron's formula,

Area of Δ$ADC$ = $\sqrt{s(s-AD)(s-CD)(s-AC)}$

                         = $\sqrt{13.5(13.5-6)(13.5-12)(13.5-9)}$

                         = $\sqrt{13.5(7.5)(1.5)(4.5)}$

                         = $\sqrt{683.4375}$

                         = $26.142$ $cm^2$

Similarly,

In Δ$ABC$,

Semi perimeter, $s=\frac{AB+BC+AC}{2}$

⇒ $s=\frac{7+15+9}{2}$

      $=\frac{31}{2}$

      $=15.5$ cm

Using heron's formula,

Area of Δ$ABC$ = $\sqrt{s(s-AB)(s-BC)(s-AC)}$

                         = $\sqrt{15.5(15.5-7)(15.5-15)(15.5-9)}$

                         = $\sqrt{15.5(8.5)(0.5)(3.5)}$

                         = $\sqrt{230.6525}$

                         = $15.184$ $cm^2$

Thus, area of quadrilateral $ABCD$ = Area of Δ$ADC$ + Area of Δ$ABC$

                                                          = $26.142+15.184$

                                                          = $41.326\ cm^2$.

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