Question

find the area of
a quadrilateral ABCD in which AB = 3cm
BC = 4cm , CD = 4 cm , DA = 5 cm
and AC = 5cm​

Answer 1

Answer:

Area of the quadrilateral = $\sf{{15.16cm}^{2}}$

Step-by-step explanation:

Given :

AB = 3 cm

BC = 4 cm

CD = 4 cm

DA = 5 cm

AC = 5 cm

To find :

The area of the quadrilateral

Concept :

To find the area of the quadrilateral . We have to do the sum of the area of $\triangle\sf{ABC}$ with the area of $\triangle\sf{ACD}$ .

So , the $\triangle\sf{ABC}$ is a triangle . To find their area use this formula -:

First -:

Let , AB = a

BC = b

AC = c

And also -:

Let , DA = a

CD = b

AC = c

So ,

$\boxed{\sf{s=\dfrac{a+b+c}{2}}}$

After that find the area of the triangle use this formula -:

$\boxed{\sf{A=\sqrt{s(s-a)(s-b)(s-c)}}}$

Where,

A = Area of the triangle

Solution :

Area of $\triangle\sf{ABC}$ -:

First -:

$:\implies\sf{s=\dfrac{3cm+4cm+5cm}{2}}$

$:\implies\sf{s=\dfrac{\cancel{12}cm}{\cancel{2}}}$

$:\implies\sf{s=6cm}$

Now , area -:

$:\implies\triangle\sf{ABC=\sqrt{6cm(6cm-3cm)(6cm-4cm)(6cm-5cm)}}$

$:\implies\triangle\sf{ABC=\sqrt{6\times3\times2\times1}{cm}^{2}}$

$:\implies\triangle\sf{ABC=\sqrt{(2\times3)\times3\times2}{cm}^{2}}$

$:\implies\triangle\sf{ABC=\sqrt{{3}^{2}\times{2}^{2}}{cm}^{2}}$

$:\implies\triangle\sf{ABC=3\times2\:\:{cm}^{2}}$

$:\implies\triangle\sf{ABC=6{cm}^{2}}$

So , the area of $\triangle\sf{ABC}$ is 6 cm² .

Area of $\triangle\sf{ACD}$ -:

First -:

$:\implies\sf{s=\dfrac{5cm+4cm+5cm}{2}}$

$:\implies\sf{s=\dfrac{\cancel{14}cm}{\cancel{2}}}$

$:\implies\sf{s=7cm}$

Now , area -:

$:\implies\triangle\sf{ACD=\sqrt{7cm(7cm-5cm)(7cm-4cm)(7cm-5cm)}}$

$:\implies\triangle\sf{ACD=\sqrt{7\times2\times3\times2}{cm}^{2}}$

$:\implies\triangle\sf{ACD=\sqrt{{2}^{2}}\sqrt{7\times3}{cm}^{2}}$

$:\implies\triangle\sf{ACD=2\sqrt{21}{cm}^{2}}$

$:\implies\triangle\sf{ACD=(2\times4.58){cm}^{2}}$

$:\implies\triangle\sf{ACD=9.16{cm}^{2}}$

Area of the quadrilateral -:

$:\implies\triangle\sf{ABC+\triangle{ACD}}$

$:\implies\sf{{6cm}^{2}+{9.16cm}^{2}}$

$:\implies\sf{{15.16cm}^{2}}$

So , the area of the quadrilateral is $\sf{{15.16cm}^{2}}$