Question

Find the area of a quadrilateral ABCD in which AB=3cm,BC=4cm,CD=4cm,DA=5cm and ac=5cm.Find it through heron formula
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Answer 1

Answer:

here's your answer..

Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC

Area of ∆ABC ::•• •●

heron's formula

\begin{lgathered}= \sqrt{s(s - a)(s - b)(s - c)} \\\end{lgathered}

=

s(s−a)(s−b)(s−c)

{where, s is semi perimeter}

\begin{lgathered}s = \frac{a + b + c}{2} \\ \: = \frac{3 + 4 + 5}{2 } \\ = \frac{12}{2} \\ = 6\end{lgathered}

s=

2

a+b+c

=

2

3+4+5

=

2

12

=6

area of ∆ABC

\begin{lgathered}= \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} \\ = \sqrt{6 \times 3 \times 2 \times 1 } \: {cm}^{2} \\ = \sqrt{6 \times 6} \: {cm}^{2} \\ = 6 {cm}^{2}\end{lgathered}

=

s(s−a)(s−b)(s−c)

=

6(6−3)(6−4)(6−5)

=

6×3×2×1

cm

2

=

6×6

cm

2

=6cm

2

Area of∆ADC ::•• •●

\begin{lgathered}= \sqrt{s(s - a)(s - b)(s - c)} \\ \: s = \frac{a + b + c}{2} \\ \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7cm \\ \\ ar \: of \: triangle \: \\ = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \\ = \sqrt{7 \times 2 \times 3 \times 2} \\ = \sqrt{2 \times 2 \times 7 \times 3} \\ = 2\sqrt{21} \\ = 2 \times 4.58 = 9.16 {cm}^{2}\end{lgathered}

=

s(s−a)(s−b)(s−c)

s=

2

a+b+c

2

5+4+5

=

2

14

=7cm

aroftriangle

=

7(7−5)(7−4)(7−5)

=

7×2×3×2

=

2×2×7×3

=2

21

=2×4.58=9.16cm

2

::::::::::::::::::::::: ◆

Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC

\begin{lgathered}= 6 + 9.16 \\ = 15.16 {cm}^{2} \\ = 15.2 {cm}^{2} \: (approx.)\end{lgathered}

=6+9.16

=15.16cm

2

=15.2cm

2

(approx.)

So the Area of Quadrilateral ABCD = 15.2cm .sq

I hope

this helps

you