find the quadrilateral ABCD in which AB=3 cm,BC=4 cm CD=4cm ,DA=5cm and AC=5cm
Answers
Step-by-step explanation:
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here's your answer..
Area of Quadrilateral = ar of ∆ABC + ar of ∆ADC
Area of ∆ABC ::•• •●
heron's formula
\begin{gathered}= \sqrt{s(s - a)(s - b)(s - c)} \\\end{gathered}
=
s(s−a)(s−b)(s−c)
{where, s is semi perimeter}
\begin{gathered}s = \frac{a + b + c}{2} \\ \: = \frac{3 + 4 + 5}{2 } \\ = \frac{12}{2} \\ = 6\end{gathered}
s=
2
a+b+c
=
2
3+4+5
=
2
12
=6
area of ∆ABC
\begin{gathered}= \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{gathered}
=
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{gathered}= \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{gathered}
=
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{gathered}= 6 + 9.16 \\ = 15.16 {cm}^{2} \\ = 15.2 {cm}^{2} \: (approx.)\end{gathered}
=6+9.16
=15.16cm
2
=15.2cm
2
(approx.)
So the Area of Quadrilateral ABCD = 15.2cm .sq
I hope
this helps
you
