Find the area of quadrilateral, where AB=4cm, BC=5cm, CD=3cm and DA=4cm and one of its diagonals AC=5cm.
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Answers
•••••• ::::::: SOLUTION ::::::::••••••
Area of Quadrilateral = Ar of ∆ABC + Ar of ∆ADC
◆ • • Area of ∆ ABC
by using Heron's formula :
\begin{gathered} = \sqrt{s(s - a)(s - b)(s - c} \\ s = \frac{a + b + c}{2} = \frac{3 + 4 + 5}{2} = \frac{12}{2} \\ = 6 \\ \\ ar \: of \: triangle \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} {cm}^{2} \\ = \sqrt{6 \times 3 \times 2 \times 1} {cm}^{2} \\ = \sqrt{6 \times 6} = 6 {cm}^{2} \end{gathered}
=
s(s−a)(s−b)(s−c
s=
2
a+b+c
=
2
3+4+5
=
2
12
=6
aroftriangle
=
s(s−a)(s−b)(s−c)
=
6(6−3)(6−4)(6−5)
cm
2
=
6×3×2×1
cm
2
=
6×6
=6cm
2
◆ • • Area of ∆ ADC
by using Heron's formula
\begin{gathered}s = \frac{a + b + c}{2} \\ = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7 \\ \\ ar \: of \: triangle \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \\ = \sqrt{7 \times 2 \times 3 \times 2} \\ = 2 \sqrt{21} \ {cm}^{2} \\ = 2 \times 4.58 = 9.16 {cm}^{2} \end{gathered}
s=
2
a+b+c
=
2
5+4+5
=
2
14
=7
aroftriangle
=
s(s−a)(s−b)(s−c)
=
7(7−5)(7−4)(7−5)
=
7×2×3×2
=2
21
cm
2
=2×4.58=9.16cm
2
now ,
AREA of Quadrilateral = ar of ∆ABC + ar of ∆ADC
\begin{gathered} =( 6 + 9.16) {cm}^{2} \\ = 15.16 {cm}^{2} \\ = 15.2 {cm}^{2} (approx.)\end{gathered}
=(6+9.16)cm
2
=15.16cm
2
=15.2cm
2
(approx.)
So , the area of Quadrilateral ABCD is 15.2cm sq.
I hope
this helps
you.
Answer:
hey mate ur answer is 15.2cm²