Question

find the area of a quadrilateral ABCD in which AB=3cm,BC4cm ,DA=5cm and AC=5cm

Answer 1

Here's your answer!!

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In ∆ABC,
$= > {ab}^{2} + {bc}^{2} = {ac}^{2} \\ = > 9 + 16 = 25$
It means it is an right- angle triangle at B.
So,

Area of ∆ ABC =
$= > \frac{1}{2} \times base \times height$
$= > \frac{1}{2} \times 3 \times 4 \\ = > 6 {cm}^{2}$
Now,
In ∆ACD, we have ,
$= > a = 5cm \: \\ = > b = 4cm \\ = > c = 5cm$
Therefore,
$= > s = \frac{a + b + c}{2} \\ = > s = \frac{5 + 4 + 5}{2} \\ = > s = \frac{14}{2} \\ = > s = 7$
So,
Area of ∆ ACD=
$= > \sqrt{s(s - a)(s - b)(s - c)} \\ = > \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \\ = > \sqrt{7 \times 2 \times 3 \times2} \\ = > \sqrt{84}$
$= > 9.2 {cm}^{2}$
Hence,
Area of quadrilateral ABCD =
=>Area of ∆ ABC +Area of ∆ ACD
$= > 6 + 9.2 \\ = > 15.2 {cm}^{2}$(approx)
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See the attachment for figure !!

Hope it helps you!! :)

Answer 2

first divide the quadrilateral into two triangles

∆ABC & ∆ACD
by applying herons formula we can find the area of the two triangles
ar(∆ABC) =
S= ab + bc + ca ÷ 2
3+4+5 ÷ 2
s = 12÷2
s = 6
area = √S(S-A) (S-B) (S-C)
= √6(6-3) (6-4) (6-5)
= √6×3×2×1
= √36
= √6×6
area = 6

for ∆ACD
S = ac + dc + ad ÷ 2
5+4+5 ÷ 2
s = 14÷2
s = 7
area = √S(S-A) (S-B) (S-C)
= √7(7-5) (7-4) (7-5)
= √7×2×3×2
= 2√21

area = 2√21

area of quadrilateral = ar(∆ABC) + ar(∆ACD)
= 6 + 2√21
= 6 + 2×4.58
= 6 + 9.16
= 15.16
area = 15.16cm²