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Question 1 Heron’s Formula - Exercise 12.2

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m

and AD = 8 m. How much area does it occupy?

Answer:

First, construct a quadrilateral ABCD and join BD.

We know that

∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ΔBCD

\begin{aligned} &B D^{2}=B C^{2}+C D^{2}\\ &\Rightarrow B D^{2}=12^{2}+5^{2}\\ &\Rightarrow B D^{2}=169\\ &\Rightarrow B D=13 \mathrm{m}\\ &\text { Now, the area of } \Delta B C D=(1 / 2 \times 12 \times 5)=30 \mathrm{m}^{2} \end{aligned}

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8 + 9 + 13)/2 m

= 30/2 m = 15 m

Using Heron's formula,

Area of ΔABD

\begin{array}{l} \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{15(15-13)(15-9)(15-8)} m^{2} \\ =\sqrt{15 \times 2 \times 6 \times 7 m^{2}} \\ =6 v 35 \mathrm{m}^{2}=35.5 \mathrm{m}^{2} \text { (approximately) } \end{array}

=30 \mathrm{m}^{2}+35.5 \mathrm{m}^{2}=65.5 \mathrm{m}^{2}

"hello students welcome to lido q a video
session
i am seth your math tutor and question
for today is
a part in a shape of quadrilateral abcd
has angle c is equal to 90 degree a b is
equal to 9 meter
dc is equal to 12 meter cd is equal to 5
meter and 80 is equal to
8 meter how much area does it occupy
so the park layout is shown here below
abcd is the layout we know
from the question that angle c is 90
so
angle c is equal to 90.
a b is equal to 9 meter
bc is equal to 12 meter
cd is equal to 5 meter
and 80
equals to 8 meter
now you have to construct the
quadrilateral as shown here
that is body little abcd in this
quadrilateral you have to join
the diagonal bd
so here we have joined the bd
now
apply pythagoras theorem in triangle bcd
triangle bcd is a right angle triangle
because it is given that angle c is 90
degree
so as per pythagoras theorem
you can say that bd square is equal to
bc square plus cd square
we need to find bd so bd square
will be equal to 12 square plus y square
and which is equal to 169 root of 169 is
bd
hence bd is equal to 13 meter
now area of triangle bcd can be given by
half into base into height
base here is pc
and height is dc
so it is half into 12 into
phi so we will continue here
that is area of triangle bcd
that is a is equal to half into 12 into
5 that is 30 meter square
now semi perimeter of triangle abd will
be
semi perimeter
s of triangle abd
s is equal to perimeter upon 2
that is semi perimeter 8 plus 9
plus 13 upon 2
and that is 30 upon 2
so s is equal to 15 meter
now once you have found the semi
perimeter you can use the herons formula
to find the area
so as per the herons formula
area of triangle
area of triangle abd
can be given by under root of
as s minus a s minus b
s minus c
so 15 under root of 15
15 minus 8 15 minus
9 and 15 minus 13
so that is under root of
15 into 2 into 6 into
7 and that is meter square
and that comes out to be 6 root 35
approx
will be
so your area of triangle
abd will be equal to
6 root 35
and that is equal to area will be
35.5 meter square approx
now what you need to do is add both the
triangles area
so area of quadrilateral a
b c b that is the area of the park
will be equal to area of
triangle abd
plus area of triangle
bcd previously we have found the area of
triangle bcd to be
30 meter square and area of triangle abd
just now we have found as 35.5
meter square so abd plus bcd will be
equal to 35.5 plus
30
and hence this will give you area of the
part
which is equal to 65.5 meter square
so that is our required area if you have
any query you can drop it in our comment
section
and subscribe to lido for more such q a
thank you for watching
"

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