Question

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23 question

Answer 1

In this question at first we will find the area of triangle ABC then we find the area of triangle DBC then we will Subtract Area of DBC from Area of ABC.

Now, For Triangle ABC,

=> S = a+b+c/2 = 120+122+22/2=264/2 =132 cm

$= > Ar(ABC) = \sqrt{s(s - a)(s - b)(s - c)}$
$= > Ar(ABC) = \sqrt{132(132 - 120)(132 - 122)(132 - 22)}$
$= > Ar(ABC) = \sqrt{132 \times 12 \times 10 \times 110}$
$= > Ar(ABC) = \sqrt{11 \times 12 \times 12 \times 10 \times 10 \times 11}$
$= > Ar(ABC) = (11 \times 12 \times 10) \: {cm}^{2}$

$= > Ar(ABC) = 1320 \: {cm}^{2}$

Again,

For Triangle DBC,

=> S = a+b+c/2 = 24+26+22/2 = 36

$= > Ar(DBC) = \sqrt{s(s - a)(s - b)(s - c)}$
$= > Ar(DBC) = \sqrt{36(36 - 24)(36 - 26)(36 - 22)}$

$= > Ar(DBC) = \sqrt{36 \times 12 \times 10 \times 14}$
$= > Ar(DBC) = \sqrt{60480}$

$= > Ar(DBC) = 245.9 \: {cm}^{2}$

Hence,

=> Area of Shaded region = Area of ABC - Area of DBC

=> Area of Shaded region = (1320 - 245.9) cm^2

=> Area of Shaded Region = 1074.1 cm^2

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