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Question
Find the area of a quadrilateral field whose sides are 12m, 9m, 18m and 21m respectively and the angle between the first two sides is a right angle. Take the value of `sqrt(6)` as 2.5.
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Solution
In the given quadrilateral ABCD, join diagonal AC
ABC is a right triangle
We know that, Area of a Triangle = `(1)/(2)"b.h" "i.e" (1)/(2)("Base" xx "Height")`
Area of a Triangle ABC = `(1)/(2)9.12` = 54m2
AC is the hypotenuse, AC
= `sqrt(12^2 + 9^2)`
= `sqrt(225)`
= 15m
Triangle ACD has sides 15m, 18m, 21m
We know that, Area of a Triangle whose sides are a, b, and c and semiperimeter is s is given by `sqrt("s"("s" - "a")("s" - "b")("s" -"c")); "s" = ("a" + "b" + "c")/(2)`
For a triangle whose sides are cm, cm and cm
i.e a = 15, b = 18 and c = 21,
s = `(15 + 18 + 21)/(2)`
= `(54)/(2)`
= 27
Area
= `sqrt(27(27 - 15)(27 - 18)(27 - 21)`
= `sqrt(27(12)(9)(6)`
= `sqrt(9 xx 3(6 xx 2)(9)(6)`
= `9 xx 6sqrt(6)`
= `54sqrt(6)`
= 54(2.5)
Area(Quad ABCD) = Ar(Triangle ABC) + Ar(Triangle ADC)
54 + 54(2.5)
= 54(1 + 2.5)
= 54(3.5)
= 189m2.
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