Find the area of quadrilateral ABCD in which AB=17 cm ,AD=9 cm,CD=12 cm,angle ACB=90 and AC=15 cm
COULD YU EXPLAIN ME THAT IN THESE TYPE OF QUESTIONS IN WHICH ORDER SHALL THE FIGURE MUST BE DRAWN
mhesh1:
The figure must be drawn in the order which the quadrilateral is mentioned. AB BC CD DA. But are you sure if the question is correct? I think it is angle ABC and not ACB.
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the ans is 144 cm² I hope this and would be helpful!!!
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by Pythagoras theorem,
BC = √AB²−√AC²
=√17² - √15²
=√289 - 225
BC = √64 = 8 cm
Perimeter of quadrilateral ABCD = 17 + 9 + 12 + 8 = 46 cm
Area of triangle △ABC=1/2×base×height
△ABC=12×8 x 15
= 60 cm²
NOW, for the area of the triangle, ACD
we will take a = 15 cm, b = 12 cm and c = 9 cm
S = 12(a+b+c)
S = 12(15+12+9)=36/2=18cm
Area of one triangular =S√(S−a)(S−b)(S−c) (Heron's formula)
=18√(18−15)(18−12)(18−9)
=√18×3×6×9
=√18×18×3×3
= 18×3=54cm²
area of quadrilateral ABCD= area of △ABC + area of △AC
= 60 + 54 = 114 cm²
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