Find the area of a quadrilateral ABCD in which AB=3cm,BC=3cm ,CD 4cm , CD= 4cm ,DA=5cm and AC=5cm
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Figure:-
Given:-
- AB=3cm,BC=3cm ,CD 4cm , CD= 4cm ,DA=5cm and AC=5cm.
To find:-
- Find the area of a quadrilateral ABCD.
Solutions:-
In ∆ABC,
AC² = AB² + BC²
(5)² = (3)² +(4)²
Therefore,
∆ABC is a right angle triangle at point B.
Area of ∆ABC = 1/2 × AB × BC
= 1/2 × 3 × 4
= 3 × 2
= 6cm
In ∆ADC,
S = perimeter/2
S = (5 + 4 + 5)/2
S = 14/2
S = 7cm
By heroe's formula,
Area of triangle = √s(s - a)(s - b)(s - c)
Area of ∆ADC = √7(7 - 5)(7 - 4)(7 - 5)
= √7 × 2 × 3 × 2
= √84
= 2√21
= 2 × 4.583
= 9.166cm²
Area of ABCD = Area of ∆ABC + Area of ∆ACD
= 6 + 9.166
= 15.166
= 15.2cm²
Hence, the area of a quadrilateral ABCD is 15.2cm²
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