Find area of quadrilateral a b c d in which a b 3cm
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In ∆ABC, we have
AB2 + BC2 = 9 + 16 = 25
= AC2
Hence, ABC is a right triangle, right angled at B
[By converse of Pythagoras theorem]
∴ Area of ∆ABC = 1/ 2 × base × height
= 1/ 2 × 3 × 4 cm2 = 6 cm2.
In ∆ACD, a = 5 cm, b = 4 cm, c = 5 cm.
∴ s = a +b + c+2 = 5+ 4+5 / 2 cm = 7cm
Area of ΔABCD=√s(s-a)(s-b)(s-c)
=√7*(7-5)(7-4)(7-5) cm^2
=√7*2*3*2
=√84cm^2
=9.2cm^2
Area of the quadrilateral=Area ofΔABC +Area of ACD
=(6+9.2)cm^2
=15.2cm^2
AB2 + BC2 = 9 + 16 = 25
= AC2
Hence, ABC is a right triangle, right angled at B
[By converse of Pythagoras theorem]
∴ Area of ∆ABC = 1/ 2 × base × height
= 1/ 2 × 3 × 4 cm2 = 6 cm2.
In ∆ACD, a = 5 cm, b = 4 cm, c = 5 cm.
∴ s = a +b + c+2 = 5+ 4+5 / 2 cm = 7cm
Area of ΔABCD=√s(s-a)(s-b)(s-c)
=√7*(7-5)(7-4)(7-5) cm^2
=√7*2*3*2
=√84cm^2
=9.2cm^2
Area of the quadrilateral=Area ofΔABC +Area of ACD
=(6+9.2)cm^2
=15.2cm^2
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