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Asked by Rasneet Last Modified
Answer 
Tasneem
For ΔABC,
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
Therefore, ΔABC is a right-angled triangle, right-angled at point B.
Area of ΔABC
For ΔADC,
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron’s formula,
Area of triangle
Area of ABCD = Area of ΔABC + Area of ΔACD
= (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 ( Approximate Value)
read lessArea of triangle ADC
= √7 ( 7 - 5 ) ( 7 - 5 ) ( 7 - 4 )
= √ 7 × 2 × 2 × 3 = 2 √21cm² = 2 × 4.6 = 9.2cm².
Area of triangle ABC
= √ 6 ( 6 - 5 ) ( 6 - 4 ) ( 6 - 3 ) = √6 × 1 × 2 × 3
= √ 3 × 2 × 2 × 3 = 6cm².
Area of quadrilateral ABCD
= area of ABC + area of ACD
= 6cm² + 9.2cm² = 15.2cm².
read less
Area of triangle ABC+ Area of triangle ACD
=Area of Right angle triangle ABC+area of isosceles triangle ACD
= (1÷2×base ×height)+ {base ÷2(sqrt(side^2)-((base^2)/4))}
=(1÷2(AB×AC))+{DC÷2(SQRT( (AC^2)-(DC^2/4)))}
= (1÷2(4×3))+{4/2(SQRT (5^2)-(4^2/4)))}
=6+{2(SQRT (25-4))}
=6+2(SQRT21)
2{3+sqrt (21)}
read less
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Related Questions
A park, in the shape of a quadrilateral ABCD, has ∠ C = AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
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