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Hans Raj 8 years, 6 months ago
area of quadrilateral abcd = areas of triangles abc + acd
= root 18(18-12)(18-15)(18-9) + root 11(11-6)(11-7)(11-9)
= root 18x6x3x9 + root 11x5x4x2
= root 9x9x9x2x2 + root 4x110
= 9x2x3 + 2 root 110
= 54 + 2 {tex} \sqrt{110}{/tex} unit2
Hans Raj 8 years, 6 months ago
area of quadrilateral abcd = areas of triangles abc + acd
= root 18(18-12)(18-15)(18-9) + root 11(11-6)(11-7)(11-9)
= root 18x6x3x9 + root 11x5x4x2
= root 9x9x9x2x2 + root 4x110
= 9x2x3 + 2 root 110
= 54 + 2 root 110 unit2
Posted by Dev Shukla 8 years, 6 months ago
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Posted by Krishna Goel 8 years, 6 months ago
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Hans Raj 8 years, 6 months ago
s=5+5+5/2=15/2=7.5
area = root 7.5(7.5-5)(7.5-5)(7.5-5)
= root 7.5x2.5x2.5x2.5
= 2.5 x 2.5 root 3
= 6.25 {tex}\sqrt{3}{/tex} cm2
Agni Roy 8 years, 6 months ago
Area of an equilateral triangle={tex}√ 3 /2\times (side)^2 {/tex}=1.732/2 x (5x5)=21.65cm2
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S Singh 8 years, 6 months ago
1Thank You