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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Let us join BD. In ΔBCD, applying Pythagoras theorem, BD2 = BC2 + CD2 = (12)2 + (5)2 = 144 + 25 BD2 = 169 BD = 13 m Area of ΔBCD For ΔABD, By Heron’s formula, Area of triangle Area of ΔABD Area of the park = Area of ΔABD + Area of ΔBCD = 35.496 + 30 m2 =...
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Let us join BD. In ΔBCD, applying Pythagoras theorem, BD2= BC2+ CD2 = (12)2+ (5)2 = 144 + 25 BD2= 169 BD = 13 m Area of ΔBCD For ΔABD, By Heron’s formula, Area of triangle Area of ΔABD Area of the park = Area of ΔABD + Area of ΔBCD = 35.496 + 30 m2= 65.496 m2= 65.5 m2( Approximate Value) read less
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Let us join BD. Now BCD is a right triangle, right angled at C. ∴ area of triangle BCD = ½bh = 1/2 × 12 × 5 = 6 × 5 = 30m². By Pythagoras theorem, BD² = 12² + 5² ⇒ BD² = 144 + 25 = 169 ∴ BD = √169 = 13m. Now, area of triangle...
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Let us join BD. Now BCD is a right triangle, right angled at C. ∴ area of triangle BCD = ½bh = 1/2 × 12 × 5 = 6 × 5 = 30m². By Pythagoras theorem, BD² = 12² + 5² ⇒ BD² = 144 + 25 = 169 ∴ BD = √169 = 13m. Now, area of triangle ABD = √ 15 ( 15 - 13 ) ( 15 - 8 ) ( 15 - 9 = √ 15 × 6 × 2 × 7 = √ 5 × 3 × 3 × 2 × 2 × 7 = 6√35m² = 6 × 5.9 = 35.4m². ∴ the area of the park ABCD = area of ABD + area of BCD = 30m² + 35.4m² = 65.4m². read less
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Heron's formula = ^1/2,where s=semi perimeter =(a+b+c+d)/2 Area=271meter square
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Heron's formula = ^1/2,where s=semi perimeter =(a+b+c+d)/2 Area=271meter square
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Mca Passesd Home/Online Tutor With 14 Years Experience.

By using Heron's formula 60.54
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30% of questions those all based on ncert text book only
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