In Fig. 15.74, compute the aresa of quadrilateral ABCD.
Answers
Given : From the figure , DC = 17 cm , AD = 9 cm and BC = 8 cm
In Δ BCD, we have by using Pythagoras theorem :
CD² = BD² + BC²
(17)² = BD² + (8)²
289 = BD² + 64
BD² = 289 - 64
BD² = 225
BD = 15 cm
Area of ∆ = ½ × base × height
Area of ∆ BCD = ½ × BC × CD
Area of ∆ BCD = ½ × 8 × 17
Area of ∆ BCD = 4 × 17
Area of ∆ BCD = 68 cm²
In Δ ABD, we have by using Pythagoras theorem :
BD² = AB² + AD²
(15)² = AB² + (9)²
AB² = 225 – 81
AB² = 144
AB = √144
AB = 12 cm
Area of ∆ ABD = ½ × AB × AD
Area of ∆ ABD = ½ × 12 × 9
Area of ∆ ABD = 6 × 9
Area of ∆ ABD = 54 cm²
Therefore,
Area of Quadrilateral ABCD = Area (Δ ABD) + Area (Δ BCD)
Area of quadrilateral ABCD = 54 + 68
Area of quadrilateral ABCD = 112 cm²
Hence the area of quadrilateral ABCD is 112 cm².
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Answer:-
Given :
From the figure ,
DC = 17 cm , AD = 9 cm and BC = 8 cm
In Δ BCD, we have by using Pythagoras theorem :
CD² = BD² + BC²
(17)² = BD² + (8)²
289 = BD² + 64
BD² = 289 - 64
BD² = 225
BD = 15 cm
Area of ∆ = ½ × base × height
Area of ∆ BCD = ½ × BC × CD
Area of ∆ BCD = ½ × 8 × 17
Area of ∆ BCD = 4 × 17
Area of ∆ BCD = 68 cm²
In Δ ABD, we have by using Pythagoras theorem :
BD² = AB² + AD²
(15)² = AB² + (9)²
AB² = 225 – 81
AB² = 144
AB = √144
AB = 12 cm
Area of ∆ ABD = ½ × AB × AD
Area of ∆ ABD = ½ × 12 × 9
Area of ∆ ABD = 6 × 9
Area of ∆ ABD = 54 cm²
Therefore,
Area of Quadrilateral ABCD = Area (Δ ABD) + Area (Δ BCD)
Area of quadrilateral ABCD = 54 + 68
Area of quadrilateral ABCD = 112 cm²
Hence the area of quadrilateral ABCD is 112 cm².
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