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05 Jun 2020, 01:39
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[GMAT math practice question]
The figure shows \(□ABCD\) with \(AB = 8, BC = 17\) and \(CD = 9\) and \(∠BAC = ∠ADC = 90^o\). What is the area of \(□ABCD\)?
6.5PS.png [ 20.09 KiB | Viewed 2374 times ]
A. \(81\)
B. \(92\)
C. \(103\)
D. \(114\)
E. \(125\)
The figure shows \(□ABCD\) with \(AB = 8, BC = 17\) and \(CD = 9\) and \(∠BAC = ∠ADC = 90^o\). What is the area of \(□ABCD\)?
Attachment:
6.5PS.png [ 20.09 KiB | Viewed 2374 times ]
A. \(81\)
B. \(92\)
C. \(103\)
D. \(114\)
E. \(125\)
ArunSharma12
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05 Jun 2020, 02:06
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MathRevolution
In triangle ABC,
\(AC^2 = 17^2 - 8^2\)
\(AC = 15\)
In triangle ADC,
\(AD^2 = 15^2 - 9^2\)
\(AD = 12\)
area of ABCD = area of ABC + area of ADC
area of ABCD = \(\frac{8*15 + 9*12}{2} = 114\)
yashikaaggarwal
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05 Jun 2020, 03:27
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Area of Quadrilateral ABCD is = Area of ABC + Area of ADC
=> Area of ABC = 1/2*base*height
=> 1/2*AB*AC
=> AB = 8
=> AC = (AC^2+AB^2=BC^2)
=> AC^2+8^2=17^2
=> AC^2 = 289-64 => AC^2 = 225
=> AC = 15
Area of ABC = 1/2*8*15 = 60 -----------(1)
Area of ACD = 1/2*base*height
=> 1/2*AD*CD
=> CD =9
=> AD^2+CD^2=AC^2
=> AD^2+9^2=15^2
=> AD^2 = 225-81
=> AD^2 = 144
=> AD = 12
Area of ADC = 1/2*12*9 => 54 ---------(2)
Area of ABCD = 54+60 => 114
Answer is D
Posted from my mobile device
=> Area of ABC = 1/2*base*height
=> 1/2*AB*AC
=> AB = 8
=> AC = (AC^2+AB^2=BC^2)
=> AC^2+8^2=17^2
=> AC^2 = 289-64 => AC^2 = 225
=> AC = 15
Area of ABC = 1/2*8*15 = 60 -----------(1)
Area of ACD = 1/2*base*height
=> 1/2*AD*CD
=> CD =9
=> AD^2+CD^2=AC^2
=> AD^2+9^2=15^2
=> AD^2 = 225-81
=> AD^2 = 144
=> AD = 12
Area of ADC = 1/2*12*9 => 54 ---------(2)
Area of ABCD = 54+60 => 114
Answer is D
Posted from my mobile device
