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02 Apr 2020, 01:24
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[GMAT math practice question]
The figure shows a right triangle \(ABC\) with \(∠A = 90^o, BH = 8,\) and \(CH = 4.\) Moreover, \(AH\) and \(BC\) are perpendicular to each other. What is the length of \(AH\)?
4.2PS.png [ 5.55 KiB | Viewed 1493 times ]
A. \(\sqrt{2}\)
B. \(3\sqrt{2}\)
C. \(4\sqrt{2}\)
D. \(3\)
E. \(4\)
The figure shows a right triangle \(ABC\) with \(∠A = 90^o, BH = 8,\) and \(CH = 4.\) Moreover, \(AH\) and \(BC\) are perpendicular to each other. What is the length of \(AH\)?
Attachment:
4.2PS.png [ 5.55 KiB | Viewed 1493 times ]
A. \(\sqrt{2}\)
B. \(3\sqrt{2}\)
C. \(4\sqrt{2}\)
D. \(3\)
E. \(4\)
02 Apr 2020, 01:41
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Bookmarks
Consider side AC as ‘a’ and side AB as ‘b’
Using Pythagoras theorm:
1- triangle AHC ; x^2+16= a^2
2- Triangle AHB ; x^2+ 64= b^2
3- Triangle ABC ; a^2 + b^2 = 144
Equate these 3 ie equations 1+2 in 3
2x^2+80=144
x^2=32
Hence x=4root2
C is the answer
Posted from my mobile device
Using Pythagoras theorm:
1- triangle AHC ; x^2+16= a^2
2- Triangle AHB ; x^2+ 64= b^2
3- Triangle ABC ; a^2 + b^2 = 144
Equate these 3 ie equations 1+2 in 3
2x^2+80=144
x^2=32
Hence x=4root2
C is the answer
Posted from my mobile device
02 Apr 2020, 13:21
Kudos
Bookmarks
BAC ~ BHA
\(\frac{AB}{BC} = \frac{BH}{AB}\)
\(AB^2 = BC*BH\)
\(x^2+8^2 = 12*8 \)
\(x^2 = 32\)
\(x = 4\sqrt{2}\)
\(\frac{AB}{BC} = \frac{BH}{AB}\)
\(AB^2 = BC*BH\)
\(x^2+8^2 = 12*8 \)
\(x^2 = 32\)
\(x = 4\sqrt{2}\)
MathRevolution
[GMAT math practice question]
The figure shows a right triangle \(ABC\) with \(∠A = 90^o, BH = 8,\) and \(CH = 4.\) Moreover, \(AH\) and \(BC\) are perpendicular to each other. What is the length of \(AH\)?
A. \(\sqrt{2}\)
B. \(3\sqrt{2}\)
C. \(4\sqrt{2}\)
D. \(3\)
E. \(4\)
The figure shows a right triangle \(ABC\) with \(∠A = 90^o, BH = 8,\) and \(CH = 4.\) Moreover, \(AH\) and \(BC\) are perpendicular to each other. What is the length of \(AH\)?
Attachment:
4.2PS.png
A. \(\sqrt{2}\)
B. \(3\sqrt{2}\)
C. \(4\sqrt{2}\)
D. \(3\)
E. \(4\)
