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12 Jun 2020, 06:54
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
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Question Stats:
77% (03:00) correct
23%
(03:15)
wrong
based on 53
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In triangle ABC above, the altitude from A to BC meets BC at D and the altitude from B to CA meets AD at H. If AD = 4 cm, BD = 3 cm and CD = 2 cm and if BD/AB = HD/AH, then what is the the length of HD ?
A. \(\frac{\sqrt{5}}{2}\) cm
B. 3/2 cm
C. \(\sqrt{5}\) cm
D. 2.5 cm
E. 3 cm
Attachment:
1.png [ 5.74 KiB | Viewed 4811 times ]
yashikaaggarwal
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Joined: 19 Jan 2020
Last visit: 29 Mar 2026
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12 Jun 2020, 07:01
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Since AD is altitude
AB^2 = AD^2+BD^2
AB^2 = 4^2+3^2
AB^2 = 16+9
AB^2 = 25
AB = 5
=> Also, BD/AB = HD/AH
3/5 = HD/4-HD
12-3HD = 5HD
12 = 8HD
12/8 = HD
3/2 = HD
Answer is B
Posted from my mobile device
AB^2 = AD^2+BD^2
AB^2 = 4^2+3^2
AB^2 = 16+9
AB^2 = 25
AB = 5
=> Also, BD/AB = HD/AH
3/5 = HD/4-HD
12-3HD = 5HD
12 = 8HD
12/8 = HD
3/2 = HD
Answer is B
Posted from my mobile device
12 Jun 2020, 07:14
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Since ADB is a right-angled triangle, AB will be equal to 5.
—> \(\frac{HD}{AH }= \frac{3}{5 }\)
\(HD + AH = 4\)
\((\frac{5}{3}) HD + HD = 4 \)
\(HD =\frac{ 4 *3}{ 8 }= \frac{12}{8 }= 1.5 \)
Answer ( B)
Posted from my mobile device
—> \(\frac{HD}{AH }= \frac{3}{5 }\)
\(HD + AH = 4\)
\((\frac{5}{3}) HD + HD = 4 \)
\(HD =\frac{ 4 *3}{ 8 }= \frac{12}{8 }= 1.5 \)
Answer ( B)
Posted from my mobile device
