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555-605 (Medium)|   Geometry|      
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Bunuel
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In triangle ABC above, the altitude from A to BC meets BC at D and the 12 Jun 2020, 06:54
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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


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In triangle ABC above, the altitude from A to BC meets BC at D and the 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

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In triangle ABC above, the altitude from A to BC meets BC at D and the 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)

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In triangle ABC above, the altitude from A to BC meets BC at D and the 12 Jun 2020, 07:32
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Bunuel

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

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Solution


    • Based on the given information, let’s redraw the figure as below:

    • In right angled triangle ADB , \(AB =\sqrt{3^2+4^2} = 5 \)
    • We have \(\frac{BD}{AB} = \frac{HD}{AH}\)
      o Using, AD = AH +HD, we can write the above equation as,
         \(\frac{AB}{BD} =\frac{AD -HD}{HD}\)
        \(⟹\frac{5}{3} = \frac{4}{HD} – 1\)
        \(⟹\frac{8}{3} = \frac{4}{HD}\)
        \(⟹ HD = \frac{3}{2}\)
Thus, the correct answer is Option B.
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In triangle ABC above, the altitude from A to BC meets BC at D and the 12 Jun 2020, 07:53
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In right angled triangle ABD, hypothenuse AB=5 (from pythagoras theorem, the other sides are 3 and 4 - given in question stem)

\(\frac{BD}{AB} = \frac{HD}{AH}\)

or, \(\frac{3}{AB} = \frac{HD}{4-HD}\)

Solving for HD,
HD = \(\frac{12}{3+AB} \)
Substituting value of AB in the equation
\(HD = \frac{3}{2}\)
Answer: B
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In triangle ABC above, the altitude from A to BC meets BC at D and the 12 Jun 2020, 22:32
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AD=4; BD=3; CD=2
BD / AB = HD / AH
HD=?
AH = 4- HD

AD is altitude, AB^2 = AD^2 + BD^2
AB^2 = 4^2 + 3^2 = 16 + 9 = 25
AB = 5

Now use the ratio, as given in question stem:
BD / AB = HD / AH
3 / 5 = HD / [ 4- HD]
12- 3*HD = 5*HD
12 = 8*HD
Thus HD = 12/8 = 6/4 = 3/2

Answer : B
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In triangle ABC above, the altitude from A to BC meets BC at D and the 13 Jun 2020, 23:02
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Kindly see the attachment
IMO B
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In triangle ABC above, the altitude from A to BC meets BC at D and the 14 Jun 2020, 21:08
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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 ?



in ABD triangle

Since AD is altitude
\(AB^2 = AD^2+BD^2\\
\)

\(AB^2 = 16+9\\
\)

AB = 5


\(\frac{ BD}{AB }\) = \(\frac{ HD}{AH }\)

\(\frac{3}{5 }\) = \(\frac{ HD}{AH}\)


\(\frac{(3+5) }{ 5 }\) = \(\frac{ (HD + AH) }{ AH }\)


\(\frac{8}{5 }\) = \(\frac{ AB }{ AH }\)


AH = \(\frac{(AD * 5) }{ 8 }\) = 2.5

so HD = 4-2.5 = 1.5

Answer is B
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