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In the figure above, line AC represents a seesaw that is [#permalink]

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03 Oct 2006, 05:34

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In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C?

(1) x = 30 (2) Point B is 5 feet above the ground.

20. In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C?
(1) x = 30
(2) Point B is 5 feet above the ground.

A & B are unsufficient.

Combining

Sin 30(degree) = 5/AB
so AB can be determine and hence AC, since AC=2AB.

If we draw a perpendicula from line A which touches point C and call that point (on line A) D.

Answer is B.
Statement 2 alone is suff.
lets say: BA= x
then CA = 2x (since B is the mid point of CA)
Triangle ACD(D is the point where the perpendicular dropped from C touches the ground) is similar to triangle ABE(E is the point where the perpendicular dropped from C touches the ground).
Therefore x/5=2x/z
or z=10

Re: In the figure above, line AC represents a seesaw that is [#permalink]

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28 Nov 2013, 14:57

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Raghavender wrote:

20. In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C? (1) x = 30 (2) Point B is 5 feet above the ground.

Is there anyway someone can put the picture on the problem so we don't have to open it every time?

20. In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C? (1) x = 30 (2) Point B is 5 feet above the ground.

Is there anyway someone can put the picture on the problem so we don't have to open it every time?

Re: In the figure above, line AC represents a seesaw that is [#permalink]

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05 May 2015, 09:10

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For those of you who are not comfortable with the concept of similar triangles and its application, this question can be solved using application of simple trigonometric ratios.

In triangle ABD which is right angled at D, we can use \(Sin x = \frac{BD}{AB.}\)

Similarly, in triangle ACE which is right angled at E, we can use \(Sin x = \frac{CE}{AC}\)

From the above two equations, we can write \(\frac{BD}{AB} = \frac{CE}{AC}\) i.e. \(CE = \frac{(BD * AC)}{AB.}\)

Since \(AC = 2AB\), we can write \(CE = 2BD\).

So, for finding the height of point C from the ground we just need to know the height of point B from the ground.

We see that st-II provides us the height of B. Thus statement-II alone is sufficient to answer the question.

What i don't get is why we can assume that the two triangles are similar just because AB = BC?

I mean i looks similar, but why can we apply that?

Refer to the attached figure for description of the points.

BD and CE are perpendicular to AE.

So, in triangles ABD and ACE, angle A is common angle to both the triangles, \(\angle{ADB} = \angle{AEC} = 90\) and \(\angle {ABD} = \angle{ACE}\) (BD || CE)

Thus triangles ABD ad ACE are similar by AA (or angle -angle similarity theorem)

Thus, by similarity

AB/ AC = BD / CE

Given BD = 5 and AB = 0.5*AC

Thus CE = 10. Hence, Statement 2 is sufficient.

Per statement 1, x =30 does not provide us any other useful information.

Thus B is the correct answer.

Hope this helps.

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Similar triangles.jpg [ 11.3 KiB | Viewed 1689 times ]

Re: In the figure above, line AC represents a seesaw that is [#permalink]

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25 Jul 2015, 06:33

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noTh1ng wrote:

What i don't get is why we can assume that the two triangles are similar just because AB = BC?

I mean it looks similar, but why can we apply that?

Attachment:

Untitled.png [ 21.95 KiB | Viewed 1682 times ]

You can use three rules to prove that the triangles are similar. 1. AA- two angles are equal. 2. SSS- All three sides are proportional to each other. 3. SAS- One angle is equal and the two adjacent sides are proportional.

In this case, you know two angles are equal, x and the 90. Since the height of a point is being measured.

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