In the figure above, line AC represents a seesaw that is

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

Hi All,

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.

You can try out a similar question at a-squirrel-climbs-a-straight-wire-from-point-a-to-point-c-if-b-is-the-195315.html#p1517662

Hope its clear!

Regards
Harsh
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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.

So, Sin 30 = AD/2AB
Answer is C.

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?

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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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.

Hope it helps.

chetan2u VeritasKarishma i dont get how the angles below are equal

\(\angle {ABD} = \angle{ACE}\) (BD || CE)

is it only because BD || CE the line AC as it goes up the angle degree is changing so logically angle ABD should be smaller than that of ACE .... is my reasoning incorrect ? if yes how can these angles be equal

ADB and ACE are similar triangles do all angles are equal.

Now why do?
Angle ADB and ACE are perpendicular to the ground, so both are 90.
Angle BAD=Angle CAE... both are same angle.
Thus the third angle has to be same
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"The midsegment of a triangle is always half the length of the third side."

/Bunuel


Therefore B.

Can I trust that the line is straight? It is not mentioned

Check the instructions you get before the exam:







I'd advice to familiarize yourselves with the above, especially pay attention to the parts in red boxes.

Here is a part you are interested in:




Hope it helps.
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