Triangle ABC is inscribed in a rectangle ABEF forming two right triang

The altitude of equilateral triangle is √3/2 a, where a is the length of the equilateral triangle. Based on this rule, lets look at the two statements.

1) BE = (\(\sqrt{3}AB)/2\) - The altitude is defined, however we don't know if BE is the altitude.
2) Point C is the midpoint of EF - Insufficient, does not define the altitude.

Combining 1+2, the altitude has a length of √3a/2. Hence sufficient. C.

Here in this explanation, in (1) we not assuming that C lies on BF. But if C does not lie on BF, then triangle ABC can not form two RATs in rectangle ABEF. So, it goes without saying that C lies on EF and thus BE is its altitude. (since AB is the base). So statement 1 is alone sufficient.
Please correct me if i am wrong.

Hey Santhosh,

You're partly right! C has to lie on EF. Is that alone sufficient to say whether the triangle is equilateral? Certainly no! So this is not sufficient. See if the below explanation helps.

To prove the triangle is an equilateral, we need to prove all 3 sides of triangle are equal to the length of the rectangle.

Is AC=AB?

(1) BE = ( 3‾√AB)/2 - Nothing can be said as we don't know where the point C lies on EF. Not sufficient.

(2) Point C is the midpoint of EF

By Pythagoras Theorem, AC^2 = (AB/2)^2 + (BE)^2

AC^2 = AB^2/4 + BE^2 - We still can't tell whether AB=AC. So not sufficient.

From (1) lets substitute the value of BE in above equation and see where it takes us, we end up with AC^2 = AB^2. Hence AC=AB. If we take the other side of the triangle BC, we'll end up in the same result. Hence AC=BC=AB.

So the triangle is an equilateral one.

Hence C it is!

Cheers!

Do we really need trigonometry here to solve this? I think not
We can easily conclude that either of 2 given statements is not sufficient. Combining 1 and 2, we get in triangle AFC:
(root3/2*AB)^2 +1/2 (AB)^2 = AC^2
This gives AC = AB. From there on we can prove ABC is equilateral.

Why couldn't answer be (B)

1) Not Sufficient

2) C is the midpoint of EF.
Let's suppose EC = AB = b,
and AF = BE = a

Since rectangle ABEF forming two right triangles: AFC and BEC

So CA = CB = root(a2 + (b2/4)

it means the triangle would be an isosceles triangle. Sufficient.

So , the answer would be 'B'

Mehn...! This explanation is phenomenal. Thanks a lot for sharing.

With Time a luxury, such a long approach as given in official solution is unnecessary.

Condition 1: BE= Sqrt(3)/2 AB is insufficient to prove that AB=AC=BC( equilateral triangle)
Condition 2: C is the midpoint of EF. This doesn't tell anything about length of BE. So again insufficient to prove AB=AC=BC.

Combine 1 & 2: BE= Sqrt(3)/2 AB and EC=AB/2 we get, BC= AB
similarly, AC=AB.
Therefore, AB=AC=BC (equilateral triangle)
Therefore Answer is C.

VeritasKarishma

Can you pls provide a better explanation to this answer w/o using trigonometry

Thanks in advance!

There are many ways of inscribing ABC triangle in rectangle ABEF. We draw some of them here.

Now when will ABC be an equilateral triangle? First of all, C should be the mid-point of EF. Only then will AC = BC which is necessary for equilateral triangle.
Also, the altitude of an equilateral triangle has a fixed relation with the side. It is sqrt(3)*s/2.

Since length BE will be the altitude of the triangle, then BE must be sqrt(3)*AB/2.

Are these two enough to make an equilateral triangle? Yes. Think.
I draw a random length AB. At its mid-point, I draw an altitude of sqrt(3)*AB/2. Now will the triangle so formed be equilateral. Yes. Then these two are sufficient.

Both statements together give us this information.

Answer (C)

*Edited

VeritasKarishma
Why are we equating BE to BC?
The altitude of triangle ABC will be equal to BE (because parallel), but BC is not parallel

According to statement 1, BE=\(\sqrt{3} AB/2\). Now point C has to lie on side EF, otherwise we wouldn't get 2 right triangles. Also irrespective of assuming the position of C initially, altitude of triangle will be parallel to BE & so, area of triangle will be = 1/2*AB (which is base)*\(\sqrt{3} AB/2\)(which is altitude)= \(\sqrt{3} AB^2/4\) = Formula of area for equilateral triangle

Pls clarify

That is a typo. BE is the altitude and BE = sqrt(3)*AB/2
BC is a side of the triangle.

VeritasKarishma

Can you pls explain if the following explanation has any shortcoming in proving that statement 1 is sufficient

According to statement 1, BE=\(\sqrt{3} AB/2\). Now point C has to lie on side EF, otherwise we wouldn't get 2 right triangles. Also irrespective of assuming the position of C initially, altitude of triangle will be parallel to BE & so, area of triangle will be = 1/2*AB (which is base)*\(\sqrt{3} AB/2\)(which is altitude)= \(\sqrt{3} AB^2/4\) = Formula of area for equilateral triangle

Pls clarify

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