Find the Area of ∆ABC? 1) Two sides of the triangle have lengths 3 an

For a triangle to be a right angled triangle, it needs to satisfy the Pythagorean Theorem. So the sum of the squares of the base and height should be equal to the square of the hypotenuse.

Numbers that satisfy the theorem are usually referred to as the pythagorean Triplets.

As far as the above question is concerned, the only triplet that satisfies the options is (3,4,5).

Statement 1: Just lengths of two sides is provided, with no information being provided for which side is which. Not sufficient.

Statement 2: It is provided that the triangle is a right angled triangle, hence the theorem mentioned above would apply. Also, perimeter would be less that 15. (3+4+5=11). But statement 2 by itself would be insufficient.

Both the statements, if used together are sufficient.

Hence, option C.

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From Statement 1: third side s can be anything between 2 and 8. i.e.. 3,4,5,6 & 7 and any other real numbers be 2 and 8.

From Statement 2: third side s can be anything less than 7, though ABC is a right angle triangle

From Statements 1 and 2: third side can be any real number between 2 and 7.

Hence, OA: E

Hi, GMATinsight didn't mention anywhere that ABC needs to be a right angle triangle. So, the value of the third side can be anywhere between 2 and 7.
Hence, the area cannot be fixed.

nkme2007

Second statement states that ∆ABC is a Right ∆ABC

I hope you noticed that. I am not endorsing any answer for this question as of now though

Do not miss out taht stmt 2 says its right angled triangle.

The only right angled triangle with perimeter less than or equal to 15 is 3 4 5 triplet.

We can definitely find area based on this .

IMO C

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1) Two sides of the triangle have lengths 3 and 5

No info about the third side

1 is not sufficient

2) Right ∆ABC has Perimeter ≤15

We know that the sides are in the ratio \(1:1:\sqrt2\) in a \(45-45-90\) right triangle and \(1:\sqrt3:2\) in a \(30-60-90\) right triangle

So it is possible for ABC to have side lengths \(1,1,\sqrt2\) and \(1,\sqrt3,2\) which means we can have multiple values for the area

2 is not sufficient

(1)+(2)

If the legs are \(3\) and \(5\), the area is \(\frac{15}{2}\)

If one leg is \(3\) and hypotenuse \(5\), the other leg is \(4\) and the area is \(6\)

(1)+(2) is not sufficient

Answer is (E)

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Find the Area of ∆ABC?

1) Two sides of the triangle have lengths 3 and 5
Clearly insufficient. The last side could be anything. We also don't know what the base or height is.

2) Right ∆ABC has Perimeter ≤15
Clearly insufficient since there are multiple possible combinations for the side lengths. Even if we were able to figure out what that third side is (we can't), we still wouldn't know what the base or height is.
Insufficient.

Combined:
Same issue exists.
Insufficient.

E

why is C incorrect?

we can deduce from (2) that third side is 4 - and therefore in a right triangle, the legs would be 3 and 4 (given the 3-4-5 tripet). Please explain