For (1):
Let the legs be \(x\) and \(y\).

Given: \(h=10\) and \(\frac{xy}{2}=25\) (area of the right triangle \(area=\frac{leg_1*leg_2}{2}\)) --> \(h^2=100=x^2+y^2\) (Pythagoras) and \(xy=50\).

Q: \(P=x+y+10=?\) So we have to calculate the value \(x+y\).

Square \(x+y\) --> \((x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200\) --> \(x+y=\sqrt{200}\).

Thus \(P=x+y+10=\sqrt{200}+10\).

Sufficient.

Bunuel please let me know if this method I used below is correct for statement 1.

1. 1/2 b*h = 25
So, b= 50/h

Also,
b^2+h^2=100
putting b=50/h in this equation we get,

h^2(100-h^2)=2500

Therefore h=b=50^1/2.

Hence sufficient.

Hi Bunuel
Quick question,can i apply the formula 1:square root3:2
I got answer of 5+10+5squareroot of3

Posted from my mobile device

You are assuming with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=8\).

For example: \(a=1\) and \(b=\sqrt{99}\) or \(a=2\) and \(b=\sqrt{96}\) ...

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

Hello Bunuel
Can we consider hypotenuse=10; and height=10 in statement 2? I asked this question because there is no strict condition that the base and height MUST be the equal legs.
Thanks__

We are given that the hypotenuse of a right triangle is 10 cm and we need to determine the perimeter of the triangle. We can let the other two sides (i.e., the two legs) of the right triangle be a and b. Since it’s a right triangle, by the Pythagorean theorem, we have a^2 + b^2 = 10^2, or a^2 + b^2 = 100. If we can find the values of a and b, then we can determine the perimeter of the triangle, since it will be a + b + 10.

Statement One Alone:

The area of the triangle is 25 square centimeters.

We are given a right triangle and we’ve let the two non-hypotenuse sides be a and b. Recall that the two non-hypotenuse sides of a right triangle are actually the base and height of the triangle, so the area of this triangle is A = ab/2. From the aforementioned equation a^2 + b^2 = 100, we can solve b as b = √(100 - a^2). Since we are given that A = 25, and substituting b = √(100 - a^2) in A = ab/2, we can say:

25 = [a√(100 - a^2)]/2

We see that we can solve for a (though we don’t have to actually solve for it). And once we’ve solved for a, we can determine the value of b, since b = √(100 - a^2). Therefore, we can determine the perimeter of the triangle, since we can determine the values of both a and b. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The 2 legs of the triangle are of equal length.

Since a^2 + b^2 = 100 and we are given that a = b, we can say:

a^2 + a^2 = 100

We see that we can solve for a (though we don’t have to actually solve for it). And once we’ve solved for a, we can determine the value of b, since b = a. Therefore, we can determine the perimeter of the triangle, since we can determine the values of both a and b. Statement two alone is sufficient to answer the question.

Answer: D

I think there is a faster way to evaluate statement 1.
Area=25 => b*h=50 => b and h are factors of 50. Possible pairs: (1,50); (2,25); (5,10).
Remember: this triangle is a right triangle, which means the hypotenuse is supposed to be the largest side of this triangle.
So out of these three pairs only one pair (5,10). Sufficient. Bunuel VeritasKarishma what do you think?