marcusaurelius wrote:
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.
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
Something I always struggle with when given an equation with exponents such as the one Scott solved for in statement (1) (25 = [a√(100 - a^2)]/2) is how do we know that there aren't multiple solutions to the equation? (i.e. like when you're factoring the form of aX^2+bx+c). Is there any indication that there will NOT be multiple possible solutions from the equation without fully factoring out/solving?