What is the area of the triangle ABC? (1) y^2 - 14y + 48 = 0

Question

https://gmatclub.com/forum/download/file.php?id=64141 What is the area of the triangle ABC? (1) y^2 - 14y + 48 = 0 (2) y^2 - 16y + 60 = 0 1.jpg

sanjitscorps18 · Answer

For this figure to be a triangle, we have 0 < y < 10

(1) y^2−14y+48=0
Solving this we get y = 8 or 6. Two values hence insufficient

(2) y^2−16y+60=0
Solving this we get y = 10 or 6. But y < 10 hence we get value of y = 6. Sufficient

Area can be found out by the Heron's formula

Option B

IanStewart · Answer

This solution is perfect -- I'd just point out that there's no need to use Heron's formula to find the area here (not that we need to calculate the area in DS, but if it were PS, we might need to). Since the triangle is isosceles, a height drawn between the two equal sides will cut the base in half. Since the base is 6, the height will divide the isosceles triangle into two right triangles, with hypotenuse 5 and base 3, so in this case it will divide the triangle into two 3-4-5 triangles, and the height will be 4. Since the base is 6 the area will be 4*6/2 = 12.

What is the area of the triangle ABC? (1) y^2 - 14y + 48 = 0

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