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Re: Right triangle ABC and rectangle EFGH have the same perimete
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17 Apr 2020, 05:04
shadowmr20 wrote:
why is the answer 4?
If we assume that side BC is the hypotenuse of the triangle, then we can apply the Pythagorean theorem to determine that BC = 5 So the perimeter of the triangle = 3 + 4 + 5 = 12
The perimeter of the rectangle = x + x + 2 + 2 = 2x + 4
Since we're told that triangle ABC and rectangle EFGH have the same perimeter, we can write: 2x + 4 = 12 Solve to get x = 4
Re: Right triangle ABC and rectangle EFGH have the same perimete
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15 Jun 2021, 22:29
2
Triangle ABC is a right triangle, so this must be a 3–4–5 triangle, and the length of side BC is 5. That means the perimeter of triangle ABC is 3 + 4 + 5 = 12. Thus, the perimeter of rectangle EFGH is also 12. Using that information, find x: 2 × (2 + x) = 12. 4 + 2x = 12 2x = 8 x = 4
Re: Right triangle ABC and rectangle EFGH have the same perimete
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15 Jun 2021, 23:46
1
Assuming that triangle BAC is a right-angled triangle, then by the Pythagorean theorem, BC = 5. With all sides completed now, Perimeter (triangle BAC) = 3+4+5 =12.
Now, equate this (12) with the perimeter of rectangle EFGH (2L+2W)= 2x + 2 + 2 = 2x+4
12 = 2x + 4 (subtract 4 from each side so as to balance the equation) 12-4 = 2x+4-4 8=2x (dividing by 2 to solve for x yields the following)
2x/2= 8/2 x=4
gmatclubot
Re: Right triangle ABC and rectangle EFGH have the same perimete [#permalink]