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Re: Right triangle ABC and rectangle EFGH have the same perimete [#permalink]
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

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Re: Right triangle ABC and rectangle EFGH have the same perimete [#permalink]
so if this question appears as a QC , should the option D be the answer ?
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Re: Right triangle ABC and rectangle EFGH have the same perimete [#permalink]
2
fabiha22 wrote:
so if this question appears as a QC , should the option D be the answer ?


It all depends on how the QC question was worded.

If the question were:
QUANTITY A: length of side AB
QUANTITY B: length of side EF
Then the correct answer would be A

If the question were:
QUANTITY A: length of side AB
QUANTITY B: length of side FG
Then the correct answer would be B
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Re: Right triangle ABC and rectangle EFGH have the same perimete [#permalink]
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
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Re: Right triangle ABC and rectangle EFGH have the same perimete [#permalink]
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
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