A small rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?
a/ 19200
b/ 19600
c/ 20000
d/ 20400
e/ 20800
Correct answer: A
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PS: Diagonal of a Rectangle / Area
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Brent@GMATPrepNow
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Let L and W equal the length and width of the rectangle respectively.melanie.espeland wrote:A small rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?
a/ 19200
b/ 19600
c/ 20000
d/ 20400
e/ 20800
perimeter = 560
So, L + L + W + W = 560
Simplify: 2L + 2W = 560
Divide both sides by 2 to get: L + W = 280
diagonal = 200
The diagonal divides the rectangle into two RIGHT TRIANGLES.
Since we have right triangles, we can apply the Pythagorean Theorem.
We get L² + W² = 200²
NOTE: Our goal is to find the value of LW [since this equals the AREA of the rectangle]
If we take L + W = 280 and square both sides we get (L + W)² = 280²
If we expand this, we get: L² + 2LW + W² = 280²
Notice that we have L² + W² "hiding" in this expression.
That is, we get: L² + 2LW + W² = 280²
We already know that L² + W² = 200², so, we'll take L² + 2LW + W² = 280² and replace L² + W² with 200² to get:
2LW + 200² = 280²
So, 2LW = 280² - 200²
Evaluate: 2LW = 38,400
Solve: LW = [spoiler]19,200 = A[/spoiler]
For extra practice, here's a similar question: https://www.beatthegmat.com/area-of-a-re ... 00119.html
Cheers,
Brent
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GMATGuruNY
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ALWAYS LOOK FOR SPECIAL TRIANGLES.melanie.espeland wrote:A small rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?
a/ 19200
b/ 19600
c/ 20000
d/ 20400
e/ 20800
Draw the rectangle and its diagonal:
Since diagonal AD is a multiple of 5, check whether ∆ABD is a multiple of a 3:4:5 triangle.
If each side of a 3:4:5 triangle is multiplied by 40, we get::
(40*3):(40*4):(40*5) = 120:160:200.
The following figure is implied:
Check whether the resulting perimeter for rectangle ABCD is 560:
120+160+120+160 = 560.
Success!
Implication:
For the perimeter of rectangle ABCD to be 560, ∆ABD must be a multiple of a 3:4:5 triangle with sides 120, 160 and 200.
Thus:
Area of rectangle ABCD = L * W = 160 * 120 = 19200.
The correct answer is A.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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Mathsbuddy
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Both solutions given so far work well.
However, you could do them faster by scaling everything down to more manageable figures:
Perhaps perimeter = 56 and diagonal = 20
Perhaps perimeter = 28 and diagonal = 10
Perhaps perimeter = 14 and diagonal = 5
The last one here is simplest, by dividing all by 40
Now the solution seems obvious:
P = 14 = 7*2
so length + width = 7
Easy 3-4-5 triangle is evident
Area = 3 x 4 x 40 x 40
Consider only the last non-zero digits: 3 x 4 x 4 x 4 = ...2
Only answer A bears a "2" as a last non-zero digit.
However, you could do them faster by scaling everything down to more manageable figures:
Perhaps perimeter = 56 and diagonal = 20
Perhaps perimeter = 28 and diagonal = 10
Perhaps perimeter = 14 and diagonal = 5
The last one here is simplest, by dividing all by 40
Now the solution seems obvious:
P = 14 = 7*2
so length + width = 7
Easy 3-4-5 triangle is evident
Area = 3 x 4 x 40 x 40
Consider only the last non-zero digits: 3 x 4 x 4 x 4 = ...2
Only answer A bears a "2" as a last non-zero digit.
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Mathsbuddy
- Master | Next Rank: 500 Posts
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Both solutions given so far work well.
However, you could do them faster by scaling everything down to more manageable figures:
Perhaps perimeter = 56 and diagonal = 20
Perhaps perimeter = 28 and diagonal = 10
Perhaps perimeter = 14 and diagonal = 5
The last one here is simplest, by dividing all by 40
Now the solution seems obvious:
P = 14 = 7*2
so length + width = 7
Easy 3-4-5 triangle is evident
Area = 3 x 4 x 40 x 40
Consider only the last non-zero digits: 3 x 4 x 4 x 4 = ...2
Only answer A bears a "2" as a last non-zero digit.
However, you could do them faster by scaling everything down to more manageable figures:
Perhaps perimeter = 56 and diagonal = 20
Perhaps perimeter = 28 and diagonal = 10
Perhaps perimeter = 14 and diagonal = 5
The last one here is simplest, by dividing all by 40
Now the solution seems obvious:
P = 14 = 7*2
so length + width = 7
Easy 3-4-5 triangle is evident
Area = 3 x 4 x 40 x 40
Consider only the last non-zero digits: 3 x 4 x 4 x 4 = ...2
Only answer A bears a "2" as a last non-zero digit.
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Matt@VeritasPrep
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We could also do some fancypants algebra here.
We know that 2(a+b) = 560 and that a² + b² = 200². We want to find a*b.
If we take (a+b) and square it, we get (a+b)², or a² + 2ab + b². That gives us most of what we want ... to get to ab, we just need to subtract a² + b². Hence we know that (a+b)² - (a² + b²) will give us 2ab.
That means that
(a+b)² - (a² + b²) = 2ab
or
280² - 200² = 2ab
or
(280+200)(280-200) = 2ab
or
480 * 80 = 2ab
or
240 * 80 = ab
Hence ab = 19200, and we're done!
The identity (a+b)² - (a² + b²) = 2ab is useful to remember, and could show up again on test day.
We know that 2(a+b) = 560 and that a² + b² = 200². We want to find a*b.
If we take (a+b) and square it, we get (a+b)², or a² + 2ab + b². That gives us most of what we want ... to get to ab, we just need to subtract a² + b². Hence we know that (a+b)² - (a² + b²) will give us 2ab.
That means that
(a+b)² - (a² + b²) = 2ab
or
280² - 200² = 2ab
or
(280+200)(280-200) = 2ab
or
480 * 80 = 2ab
or
240 * 80 = ab
Hence ab = 19200, and we're done!
The identity (a+b)² - (a² + b²) = 2ab is useful to remember, and could show up again on test day.
