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area

This topic has 8 expert replies and 3 member replies
lukaswelker Senior | Next Rank: 100 Posts Default Avatar
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area

Post Thu Apr 17, 2014 8:42 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Hello guys

    I'm going to far in calculations and I get it wrong. There must be an easier way.

    Here goes the question,

    A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

    19200 : 19600 : 20000 : 20400 : 20800.

    Please let me know if you see an easier way.

    Many thanks
    Lukas

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    theCodeToGMAT Legendary Member
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    Post Thu Apr 17, 2014 9:14 am
    560 = 2 * (L + B)

    L + B = 280

    To find: L*B

    L^2 + B^2 = (200)^2
    (L+B)^2 -2*L*B = 200^2
    280^2 - 200^2 = 2 * L * B
    480 * 80 = 2 * L * B
    L * B = 19200

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    Post Thu Apr 17, 2014 9:26 am
    lukaswelker 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.
    Let L and W equal the length and width of the rectangle respectively.

    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: + 2LW + = 280²

    We already know that L² + W² = 200², so, we'll take + 2LW + = 280² and replace L² + W² with 200² to get:
    2LW + 200² = 280²
    So, 2LW = 280² - 200²
    Evaluate: 2LW = 38,400
    Solve: LW = 19,200 = A

    For extra practice, here's a similar question: http://www.beatthegmat.com/area-of-a-rectangle-t100119.html

    Cheers,
    Brent

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    osama_salah Newbie | Next Rank: 10 Posts Default Avatar
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    8 messages
    Post Mon Jun 27, 2016 5:29 am
    How did you calculate 280² - 200² = 38,400 ?

    Brent@GMATPrepNow wrote:
    lukaswelker 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.
    Let L and W equal the length and width of the rectangle respectively.

    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: + 2LW + = 280²

    We already know that L² + W² = 200², so, we'll take + 2LW + = 280² and replace L² + W² with 200² to get:
    2LW + 200² = 280²
    So, 2LW = 280² - 200²
    Evaluate: 2LW = 38,400
    Solve: LW = 19,200 = A

    For extra practice, here's a similar question: http://www.beatthegmat.com/area-of-a-rectangle-t100119.html

    Cheers,
    Brent

    Post Mon Jun 27, 2016 5:32 am
    osama_salah wrote:
    How did you calculate 280² - 200² = 38,400 ?
    Aside: 28² = 784, which means 280² = 78,400

    So, 280² - 200² = 78,400 - 40,000
    = 38,400

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    setiavipul Newbie | Next Rank: 10 Posts
    Joined
    06 Jun 2017
    Posted:
    1 messages
    Post Wed Jun 21, 2017 7:15 am
    Brent@GMATPrepNow wrote:
    lukaswelker 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.
    Let L and W equal the length and width of the rectangle respectively.

    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: + 2LW + = 280²

    We already know that L² + W² = 200², so, we'll take + 2LW + = 280² and replace L² + W² with 200² to get:
    2LW + 200² = 280²
    So, 2LW = 280² - 200²
    Evaluate: 2LW = 38,400
    Solve: LW = 19,200 = A

    For extra practice, here's a similar question: http://www.beatthegmat.com/area-of-a-rectangle-t100119.html

    Cheers,
    Brent
    Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?

    Post Wed Jun 21, 2017 7:22 am
    Quote:
    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
    ALWAYS LOOK FOR SPECIAL TRIANGLES.
    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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    Post Wed Jun 21, 2017 1:57 pm
    Hi All,

    It's important to remember that nothing about a GMAT question is ever 'random' - the wording and numbers/data are always carefully chosen. Thus, you can sometimes use the 'design' of a prompt to your advantage and spot the built-in patterns that are often there.

    Here, notice how ALL of the numbers are relatively 'nice', round numbers - even the diagonal is a nice number (and that doesn't happen very often when dealing with rectangles).... Since the answer choices are also round numbers, it's likely that the triangles that are 'hidden' in this rectangle are based on one of the common right-triangle patterns (in this case, the 3/4/5 - since 200 is a multiple of 5).

    Using that knowledge to our advantage, IF we had a 3/4/5 and the diagonal was 200, then that would be '40 times' 5... so the other two sides would be 40 times 4 and 40 times 3: 160 and 120. With those two side lengths, we'd have a perimeter of 2(160) + 2(120) = 560... and that is an exact MATCH for what we were told, so this MUST be the situation that we're dealing with.

    At this point, the area can be calculated easily enough: (L)(W) = (160)(120) = 19,200

    Final Answer: A

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    Rich

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    Post Thu Jun 22, 2017 5:09 am
    setiavipul wrote:
    Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?
    That rule doesn't apply to rectangles. It does, however apply to squares

    Cheers,
    Brent

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    Post Thu Jun 22, 2017 8:41 am
    Quote:
    Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?
    Note also, that if the figure were a square, and the perimeter were 560, each side would be 560/4 = 140. If each side of a 45/45/90 triangle were 140, the hypotenuse, or diagonal, would be 140 *rt2. So the fact that the diagonal is 200 tells us that we're not dealing with a square (or 45:45:90 triangles.)

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    Post Thu Jun 22, 2017 8:44 am
    DavidG@VeritasPrep wrote:
    Quote:
    Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?
    Note also, that if the figure were a square, and the perimeter were 560, each side would be 560/4 = 140. If each side of a 45/45/90 triangle were 140, the hypotenuse, or diagonal, would be 140 *rt2. So the fact that the diagonal is 200 tells us that we're not dealing with a square (or 45:45:90 triangles.)
    One more fun note. For a set perimeter, the rectangle with the greatest area is a square. If the shape in question had been a square, and the sides had each been 140, the area would have been 140*140 = 19,600. Because we know the shape is not a square, we know the area would be less than 19,600. The answer must be A

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    Post Thu Jun 22, 2017 4:50 pm
    DavidG@VeritasPrep wrote:
    DavidG@VeritasPrep wrote:
    Quote:
    Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?
    Note also, that if the figure were a square, and the perimeter were 560, each side would be 560/4 = 140. If each side of a 45/45/90 triangle were 140, the hypotenuse, or diagonal, would be 140 *rt2. So the fact that the diagonal is 200 tells us that we're not dealing with a square (or 45:45:90 triangles.)
    One more fun note. For a set perimeter, the rectangle with the greatest area is a square. If the shape in question had been a square, and the sides had each been 140, the area would have been 140*140 = 19,600. Because we know the shape is not a square, we know the area would be less than 19,600. The answer must be A
    I love that, nice hack!

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