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shaded region perimeter and area ?
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EricLien9122
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A quick way to solve this question is just add 2+2=4, because the hypotenuse has square root (2) included in the answer choice.
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iamcste
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First part is use of pythagoras theorem
It forms a Trapezium
A=1/2( sum of parallel sides)* height
Height is slightly tricky= sq rt 2 ( Pls refer attachment to calculate height)
Diagram may not be required but in case you want to refer
Areas= 1/2* 6 sq rt 2* sq rt 2
=(1/2)*6*2
=6
It forms a Trapezium
A=1/2( sum of parallel sides)* height
Height is slightly tricky= sq rt 2 ( Pls refer attachment to calculate height)
Diagram may not be required but in case you want to refer
Areas= 1/2* 6 sq rt 2* sq rt 2
=(1/2)*6*2
=6
Last edited by iamcste on Sat Nov 08, 2008 10:30 am, edited 2 times in total.
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logitech
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Question is not asking for the areaiamcste wrote:First part is use of pythagoras theorem
Second part,
It forms a Trapezium
A=1/2( sum of parallel sides)* height
Sum of parallel slides = sum of diagonals =we calculated in the first part=2 square root 2 + 4 square root 2
Height is slightly tricky= sq rt 2 ( Pls refer attachment to calculate height)
Trick is to note that Triangle AHQ forms rt angled triangle
We know AH ( from attachment) and HQ( since its a midpoint)=2
Hence use pythagoras to find height AQ and the same =sq rt 2
Areas= 1/2* 6 sq rt 2* sq rt 2
=(1/2)*6*2
=6
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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logitech
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yeap it makes me cleariamcste wrote:Pls check the Qtn header "subject: shaded region perimeter and area ?"
and also check in the attachment after qtn for Perimeter , we have qtn for area
Hope this make you clear.
Okay here is the area:
[(4*4) - ( 2*2)]/2 = 6
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
EFGH is a square. Since P is the midpoint of side EF and Q is the midpoint of
EH we see that both segments PF and QH have length 2. The side QP is the hypotenuse of right triangle QEP. Since both segments EP and EQ have length 2 we have by the pythagorian theorem QP = sqrt(4 + 4) = 2sqrt(2).
Similarly HF is the hypotenuse ot the right triangle HGF so again by the pythagorian theorem HF = sqrt(16 + 16) = 4sqrt(2). Adding it all up we get
Perimeter QPFH = 2 + 2 = 2sqrt(2) + 4sqrt(2) = 4 + 6sqrt(2).
EH we see that both segments PF and QH have length 2. The side QP is the hypotenuse of right triangle QEP. Since both segments EP and EQ have length 2 we have by the pythagorian theorem QP = sqrt(4 + 4) = 2sqrt(2).
Similarly HF is the hypotenuse ot the right triangle HGF so again by the pythagorian theorem HF = sqrt(16 + 16) = 4sqrt(2). Adding it all up we get
Perimeter QPFH = 2 + 2 = 2sqrt(2) + 4sqrt(2) = 4 + 6sqrt(2).
Ossa Elhadary, PhD, CISA, PMP
Math Specialist
Test Prep New York
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Math Specialist
Test Prep New York
maximize your score, minimize your stress
www.testprepny.com
[email protected]
