Trapezoid
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Source: Beat The GMAT — Data Sufficiency |
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4meonly
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The area of the trapezoid = 1/2*(sum of 2 basis)*height
If ABCD is isosceles trapezoid, than x=x and y=y (see picture)
The area of the trapezoid = 1/2*(2(x+y))*height = (x+y)*h
Since we know h=6, we need only (x+y)
(1)
So, We have (x+y)^2 + 6^2 = 10^2
(x+y)^2 = 36
x+y = 6
SUFF
(2)
AB=7, x=7
nothing about y
Answer A
Where is my mistake?
If ABCD is isosceles trapezoid, than x=x and y=y (see picture)
The area of the trapezoid = 1/2*(2(x+y))*height = (x+y)*h
Since we know h=6, we need only (x+y)
(1)
So, We have (x+y)^2 + 6^2 = 10^2
(x+y)^2 = 36
x+y = 6
SUFF
(2)
AB=7, x=7
nothing about y
Answer A
Where is my mistake?
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Let us first drop two perpendiculars from AB to DC, one from point A and one from point B. Let us call the points on DC where the perpendiculars touch P and Q respectively.4meonly wrote:OA C
Area of the trapezoid is, h(AB+DC)/2.
we know h=6. We need the values of AB and DC to find the value of the Area.
From Statement 1: Since AC = 10, we can say, PC = 8 (Using pythogorus theorem for triangle APC and we know the values for AC and AP as 10 and 6 respectively). But we still don't know the value for DC and AB. So, insuffiecient.
From Statement 2: AB =7. But there is no way to figure out the value of DC. So, insufficient.
Combining statements 1 and 2: Since AB =7. PQ =7. Therefore, QC = (PC-PQ) = (8-1)=7.
QC = DP. Therefore, DP =1.
So, DC = DP+PC = 1+8 = 9.
and we know AB = 7.
So, the area = 6(7+9)/2 = 48.
--- sufficient.
Therefore, i go with C.
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4meonly
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PD = QC?Let us first drop two perpendiculars from AB to DC, one from point A and one from point B. Let us call the points on DC where the perpendiculars touch P and Q respectively.
I think yes, PD = QC
PD+QC+AB+PQ = 2PD + 2AB
My question is:
PD = QC? If yes, then answer is A
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Vemuri
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Your assumption that x=x & y=y is not correct. A trapeziod is a quadrilateral that has only 2 sides parallel. It doesn't say anything about the length of the sides.4meonly wrote:The area of the trapezoid = 1/2*(sum of 2 basis)*height
If ABCD is isosceles trapezoid, than x=x and y=y (see picture)
The area of the trapezoid = 1/2*(2(x+y))*height = (x+y)*h
Since we know h=6, we need only (x+y)
(1)
So, We have (x+y)^2 + 6^2 = 10^2
(x+y)^2 = 36
x+y = 6
SUFF
(2)
AB=7, x=7
nothing about y
Answer A
Where is my mistake?
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4meonly
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What is isosceles trapeziod?
Why my assumption is incorrect?
https://en.wikipedia.org/wiki/Isosceles_trapezoidAn isosceles trapezoid (isosceles trapezium in British English) is a quadrilateral with a line of symmetry bisecting one pair of opposite sides, making it automatically a trapezoid. Two opposite sides (bases) are parallel, the two other sides (legs) are of equal length. The diagonals are of equal length.
Why my assumption is incorrect?
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Bidisha_800
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I agree with 4meonly. The answer should be (A)
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mals24
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Ok your reasoning is fine. There is just a small error in the calculation of bold part. x+y = 8. Otherwise your reasoning is correct. Where did you get this question from btw?4meonly wrote:The area of the trapezoid = 1/2*(sum of 2 basis)*height
If ABCD is isosceles trapezoid, than x=x and y=y (see picture)
The area of the trapezoid = 1/2*(2(x+y))*height = (x+y)*h
Since we know h=6, we need only (x+y)
(1)
So, We have (x+y)^2 + 6^2 = 10^2
(x+y)^2 = 36
x+y = 6
SUFF
(2)
AB=7, x=7
nothing about y
Answer A
Where is my mistake?
