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MathRevolution
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MathRevolution
Attachment:
trapezoid ABCD.jpg
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7


* A solution will be posted in two days.

HI..
Although it should be given that O is the center, I take it as center..

Join AO.. we have an isosceles triangle with equal sides as radius or 4 and the third side as 6
so throw an altitude on AB from O as shown..


the ALTITUDE= \(\sqrt{4^2-3^2}=[m]\sqrt{7}\)[/m]..
now we know two side as 6 and 8 and ALTITUDE as \(\sqrt{7}\)..
Area= \(\frac{(6+8)}{2}*\sqrt{7}\)=\(7\sqrt{7}\)

C

How can we assume that the altitude from O to AB will bisect AB?
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MathRevolution
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7

->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.

hold up... where did u get HC = 1 and OH = 3 from ?
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aekukula
MathRevolution
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7

->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.

hold up... where did u get HC = 1 and OH = 3 from ?


From A, draw a line downwards vertically and it makes a point. Suppose the point E, which makes AB=EH=6. Then, (8-6)/2=HC=DE and OH=4-HC=4-1=3.
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two things i really wanna mention here
First =>This is a very poor Quality GMAT Question .
And i myself being a very poor GMAT question writer can actually say that.. :)

The Golden Rule of Gmat => """ IT THEY DO NOT SAY => NEVER EVER EVER ASSUME """
So O may or may not be the centre of the circle ..

Regards...
S.C.S.A

P.S => I actually solved this question for 15 minutes without taking O as the centre.. :)
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MeghaP
chetan2u
MathRevolution
Attachment:
trapezoid ABCD.jpg
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7


* A solution will be posted in two days.

HI..
Although it should be given that O is the center, I take it as center..

Join AO.. we have an isosceles triangle with equal sides as radius or 4 and the third side as 6
so throw an altitude on AB from O as shown..


the ALTITUDE= \(\sqrt{4^2-3^2}=[m]\sqrt{7}\)[/m]..
now we know two side as 6 and 8 and ALTITUDE as \(\sqrt{7}\)..
Area= \(\frac{(6+8)}{2}*\sqrt{7}\)=\(7\sqrt{7}\)

C

How can we assume that the altitude from O to AB will bisect AB?


Hey MeghaP the reason here is => The perpendicular drawn from the centre of the circle to the chord => Always Bisects the chord..
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Chiragjordan
MeghaP
chetan2u

If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7


* A solution will be posted in two days.

HI..
Although it should be given that O is the center, I take it as center..

Join AO.. we have an isosceles triangle with equal sides as radius or 4 and the third side as 6
so throw an altitude on AB from O as shown..


the ALTITUDE= \(\sqrt{4^2-3^2}=[m]\sqrt{7}\)[/m]..
now we know two side as 6 and 8 and ALTITUDE as \(\sqrt{7}\)..
Area= \(\frac{(6+8)}{2}*\sqrt{7}\)=\(7\sqrt{7}\)

C

How can we assume that the altitude from O to AB will bisect AB?


Hey MeghaP the reason here is => The perpendicular drawn from the centre of the circle to the chord => Always Bisects the chord..[/quote]


Thank you so much for this :)
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this question has to mention O is a center of a circle
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MathRevolution
aekukula
MathRevolution
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7

->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.

hold up... where did u get HC = 1 and OH = 3 from ?


From A, draw a line downwards vertically and it makes a point. Suppose the point E, which makes AB=EH=6. Then, (8-6)/2=HC=DE and OH=4-HC=4-1=3.

Casting the issue about O being the centre of the circle aside...

I am still confused how HC = 1 and OH = 3

The way I interpreted it was the base of both triangles is 1. Hence, if we take the altitude as bisecting the base, then each of two right triangles have a base of 0.5. The hypotenuse is 4.

4^2 = 0.5^2 + x^2 ...solve for x to get the height.

What did I miss?
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aekukula
MathRevolution
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7

->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.

hold up... where did u get HC = 1 and OH = 3 from ?





can someone explain the derivation of root7 clearly?
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