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AB=4,BC=24, and AD=26

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AB=4,BC=24, and AD=26 [#permalink] New post   18 Sep 2020, 10:27
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AB=4,BC=24, and AD=26

Quantity A
Quantity B
The length of segment CD
10



A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

320 GRE Math Questions
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B
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Re: AB=4,BC=24, and AD=26 [#permalink] New post   18 Sep 2020, 10:28
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Re: AB=4,BC=24, and AD=26 [#permalink] New post   18 Sep 2020, 11:13
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ans: B

let, the intersect point between AD and BC is E

now, <ABE=90 and <BAE=<BEA=45

the opposite side length (AB) of <BEA is 4

now <BEA=<CED=45 [corresponding angle]

so CD must be 4, because the opposite side length of 45 degree is 4.

so quantity B is greater
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Re: AB=4,BC=24, and AD=26 [#permalink] New post   19 Sep 2020, 06:45
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If you create an imaginary rectangle of vertices ABCX (with X the imaginary vertex), you get a triangle AXD. Then, you can use Pythagoras:

\((AX)^2 + (XD)^2 = (AD)^2\)

\(24^2 + (XD)^2 = 26^2\)

doing some math:

XD = 10.

but they are asking for CD, therefore,

(remember that XC = AB = 4)
CD = XD - XC,
CD = 10 - 4
CD = 6,

In conclusion, A<B, option B.
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Re: AB=4,BC=24, and AD=26 [#permalink] New post   22 Sep 2020, 13:41
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Carcass wrote:
Attachment:
GRE AB=4,BC=24, and AD=26.png


AB=4,BC=24, and AD=26

Quantity A
Quantity B
The length of segment \(\overline{CD}\)
10



A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

320 GRE Math Questions



A quick solution is to realize that \(BC\) and \(AD\) are parts of a 5:12:13 right triangle.

To see it more clearly, drop a line down from point A and extend point C out to meet it. Essentially, what you have is \(BC\) being the height of this new triangle and \(AD\) being the new hypotenuse.

Since we have a 5:12:13 triangle, we know that new length of \(CD\) plus the added length must be 10 (10:24:26). That added length is actually the length of \(AB\), which is 4.

10 - 4 = 6, which is the length of \(CD\), giving the answer B.
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Re: AB=4,BC=24, and AD=26 [#permalink] New post   07 Jun 2021, 00:32
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cote wrote:
If you create an imaginary rectangle of vertices ABCX (with X the imaginary vertex), you get a triangle AXD. Then, you can use Pythagoras:

\((AX)^2 + (XD)^2 = (AD)^2\)

\(24^2 + (XD)^2 = 26^2\)

doing some math:

XD = 10.

but they are asking for CD, therefore,

(remember that XC = AB = 4)
CD = XD - XC,
CD = 10 - 4
CD = 6,

In conclusion, A<B, option B.


Best solution.
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Re: AB=4,BC=24, and AD=26 [#permalink] New post   02 Nov 2024, 18:26
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Re: AB=4,BC=24, and AD=26 [#permalink]
02 Nov 2024, 18:26
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