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Math Revolution GMAT Instructor V
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The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 62% (01:25) correct 38% (01:45) wrong based on 81 sessions

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[Math Revolution GMAT math practice question]

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. What is the area of the square?

A. 16
B. 25
C. 32
D. 36
E. 50

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Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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Distance between the vertices of the diagonal is given by sqrt(36+64) = 10.

If the side of a sqaure is a, then diagonal will be a*sqrt(2). Diagonal = 10. Side becomes 10/sqrt(2).

Area = 100/2 = 50.

E is the answer

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Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. What is the area of the square?

A. 16
B. 25
C. 32
D. 36
E. 50

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane

So Diagonal = $$\sqrt{(36+64)}$$==> 10

Area = $$1/2(d^2)$$

Area =1/2(10^2)

=50

Hence E
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Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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NandishSS wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. What is the area of the square?

A. 16
B. 25
C. 32
D. 36
E. 50

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane

So Diagonal = $$\sqrt{(36+64)}$$==> 10

Area = $$1/2(d^2)$$

Area =1/2(10^2)

=50

Hence E

I have a silly question. Can we not use the given co-ordinates to find the length of sides? But that will produce 2 different lengths and that will not be a square! I am missing some funda here. Please help _________________
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7360
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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=>

The length of the diagonal of the square is $$\sqrt{{ (7-1)^2 + (6-(-2))^2 }} = √100 = 10.$$
The side-length of the square is $$\frac{10}{√2} = 5 √2$$ since the side-length of a square is equal to the length of its diagonal divided by $$√2$$. Thus, the area of the square is $$(5 √2)^2 = 50.$$

Therefore, E is the answer.
Answer: E
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Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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Hey MrCleantek

Given: The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. Not vertices of sides

Quote:
I have a silly question. Can we not use the given co-ordinates to find the length of sides? But that will produce 2 different lengths and that will not be a square! I am missing some funda here. Please help _________________
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Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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NandishSS wrote:
Hey MrCleantek

Given: The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. Not vertices of sides

Quote:
I have a silly question. Can we not use the given co-ordinates to find the length of sides? But that will produce 2 different lengths and that will not be a square! I am missing some funda here. Please help But the sides can be derived from the vertices right since its a square? I mean draw perpendicular lines from there and you would have the sides.
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Re: The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. What is the area of the square?

A. 16
B. 25
C. 32
D. 36
E. 50

By the distance formula, the length of the diagonal of the square is:

d = √[(7 - 1)^2 + (-2 - 6)^2] = √[36 + 64] = √100 = 10.

Recall that the area of a square, given a diagonal of length d, is A = d^2/2. Thus, the area of the square is:

A = 10^2/2 = 100/2 = 50

Answer: E
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The points (1,6) and (7,-2) are two vertices of one diagonal of a squa  [#permalink]

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MrCleantek wrote:
NandishSS wrote:
Hey MrCleantek

Given: The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. Not vertices of sides

Quote:
I have a silly question. Can we not use the given co-ordinates to find the length of sides? But that will produce 2 different lengths and that will not be a square! I am missing some funda here. Please help But the sides can be derived from the vertices right since its a square? I mean draw perpendicular lines from there and you would have the sides.

Yes, you can easily plot the sides with help of diagonal points or you can use the ratio formula for right angle isosceles triangle
($$x : x : x\sqrt{2}$$)
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"Do not watch clock; Do what it does. KEEP GOING." The points (1,6) and (7,-2) are two vertices of one diagonal of a squa   [#permalink] 24 Nov 2018, 11:15
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# The points (1,6) and (7,-2) are two vertices of one diagonal of a squa

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