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505-555 Level|   Geometry|                  
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Bunuel
fatihaysu
If a square mirror has a 20-inch diagonal,what is the approx. perimeter of the mirror,in inches?

a)40
b)60
c)80
d)100
e)120

My aprroach is;

20= X \sqrt{2}

then X=10\sqrt{2}

4*X \sqrt{2} = 45,6

so should i round it to 40 or 60?

You made an error in calculation: \(P=4*10\sqrt{2}\approx{56.6}\) (\(\sqrt{2}\approx{1.4}\)), 56.6 is closer to 60 than to 40, thus answer is B.

If P were 45.6 (as you calculated) then the answer would be A, as 45.6 is closer to 40 than to 60.

Hope it's clear.

It is definelty clear Bunuel. I am triped up by not remembering it is 1,14 or 1,4.

thank you
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memorize the fact that diagonal of a square with side s = s * root ( 2)

u ll get the answer half a min
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I'm I doing the last step correctly?

\(10\sqrt{2}\) = 10 x 1.4 = 14 right, therefore 14x4 = 56.

Thanks.
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Bunuel
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laythesmack23
I'm I doing the last step correctly?

\(10\sqrt{2}\) = 10 x 1.4 = 14 right, therefore 14x4 = 56.

Thanks.

Yes, that's correct.
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fatihaysu
If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?

(A) 40
(B) 60
(C) 80
(D) 100
(E) 120

My aprroach is;

\(20= X \sqrt{2}\)

then \(X=10\sqrt{2}\)

\(4*X \sqrt{2} = 45,6\)

so should i round it to 40 or 60?
.

Here is my approach, i think it will be very simple and do not need any calculations.
if you draw the square and diagonal inside the square. u can see square becomes part of two triangles opposite to each other.
And We know the property of the triangle, addition of two sides of triangle must be greater than its diagonal in order to complete the triangle. And each side must be less than 20 and perimeter must be less than 80, so we can eliminate answer choice C, D and E.
so Side 1 + side 2 > 20, that means Side 1 or Side 2 must be > 10. so we can eliminate the answer choice A.
Now we are left with is B
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Question:

Diagonal of a square = x * sqrt2

20 = x * sqrt2
x = 20 / sqrt2

Why are you all using x = 10 / sqrt2 ? I got to the same result with 20 / sqrt2 as length of one side, but where is my mistake?

Thanks!!
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noTh1ng
Diagonal of a square = x * sqrt2

20 = x * sqrt2
x = 20 / sqrt2

Why are you all using x = 10 / sqrt2 ? I got to the same result with 20 / sqrt2 as length of one side, but where is my mistake?

Thanks!!

\(\frac{20}{\sqrt{2}}=\frac{20*\sqrt{2}}{\sqrt{2}*\sqrt{2}}=10\sqrt{2}\).
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Bunuel
noTh1ng
Diagonal of a square = x * sqrt2

20 = x * sqrt2
x = 20 / sqrt2

Why are you all using x = 10 / sqrt2 ? I got to the same result with 20 / sqrt2 as length of one side, but where is my mistake?

Thanks!!

\(\frac{20}{\sqrt{2}}=\frac{20*\sqrt{2}}{\sqrt{2}*\sqrt{2}}=10\sqrt{2}\).


Thank you Bunuel the calculation is clear, but I just don't get why we do that?!extending?

Why isn't 20 / sqrt2 sufficient?
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noTh1ng
Bunuel
noTh1ng
Diagonal of a square = x * sqrt2

20 = x * sqrt2
x = 20 / sqrt2

Why are you all using x = 10 / sqrt2 ? I got to the same result with 20 / sqrt2 as length of one side, but where is my mistake?

Thanks!!

\(\frac{20}{\sqrt{2}}=\frac{20*\sqrt{2}}{\sqrt{2}*\sqrt{2}}=10\sqrt{2}\).


Thank you Bunuel the calculation is clear, but I just don't get why we do that?!

Because \(10\sqrt{2}\) is considered to be more simplified (rationalized) than \(\frac{20}{\sqrt{2}}\).

This algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator.

Questions involving rationalization to practice:
if-x-0-then-106291.html
if-n-is-positive-which-of-the-following-is-equal-to-31236.html
consider-a-quarter-of-a-circle-of-radius-16-let-r-be-the-131083.html
in-the-diagram-not-drawn-to-scale-sector-pq-is-a-quarter-139282.html
in-the-diagram-what-is-the-value-of-x-129962.html
the-perimeter-of-a-right-isoscles-triangle-is-127049.html
which-of-the-following-is-equal-to-98531.html
if-x-is-positive-then-1-root-x-1-root-x-163491.html
1-2-sqrt3-64378.html

Hope it helps.
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The diagonal of a square splits it in 2 right triangles --> the area of one traingle = (B*H)/2 = 100 (Height is 50% of the diagonal)
So the square have an area of 2*100=200 which is almost 14^2 ----> 4*14 = 56 nearest to 60 (B)
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fatihaysu
If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?

(A) 40
(B) 60
(C) 80
(D) 100
(E) 120


If the Square has side = \(x\)
then the diagonal of Square = \(x\sqrt{2}\)

i.e. \(x\sqrt{2} = 20\)
i.e. \(x = 20/\sqrt{2} = 10\sqrt{2}\)

Perimeter = 4*Side = \(4*x = 4*10\sqrt{2}\)

i.e. Perimeter = \(40\sqrt{2}\) = \(40*1.4 = 56\)inch

Answer: Option B
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Can somebody solve this with Trigonometry.
Thank you.
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merwalpiyush19
Can somebody solve this with Trigonometry.
Thank you.

Sure, look below.

If you draw the diagonal in a square, it will divide the square into 2 congruent triangles such that the diagonal makes \(45^{\circ}\) with the sides.

Consider, triangle ABC,

\(\frac{AB}{AC}\) = sin (\(\angle {ABC}\)) = sin (45) = \(\frac{1}{\sqrt{2}}\) , where AC =20

Thus, AB = \(\frac{20}{\sqrt{2}}\) = \(\frac{20*\sqrt{2}}{2}\)= 10*\(\sqrt{2}\) = 10*1.4 = 14
Thus, the perimeter of the square = 4*AB = 4*14 =56.

B is the closest answer
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merwalpiyush19
Can somebody solve this with Trigonometry.
Thank you.

Since this question was pretty simple with Trigonometry as shown in the explanation above so my only curiosity is

WHY do you want to know the solution of such question with Trigonometry when your base is already not strong enough in it?

over an above that WHY trigonometry when this is not what GMAT ever expects from students?


Writing here just out of concern so that other readers don't get influenced by Solutions by trigonometry as tool of solving such questions of GMAT.

Please be careful while referring to the source of material, types of solutions etc. for GMAT preparation. If it's not what GMAT expects from you then It can only ruin your direction of healthy preparation.
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fatihaysu
If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?

(A) 40
(B) 60
(C) 80
(D) 100
(E) 120


We are given that a square has a 20-inch diagonal, and we must solve for the perimeter of the square. This means we first need to know the length of a side of the square. To determine the length of a side we can use the diagonal formula for a square. We know that:

diagonal of a square = side√2 = s√2

20 = s√2

s = 20/√2

To help with our math we can multiply 20/√2 by √2/√2. This gives us:

s = 20/√2 × √2/√2 = 20√2/√4

s = 20√2/2 = 10√2

We also should have memorized that the approximate value of √2 is 1.4. Thus one side of the square is about 10 x 1.4 = 14 inches.

Finally, the perimeter of the square is 4 x 14 = 56. Since we are asked to approximate, the closest answer is 60.

Answer B.
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I solved it using this reasoning:

When I know the diagonal I have to divide it by \(\sqrt{2}\) to get the side.
I also know that \(\sqrt{2}\) is something less than 2.

Consideration 1: 20/2 would be 10 so the perimeter would be 40
Consideration 2: If I assume that 20 is the side 20x4 gives me a perimeter of 80

So the answer must lie between 40 and 80... thus 60...
Clearly this only works because of the type of answer choices... but this is what GMAT often expects from us, right?
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Bunuel
fatihaysu
If a square mirror has a 20-inch diagonal,what is the approx. perimeter of the mirror,in inches?

a)40
b)60
c)80
d)100
e)120

My aprroach is;

20= X \sqrt{2}

then X=10\sqrt{2}

4*X \sqrt{2} = 45,6

so should i round it to 40 or 60?

You made an error in calculation: \(P=4*10\sqrt{2}\approx{56.6}\) (\(\sqrt{2}\approx{1.4}\)), 56.6 is closer to 60 than to 40, thus answer is B.

If P were 45.6 (as you calculated) then the answer would be A, as 45.6 is closer to 40 than to 60.

Hope it's clear.

Bunuel hello,

one question if one side of square is \(10\sqrt{2}\) and perometr of square is 4a

shouldnt we have as an answer \(4\) * \(10\sqrt{2}\) = \(40\sqrt{2}\) :?
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