If p is the perimeter of rectangle Q, what is the value of p?

SOLUTION

If p is the perimeter of rectangle Q, what is the value of p?

Let the length and width of the rectangle be a and b. The question asks to find \(perimeter = 2(a + b)\).

(1) Each diagonal of rectangle Q has length 10. \(d^2=a^2+b^2=100\). Not sufficient.

A right triangle with a hypotenuse of 10 does not necessarily imply that we have a (6, 8, 10) right triangle. If we are told that the lengths of all sides are integers, then yes: the only integer solution for a right triangle with a hypotenuse of 10 would be (6, 8, 10).

To elaborate, consider a right triangle with a hypotenuse of 10 inscribed in a circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

Thus, any point on the circumference of a circle with a diameter of 10 could form a right triangle with the diameter. The sides do not necessarily have to be 6 and 8. For example, we could have an isosceles right triangle, which would be a 45-45-90 triangle, and the sides would be 10/√2. Alternatively, if we have a 30-60-90 triangle with a hypotenuse of 10, the sides would be 5 and 5*√3. There could be many other combinations as well.

In an algebraic context, we can think about this problem as finding solutions for the equation a^2 + b^2 = 100. There are infinitely many solutions for a and b, with (a, b) = (6, 8) being just one of them. You can choose any number between 0 and 10 (excluding 0 and 10) for a, and there will be a corresponding value for b that satisfies the equation a^2 + b^2 = 100. For instance, if a=1, then b=√99; if a=√2, then b=√98; if a=7, then b=√51, and so forth.

(2) The area of rectangle Q is 48. \(ab=48\). Not sufficient.

(1)+(2) Square \(perimeter = 2(a + b)\) to get \(perimeter^2=4(a^2+b^2+2ab)\). Since from (1) \(a^2+b^2=100\) and from (2) \(ab=48\), then \(perimeter^2=4(a^2+b^2+2ab)=4(100+2*48)=4*196\), which simplifies to \(perimeter=\sqrt{4*196}=2*14=28\). Sufficient.

Answer: C.

Similar questions to practice:
https://gmatclub.com/forum/if-the-diago ... 04205.html
https://gmatclub.com/forum/what-is-the- ... 05414.html
https://gmatclub.com/forum/what-is-the- ... 96381.html

Hope it helps.
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Let x and y be two sides of the rectangle.. then p =x+y

Statement 1
Diagonal = 10
=> x^2 + y^2 = 10^2
we cannot find p i.e. x+y using this info.
=> NOT SUFFICIENT

Statement 2
Area = 48
=> xy =48
we cannot find p i.e. x+y using this info
=> NOT SUFFICIENT

Combining (1) and (2)
we will get
value of x+y = sqrt (x+y)^2 = sqrt(x^2 + y^2 + 2xy) = sqrt ( 10^2 + 2*48)
sqrt(196) = 14
=> SUFFICIENT

So, Answer will be C
Hope it helps!
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If instead, Q were a square, would 1 be sufficient?

In a rectangle, why can't we use the Isosceles Triangle to figure out the third side since the diagonals bisect each other?

If we were told that Q is a square instead of a rectangle, then the answer would be D.

As for the second question: can you please explain what you mean? Generally you cannot find the sides of a rectangle just knowing the length of its diagonal, since knowing the length of hypotenuse (diagonal) in a right triangle (created by length and width), is not enough to find the legs of it (length and width).

Hope it's clear.
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I'm sorry I'm still not seeing how this is not answer "A". I understand the logic at arriving at answer "C", I just don't understand why you NEED to combine statements "1" and "2", contradicts my entire understanding of Data Sufficiency logic.

A rectangle is comprised of 4 right angles, no?

So ultimately the "diagonal" represents the hypotenuse forming two right triangles, no?

Can you form a right triangle with a hypotenuse of 10 with any other legs besides 6 and 8? Or do I have that wrong?

(pythagorean triplet (3, 4, 5) , (6, 8, 10))

Hi Kellygrad05,

There was a similar problem I was attempting yesterday on the forum.

Basically we are told that it is a rectangle but we aren't sure if the sides are Integers or not. For ex.

Diagonal-10, sides can be 6 and 8 (because of PT) or something like Square root 99 and 1...and such other combination

When you consider the st2 with above then we can figure out sides will be 6 and 8 as only in that condition Area will be 48 and Diagonal as 10.

Thanks
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A right triangle with hypotenuse 10, doesn't mean that we have (6, 8, 10) right triangle. If we are told that the lengths of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 10 would be (6, 8, 10).

To check this: consider the right triangle with hypotenuse 10 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

So ANY point on circumference of a circle with diameter of 10 would make the right triangle with diameter. Not necessarily sides to be 6 and 8. For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be \(\frac{10}{\sqrt{2}}\). OR if we have 30-60-90 triangle and hypotenuse is \(10\), sides would be \(5\) and \(5*\sqrt{3}\). Of course there could be many other combinations.

Similar questions to practice:
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-105414.html
what-is-the-perimeter-of-rectangle-r-96381.html

Hope it helps.
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I did not use the math way this is how I did it

it is given that 2L +2w= the perimeter of a rectangle
1. each diagnal of a rectangle is length of 10 which makes the rectangle in half so it cant, but just know the triangle height
= not sufficient
2. the area of a rectangle is 48 so l*w= area not sufficent

both will tell us 2 equations and can find the length an width to get p so it is C
thanks

The question asks us to figure out the PERIMETER of rectangle Q. For that, we'll need the length (L) and width (W) of the rectangle.

1) Each diagonal of rectangle Q has length 10.

From this Fact, we can create one equation:

L^2 + W^2 = 10^2

Unfortunately, there are lots of different values for L and W here (and most of them are non-integers), so there are lots of possible perimeters. Here are two possibilities:
L = 6, W = 8... Perimeter = 28
L = 1, W = (Root99)... Perimeters = 2 + 2(Root99)
Fact 1 is INSUFFICIENT

2) The area of rectangle Q is 48.

From this Fact, we can create one equation:

(L)(W) = 48

Again though, there are lots of different values for L and W here, so there are lots of possible perimeters. Here are two possibilities:
L = 6, W = 8... Perimeter = 28
L = 1, W = 48... Perimeter = 98
Fact 2 is INSUFFICIENT

Combined, we know...
L^2 + W^2 = 10^2
(L)(W) = 48

We have a 'system' of equations here - two variables and two unique equations. Since rectangles cannot have "negative sides", there's just one solution to this system (and it happens to be 6 and 8, although you don't have to do that work).
Combined, SUFFICIENT

Final Answer:

It's certainly important to have strong basic 'math skills', but the Quant section of the GMAT is NOT a 'math test' - it's a critical thinking Test - so you should adjust your 'view' of that section accordingly.

GMAT assassins aren't born, they're made,
Rich

That's exactly right. There are infinitely many right triangles with hypotenuse 10. For instance, sqrt(13), sqrt(87), 10.
_________________

Similar questions to practice:
https://gmatclub.com/forum/if-the-diagon ... 04205.html
https://gmatclub.com/forum/what-is-the-a ... 05414.html
https://gmatclub.com/forum/what-is-the-p ... 96381.html

Hope it helps.
_________________

i chose answer A since this is a rectangle and the diagonal will bisect the right angles, forming 45-45-90 triangle with sides ratios of 1:1:root 2.
using Pythagorean theory , we can get the value of both sides

what did i do wrong here ?

The diagonals of a rectangle bisect the angle if and only the rectangle is a square. Generally, a diagonal of a rectangle can form any angle with the adjacent sides from 0 to 90, not inclusive.
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Solution:

Given:

• The perimeter of the rectangle Q = p

Working out:

We need to find out the value of p

Statement 1:

Each diagonal of the rectangle Q has length 10

Let us assume that the length of the rectangle Q is l, and the breadth of the rectangle Q is b.

From this statement, we can infer that \(\sqrt{l^2 + b^2}\) = 10

• Squaring both the sides of the equation, we get \(l^2 + b^2 = 100\)


o There can be more than one possible combination of l and b.

o And hence, the sum of l and b is not unique.

Thus, Statement 1 alone is not sufficient to answer this question.

Statement 2:

Area of the rectangle Q is 48 units.

Let us assume that the length of the rectangle Q is l, and the breadth of the rectangle Q is b.

Thus, \(l*b = 48\)

There can be more than one combination of l and b: (6,8), (12, 4), etc. and the values of p will not be unique.

Thus, statement 2 alone is not sufficient to answer this question.

Combining both the statement:

From statement 1, we have \(l^2 + b^2 = 100\)

From statement 2, we have \(l*b = 48\)

• \((l+b)^2 = l^2 + b^2 + 2l*b\)

• Or, \((l+b)^2 = 100 + 96\)

• Or, \((l+b)^2 = 196\)

• Or, \((l+b) = 14\) units.

From here, we can calculate the value of p.

Thus, combining both the statements, we got our answer.

Answer: Option C
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