A rectangle is inscribed in a circle of radius r. If the rectangle is

Question

A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?

A.  2r \sqrt 3

B.  2r (\sqrt 3 + 1)

C.  4r \sqrt 2

D.  4r \sqrt 3

E.  4r (\sqrt 3 + 1)

ID08779

Bunuel · Accepted Answer

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle . Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle . Now, since each option has \sqrt{3} in it, then let's check a right triangle where the angles are 30°, 60°, and 90° (standard triangle for the GMAT), because its sides are always in the ratio 1 : \sqrt{3}: 2 . Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). So, in this case we would have: Rectangle.png The perimeter of the rectangle is 2r\sqrt{3}+2r=2r(\sqrt{3}+1) . Answer: B.

KarishmaB · Answer

Since the radius of the circle is r, the diameter must be 2r. Now imagine the rectangle. The diameter must be the hypotenuse of the right angled triangle of the rectangle. Say, if its sides are a and b, 
 (2r)^2 = a^2 + b^2 = 4r^2

So when you square a and b and sum them, you should get 4r^2

The given options are the perimeter of the rectangle i.e. 2(a+b). So I ignore 2 of the options and try to split the leftover into a and b. 
The obvious first choices are options (B) and (E) since we can see that we can split the sum into 3 and 1.
 a + b = r\sqrt{3} + r

Now check:
 (\sqrt{3}r)^2 + r^2 = 4r^2

That is what we wanted. Hence, the answer is (B)

gurpreetsingh · Answer

Largest perimeter is possible when rectangle is Square. Since option C is when it is square and the question itself says rectangle is not a square, C as well as D and E are out as D and E are greater than C.

Our answer must be less than C.
So only A and B left. Now do reverse engineering. B looks more plausible than A as the perimeter of rectangle must of the form a+b....

gmihir · Answer

Just wondering, why is it mentioned in question that "if rectangle is not square" - will the answer be different had the rectangle inscribed in circle be a square ?

KarishmaB · Answer

They had to mention it because otherwise, the answer could have been (C) too.
 (\sqrt{2}r)^2 + (\sqrt{2}r)^2 = 4r^2

Hence  2(a+b) = 2*(\sqrt{2}r + \sqrt{2}r) = 4\sqrt{2}r

EvaJager · Answer

The perimeter of the rectangle must be greater than  4r . Imagine a "flattened/smashed" rectangle, such that one of the sides is approaching the size of a diameter ( 2r ), and the other side is approaching 0. When the two sides collapse into the diameter, the perimeter would be  4r .
Or, if the two sides of the rectangle are  a  and  b ,  a+b  must be greater than the diagonal of the rectangle, which is  2r .

Therefore, you can immediately eliminate A, since  2r\sqrt{3}<4r .

TheNona · Answer

Thanks Karishma , but could you please elaborate more ? I still do not understand

MzJavert · Answer

Bunuel your explanation is simple, clear, and didn't make my brain hurt. :idea:

KarishmaB · Answer

The method makes a guess based on the format of the expected answer. Make a circle and inscribe a rectangle in it. The diagonal of the rectangle will be the diameter of the circle. Ques3.jpg We need to find the perimeter of the rectangle i.e. 2(a + b) We know that a^2 + b^2 = (2r)^2 = 4r^2 So what can a and b be? 3r^2 + r^2 = 4r^2 (So a = \sqrt{3}r, b = r ) or 2r^2 + 2r^2 = 4r^2 (So a = \sqrt{2}r, b = \sqrt{2}r ) etc Now, looking at the options, we see that 2(a + b) can be 2(\sqrt{3}r + r) i.e. option (B)

nikhil007 · Answer

This Question came up on my prep and I got stuck as I didn't look at answer choices.
I just wrote L^2+B^2 = (2R)^2 and I had no clue how to solve it in 2 mins.
I understand that we need to work on shortcuts and tricks, but just for academic purposes can someone show me how to manipulate this equation to answer choice algebraically? 
the only next step I could think was (L+B)^2 = 4R^2 - 2LB and had no clue how to go further.

KarishmaB · Answer

Note that the question says "which of the following could be the perimeter of the rectangle"
The rectangle can be made in many ways and the perimeter would be different in these cases. The answer option gives us one such perimeter value. Hence there is no single algebraic method of obtaining the "correct answer".
You have to look at options and say which one CAN be the perimeter value.

himanshujovi · Answer

Ok Krishna so as I see it , this question can only and only be solved by the process of elimination. Right ?

KarishmaB · Answer

Yes, since you have multiple possible answers, you have to look for one which is possible. Note that you don't really need to eliminate every choice till you arrive at the correct answer. You need to intelligently guess which options can give you the correct answer.

thinkpad18 · Answer

Is this always the case? That the right triangles within a rectangle are always 30-60-90?

KarishmaB · Answer

Got it. Why is it 30-60-90 in this case? (other than looking at the answers)?

The angles can take many values. Look at the question: which of the following  could be  the perimeter ...

The reason we think of 30-60-90 triangle is that we know sides of only 45-45-90 and 30-60-90 triangles. GMAT will not ask us to find the sides of 10-80-90 triangle. Since it is not a square, we should try out the 30-60-90 triangle.

crazi4ib · Answer

Bunuel   VeritasKarishma  any thoughts on this please

KarishmaB · Answer

The operation of squaring is not the same as that of multiplication/division.

Given: a + b = c

Squaring both sides, you get,

(a + b)^2 = c^2

a^2 + b^2 + 2ab = c^2

Take an example,  3^2 = 9 ,  4^2 = 16 . 
The sum of these two is 9 + 16 = 25. It is not the same as  (3 + 4)^2 = 7^2 = 49 .

sharmaaman5598 · Answer

The question can be solved without the process of elimination. 
given:  4r^{2} = l^{2} + b^{2}  
Perimeter(P) =   2(l+b) 
i.e., P =  2(l+ \sqrt{(4r^2 - l^2)}) 
We need to find the range of the function P. 
Perimeter is maximum when the shape is a square i.e.,
l = b =  \sqrt{2}r . (You can either know this relation or obtain this by solving  dP/dl = 0 )
=> Max(P) =  4\sqrt{2}r  
Note that since, according to question, the shape is not a square, P can actually never assume the above value.
Another constraint is that  4r^{2} >= l^{2}  for the expression in square root to not be negative
=> l <=   2r 
=> Min(P) =  4r

P ranges from    i.e.,    
where '  2\sqrt{3}r   => 2*1.73r = 3.46r (Incorrect , less than lower limit of range)
B)   2(\sqrt{3} +1)  => 2*2.73r = 5.46r (Correct , within range)
C)  4\sqrt{2}r  (Beyond range, fyi this is the perimeter if the shape were a square as calculated above)
Reject option D and E because they are larger values than C and hence outside our range)

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A rectangle is inscribed in a circle of radius r. If the rectangle is

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