Is the length of the diagonal of the rectangle greater than

Question

Is the length of the diagonal of the rectangle greater than ﻿ \sqrt{50}

(1) The width of the rectangle is 6.

(2) The length of the rectangle is 7.

Abhishekgmat87 · Answer

Length of diagonal =  \sqrt{length ^ 2 + width ^ 2}

Statement (1) The width of the rectangle is 6.
Length can be smaller than width. If length is 3, answer is yes. If length is 4, answer is no. Not sufficient

(2) The length of the rectangle is 7
If width is less than 1, answer is no. If it's greater than 1, answer is yes. Not sufficient

Combining both, length of diagonal =  \sqrt{7^2 + 6 ^2} 
=  \sqrt{85}  >  \sqrt{50}

Sufficient

Option C

Ivy17 · Answer

Hi  BrentGMATPrepNow , / zhanbo  could you help me understand how the answer to this is A?

zhanbo · Answer

Like you, I believe the answer is (C).

Ivy17 · Answer

Thanks zhanbo

AbhiroopGhosh kindly confirm what the correct answer is and if required please change the OA accordingly. Posted from my mobile device

gmatophobia · Answer

The correct answer is  A . I will post the official solution shortly.

gmatophobia · Answer

Official Explanation

Step 1: Analyze Question Statement

Let ABCD be the rectangle with length ﻿'l', breadth ﻿'b' and diagonal 'd' as shown in the figure below:
 
Since ABC is a right-angled triangle, by applying Pythagoras theorem, we get:

{AC}^2  =  {AB}^2  +  {BC}^2   ﻿Hyptoenuse^2 = Perpendicular^2 + Base^2 ]

⟹ ﻿  d^2  =  l^2  +  b^2  ﻿ ……. (1)

We need to verify if ‘d’ is greater than  ﻿\sqrt{50} 
​
 i.e., if  d^2  is greater than 50 or not.

We now move on to the statements and analyze their sufficiency.

Step 2: Analyze Statements Individually

Statement 1: The width of the rectangle is 6.

From (1), we have

﻿  d^2  =  l^2  +  b^2  ﻿

⟹ ﻿ d^2  =  l^2  +  6^2  ﻿

⟹  ﻿d^2  =  l^2  + 36

Here, for all values of ﻿ l ≥6,  d^2 >50

Thus, we can answer the question with a definite YES.

Hence, statement 1 is sufficient and we eliminate the answer options B, C and E.

Note : Here we are using the understanding that length is always the longer side.

Statement 2: The length of the rectangle is 7.

From (1), we can write

⟹ ﻿ d^2 = l^2 + b^2

⟹  ﻿d^2 = 7^2 + b^2

⟹ ﻿ d^2 =49+ b^2  ﻿

Considering the following cases, we get:

Case 1: When b = 6,  ﻿d^2 = b^2 +49 = 85

Case 2: When b = 1,  ﻿d^2 = b^2 +49 = 50

Since,  d^2  can be greater than 50 or equal to 50, we cannot answer the question with a definite YES/NO.

Thus, statement 2 is not sufficient and we can eliminate the answer option D.

Hence, the correct answer is Option A.

SusantKumarNand · Answer

the length is always greater than the breadth in case of a rectangle, however it doesn't really matter while calculating the area or the parameter of rectangle because both length and breadth are considered in area and perimeter. we usually take the bigger value as length and smaller value as breadth in rectangle.

Posted from my mobile device

Posted from my mobile device

Ivy17 · Answer

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As per my understanding, unless it is mentioned in the question stem how are we assuming that length should always be greater than the width? Since even in questions relating to geometry, if a triangle looks like a right angled triangle but the same has not been stated in the question stem, we cannot assume it to be a right angled triangle.

gmatophobia · Answer

By convention / definition

liammoyn · Answer

Can you provide a source? I have never seen anyone define the length of a rectangle as strictly the larger side.

Is the length of the diagonal of the rectangle greater than

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Is the length of the diagonal of the rectangle greater than

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