If a rhombus has the perimeter of 56 cm and the area of 100 cm^2, what

Question

If a rhombus has the perimeter of 56 cm and the area of 100 cm^2, what is the sum of the lengths of its diagonals?

A. 31.20
B. 32.30
C. 33.40
D. 34.40
E. 35

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gkumar08 · Answer

The ans is D
As perimeter of rhombus =4*a
So 4a=56
a=14
Now area or rhombus =1/2×d1×d2
So d1×d2=200
2d1×d2=400
Now as the diagonals of a rhombus bisect each other at right angles so
 (d1/2)^2+(d2/2)^2=a^2
 So d1^2+d2^2=4×196=784
Now 2d1*d2=400
Adding both equations we have 
(d1+d2)^2=1184
Now we look at the options 
35^2=1225
So sum should be little less than 35
So D is the ans

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Avinash514043 · Answer

Area = product of diagonal/2
Perimeter = 4*side
Side =14
Product of diagonal=pq=200
Using Pythagoras:
p^2 +q^2 =14^2 *4=784

Now use the formula: (p+q)^2
p^2+q^2+2pq= 784+400=1184

Therefore p+q=√1184=34.40

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GMATWhizTeam · Answer

Solution 
 • Perimeter of rhombus  = 56  cm
 o So, each side of the rhombus  = \frac{56}{4} = 14  cm. 
• Let the two diagonals of the rhombus be  l_1  and  l_2 
• Based on the above information, let’s draw the below diagram: 
  https://www.dropbox.com/s/d4zcoelgrvby8gh/rhombus_diagonals.png?dl=1 
 • Now, area of  ABCD = \frac{1}{2}* l_1 * l_2 = 100 cm^2 ⟹ l_1 * l_2 = 200 cm^2 ………..Eq.(i) 
• We know that diagonals of a rhombus bisect each other at 90 degrees. 
 o So from the above diagram, we have
   OD = \frac{l_1 }{2} 
 And,  AO = \frac{l_2 }{2}  
o In triangle AOD,
   OD^2 + AO^2 = 14^2 
 Or,  (\frac{l_1 }{2})^2 + (\frac{l_2 }{2})^2 = 14^2 
 So,  (l_1 )^2 + (l_2 )^2 = (2*14)^2 ………….Eq.(ii)   
• Thus,  ( l_1 + l_2 )^2 = (l_1 )^2 + (l_2 )^2 + 2* l1 * l2  
• Substituting the values from Eq.(i) and Eq.(ii), into the above equation, we get,
 o  ( l_1 + l_2 )^2 = (2*14)^2 +2*200 = 784 + 400 = 1184 
o Now,  1156 < ( l_1 + l_2 )^2 = 1184 < 1225 
  So,  34 < ( l_1 + l_2 ) < 35    
Thus, the correct answer is  Option D

lacktutor · Answer

P = 56
--> the side of the rhombus is 56/4 = 14

If you draw the diagonals of the rhombus, It is divided in to 4 equal right-angled triangles:
-->  4(\frac{a*b}{2})= 100

2ab = 100

a^{2}+ b^{2}= 14^{2}

--> Adding together:
 (a+b)^{2}= 296

Well, we need the value of  2(a+b ) 
-->  a+b= √296 
 2(a+b)= 2√296 = √4*296= √1194 
-->  √1194  is greater than  34(34^{2}= 1156 ) and less than  35 (35^{2}=1225) .

D  is correct one

GMATinsight · Answer

The explanation is as attached here.

Answer: Option D

MHIKER · Answer

Let the diagonals of the rhombus  a  and  b

So, the area  = \frac{1}{2}*ab=100, ab=200

One side of the rhombus  = 56/4 =14

Now,  a^2+b^2=4*14^2

=(a+b)^2-2ab=4*196=784

=(a+b)^2-2*200=784

=(a+b)^2=784+400=1184

=a+b=\sqrt{1184}

a+b=34^+

The answer is D

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If a rhombus has the perimeter of 56 cm and the area of 100 cm^2, what

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