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Given

    • The perimeter of a rhombus is 44.


To Find

    • The maximum area the rhombus could have.

Approach and Working Out


    • When the perimeter is given, the maximum area a rhombus can have when it’s a square.
    • Perimeter of a square = 44
      o Side = \(\frac{44}{4}\) = 11
    • Area = \(11^2\) = 121


Correct Answer: Option A
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Bunuel
The perimeter of a rhombus is 44. What is the maximum area the rhombus could have?

A. 121
B. 144
C. 169
D. 196
E. 225

rhombus sides are equal
perimeter = 44, side = 44/4 = 11
max area rhombus when its a square
max area: 11^2=121

Ans (A)
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The perimeter of a rhombus is 44. What is the maximum area the rhombus could have?

The maximum area is the area of square.

Given:
The perimeter of a rhombus is 44. Consider the property all sides are equal and there are 4 sides. So length of a side is 11.

From the length of side, we can calculate the area of a square.

Length of a side = 11
Area = \(11^2\) = 121

Ans:A
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The perimeter of a rhombus is 44. What is the maximum area the rhombus could have?

A. 121

B. 144

C. 169

D. 196

E. 225

one side = 44/4 = 11

maximum area = 11^2 = 121

answer A
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The perimeter of a rhombus is 44. What is the maximum area the rhombus could have?

44/4 = 11 - each side


The area will be max 1. if the diagonals are equal -
2. They bisect each other.
3. they bisect at right angles
So the resulted value of a part of diagonal= \(x = 11\sqrt{1/2}\\
\)

Diagonal 1= diagonal2 = 2x

Area of the rhombus = Diagonal 1 * diagonal 2 *1/2

Area= 2 * 11 * 11 * \sqrt{1/2}* \sqrt{1/2}

Area = 121

A. 121

B. 144

C. 169

D. 196

E. 225
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Given 4A = 44

A= 11 (side of rhombus)

In a polygon, if the perimeter is fixed, we can maximize the area by making the side equal

Max area possible here is 11 * 11 = 121

IMO A
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