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The perimeter of a rectangle is 98, and the length of one of its diagonals is 41. Find the area of the rectangle.

 

Also can someone answer this question for me https://web2.0calc.com/questions/quadrilateral-problem_1. There is no explanation and I feel that it is wrong. 

 May 3, 2020
 #1
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The perimete of a rectangle is 98 and has a diagonal of length 41.

If the perimeter is 98, then   2·length + 2·width  =  98   or   length + width  =  49.

Call the length  x  then the width will be  49 - x.

 

If you draw the rectangle with one diagonal, you will have drawn two congruent right triangles.

Consider either one of them:  one side will be  x  the other side will be  49 - x  and the hypotenuse will be  41.

 

x2 + (49 - x)2  =  412     --->    x2 + 2401 - 98x + x2  =  1681     --->     2x2 - 98x + 720  =  0

                                     --->     x2 - 49x + 360  =  0     --->     (x - 9)(x - 40)  =  0     --->     x  =  9  or  x  =  40

--->     length  =  9   and   width  =  40     (or vice versa) 

 

--->     area  =  9 x 40  =  360 

 May 3, 2020
 #2
avatar+20914 
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For your second question:  https://web2.0calc.com/questions/quadrilateral-problem_1

 

Given a rhombus with diagonals  14  and  48  --  what is the perimeter?

 

Since the diagonals of a rhombus are perpendicular to each other, four congruent right triangles are formed.

Each of these triangles have bases of  7  and  24  so we can use the Pythagorean Theorem to find the side

of the rhombus.

 

c2  =  72 + 242     --->     c2  =  49 + 576     --->     c2  =  625     --->     c  =  25.

 

The perimeter will be  4 x 25  =  100

 May 3, 2020

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