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.

Guest May 3, 2020

#1**0 **

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.

x^{2} + (49 - x)^{2} = 41^{2} ---> x^{2} + 2401 - 98x + x^{2} = 1681 ---> 2x^{2} - 98x + 720 = 0

---> x^{2} - 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 **

geno3141 May 3, 2020

#2**0 **

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.

c^{2} = 7^{2} + 24^{2} ---> c^{2} = 49 + 576 ---> c^{2} = 625 ---> c = 25.

The perimeter will be 4 x 25 = 100

geno3141 May 3, 2020