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.
+23246 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
+23246 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
