In quadrilateral abcd, ae=x+10, and be=3x-18. for what value of x is abcd a rectangle? a.24 b.14 c.18 d.16

for x  = 14, the given ABCD is a RECTANGLE.

Step-by-step explanation:

Given : ABCD is  a quadrilateral.

E is the intersection points of the diagonal AC and BD.

AE  =  x+10, and BE=  3x -18

Now, as we know:

IN A RECTANGLES , DIAGONALS BISECT EACH OTHER.

⇒  E is the mid point of AC and BD respectively.

⇒ AE = BE

⇒ x+10 = 3 x - 18

or, 10 +  18 = 3 x - x

or, 2  x = 28

or, x  = 28/2 = 14   ⇒  x = 14

Hence, for x  = 14, the given ABCD is a RECTANGLE.

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

a

step-by-step explanation: