+218 In rectangle\( WXYZ, A\) is on side\(\overline{WX}\) \(such that AX = 4, B is on side \overline{YZ} such that BY = 18, and C is on side \overline{XY} such that angle ACB= 90 and CY = 2CX. Find AB.\)
Help?
Noori
+159 Is there an image? Also is it possible if you can type that out of Latex as some of the information is not there. Very hard to read xd.
+23252 AX = 4
BY = 18
Call XC = x
then YC = 2x
Triangle(AXC) is a right triangle ---> AC2 = AX2 + XC2 ---> AC2 = 42 + x2 ---> AC = sqrt(16 + x2)
Triangle(BYC) is a right triangle ---> BC2 = BY2 + YC2 ---> BC2 = 182 + (2x)2 ---> BC = sqrt(324 + 4x2)
Area( trapezoid(AXYB) ) = ½·(x + 2x)(4 + 18) = 33x
Area( triangle(AXC) ) = ½·4·x = 2x
Area( triangle(BYC) ) = ½·18·2x = 18x
Area( triangle(ABC) ) = Area( trapezoid(AXYB) ) - Area( triangle(AXC) ) - Area( triangle(BYC) )
= 33x - 2x - 18x = 13x
Also: area( triangle(ABC) ) ½·AC·BC = ½· sqrt(16 + x2)·sqrt(324 + 4x2)
= ½· sqrt( (16 + x2) · (324 + 4x2) )
= ½· sqrt( 4x4 + 388x2 + 5184 )
Therefore: ½· sqrt( 4x4 + 388x2 + 5184 ) = 13x
sqrt( 4x4 + 388x2 + 5184 ) = 26x
4x4 + 388x2 + 5184 = 676x
4x4 - 288x2 + 5184 = 0
x4 - 72x2 + 1296 = 0
(x2 - 36)(x2 - 36 = 0
(x + 6)(x - 6)(x + 6)(x - 6) = 0
---> x = 6
---> XC = x ---> XC = 6
---> YC = 2x ---> YC = 12
---> AC = sqrt(16 + x2) ---> AC = sqrt(52)
---> BC = sqrt(324 + 4x2) ---> BC = sqrt(468)
---> AB2 = AC2 + BC2 ---> AB2 = 52 + 468 = 520 ---> AB = sqrt(520)
