14. x = 7; Diagonal length = √98 = 9.89
15. x = 7; Diagonal length =√1352 = 36.8
16. x = 6; Diagonal length = √1250 = 35.4
17. x = 3; Diagonal length = √1586 = 39.8
18. x = 4; Diagonal length = √512 = 22.6
19. x = 10; Diagonal length = √5000 = 70.7
How do we find the value of x and each diagonal length?
14. HJ = x and IK = 2x - 7
HJ = IK → x = 2x−7
Now, solve for x: ⇒ 7=x
HI = JK = √(HJ² + IK²)
√(7² + (2×7 - 7)²)
√49+49 → √98 = 9.89
15. HJ = 3x + 5 and IK = 5x - 9
3x + 5 = 5x − 9
14 = 2x
x = 7
Diagonal length = HI = JK = √((3(7)+5)² + (5(7)-9)²
HI = JK = √676 + 676
HI = JK= √1352 = 36.8
16. HJ = 3x + 7 and IK = 6x - 11
3x+7 = 6x−11
18 = 3x ⇒ x = 6
Diagonal length = HI=JK= √((3(6)+7)² + (6(6)-11)²)
Diagonal length = √625 + 625
Diagonal length = √1250 = 35.4
17. HJ = 19+ 2x and IK = 3x + 22
19+ 2x = 3x + 22
x = 3
Diagonal length = √((19+2(3))² + (3(3)+22)²)
Diagonal length = √625+961
Diagonal length = √1586 = 39.8
18. HJ = 4x and IK=7x-12
4x = 7x-12
3x = 12
x = 4
Diagonal length = √((4(4))² + (7(4)-12)²)
Diagonal length = √256+256
Diagonal length = √512 = 22.6
19. HJ=x+40 and IK = 5x
x+40 = 5x
4x = 40
x = 10
Diagonal length = √((10+40)² + (5(10))²)
Diagonal length = √2500 + 2500
Diagonal length = √5000 = 70.7