Maths Problem

1)
Let BX = y cm，then XC = (8 - y) cm

In ΔABX:
AX² = BX² + AB² (Pythagorean theorem)
AX² = (y cm)² + (6 cm)²
AX = √(y² + 36) cm

AX = XC
√(y² + 36) = 8 - y
y² + 36 = 64 - 16y + y²
16y = 28
y = 7/4
y = 1.75

Hence, BX = 1.75 cm


2.
Let A, B, C and D be the 4 vertices of the quadrilateral,
where AB = 5, BC = x, CD = 7 and DA = 3

Let H be the point of intersection of the two diagonals.
let AH = p, BH = q, CH = r and DH = s

By Pythagorean theorem :
p² + q² = 5² ...... [1]
q² + r² = x² ...... [2]
r² + s² = 7² ...... [3]
s² + p² = 3² ...... [4]

[1] + [3]:
p² + q² + r² + s² = 74 ...... [5]

[2] + [4]:
p² + q² + r² + s² = x² + 9 ...... [6]

[6] = [5]:
x² + 9 = 74
x² = 65
x = √65