Answer:
C 13
Explanation:
2 right-angled triangles. so, Pythagoras applies for both.
c² = a² + b²
with c being the Hypotenuse (the side opposite of the 90° angle).
starting with the small triangle we have
5² = 3² + x-part²
x-part² = 5² - 3² = 25 - 9 = 16
x-part = 4 cm
the markers indicate that x-part as the third side of the small triangle is exactly as long as the other part of x.
that means
x = 2 × x-part = 2 × 4 = 8 cm
for the large triangle we then get
y² = x² + 15² = 8² + 225 = 64 + 225 = 289
y = 17 cm
and so,
y - x = 17 - 8 = 9
but that is not an answer option.
and now it dawned on me : x was not meant to be the whole third side of the large triangle. x is supposed to be only the upper part of that side, above the small triangle.
in essence, x is what I originally called "x-part".
so, the Pythagoras calculations remain basically the same, but we should name things differently like
5² = 3² + small-side²
small-side = 4 cm
x = small-side = 4 cm
long-side = 2 × small-side = 2 × 4 = 8 cm
y² = long-side² + 15² = 8² + 15² = 64 + 225 = 289
y = 17 cm
and the actual y - x = 17 - 4 = 13 cm