Two right-angled triangles are shown below - Junior Cycle Mathematics - Question 11 - 2015

For the triangle with height 9 and base n: $9^2 + n^2 = (n + 1)^2$ $81 + n^2 = n^2 + 2n + 1$ $81 = 2n + 1$ $80 = 2n$ So, $n = 40$.

The bases follow a quadratic pattern: 84, 112, 144. Calculating the first differences:

$112 - 84 = 28$
$144 - 112 = 32$

Now calculating the second differences:

$32 - 28 = 4$

The next first difference will be: $32 + 4 = 36$

Thus, the length of the next base is: $144 + 36 = 180$.

From h(1) = 5: $h(1) = 2(1)^2 + b(1) + c = 5$ This leads to:
Equation 1:
$2 + b + c = 5$ $b + c = 3$

From h(2) = 13: $h(2) = 2(2)^2 + b(2) + c = 13$ This leads to:
Equation 2:
$8 + 2b + c = 13$ $2b + c = 5$.

To solve: Equation 1:
b + c = 3
Equation 2:
2b + c = 5

Subtract Equation 1 from Equation 2: $2b + c - (b + c) = 5 - 3$ $b = 2$

Substituting back into Equation 1: $2 + c = 3$
Thus,
$c = 1$.