Sure, let's solve this problem step-by-step.
(i) If the hypotenuse of the triangle is denoted by 'h', then:
- One side is 1 cm less than the hypotenuse, which gives [tex]\( a = h - 1 \)[/tex].
- The other side is 8 cm less than the hypotenuse, which gives [tex]\( b = h - 8 \)[/tex].
(ii) To find the value of the hypotenuse [tex]\( h \)[/tex], we use the Pythagorean theorem, which states:
[tex]\[ a^2 + b^2 = h^2 \][/tex]
Substituting the expressions for the sides:
[tex]\[ (h - 1)^2 + (h - 8)^2 = h^2 \][/tex]
Expanding and solving the equation:
[tex]\[ (h - 1)^2 = h^2 - 2h + 1 \][/tex]
[tex]\[ (h - 8)^2 = h^2 - 16h + 64 \][/tex]
Adding these two expressions:
[tex]\[ h^2 - 2h + 1 + h^2 - 16h + 64 = h^2 \][/tex]
Simplifying:
[tex]\[ 2h^2 - 18h + 65 = h^2 \][/tex]
[tex]\[ h^2 - 18h + 65 = 0 \][/tex]
Solving this quadratic equation for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(65)}}{2(1)} \][/tex]
[tex]\[ h = \frac{18 \pm \sqrt{324 - 260}}{2} \][/tex]
[tex]\[ h = \frac{18 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ h = \frac{18 \pm 8}{2} \][/tex]
This gives us two potential values:
[tex]\[ h = \frac{26}{2} = 13 \][/tex]
[tex]\[ h = \frac{10}{2} = 5 \][/tex]
Since the hypotenuse is the longest side of the right-angled triangle, and it must be greater than the other sides, we choose:
[tex]\[ h = 13 \][/tex]
(iii) Now, we can find the sides of the right-angled triangle:
- One side is [tex]\( a = h - 1 = 13 - 1 = 12 \)[/tex] cm.
- The other side is [tex]\( b = h - 8 = 13 - 8 = 5 \)[/tex] cm.
So, the sides of the right-angled triangle are 12 cm and 5 cm, with the hypotenuse being 13 cm.
