Pythagoras’ Theorem at Extended Level

An isosceles triangle has sides 13 cm, 13 cm and 10 cm. Find its height \(h\) and its area.

The line of symmetry cuts the base into 5 cm and 5 cm. In the right-angled half, 13 is the hypotenuse:

\(h = \sqrt{13^{2} – 5^{2}} = \sqrt{169 – 25} = \sqrt{144} = 12\) cm

Area \(= \dfrac{1}{2} \times 10 \times 12 = 60\) cm²

\(A\) is \((-3, -1)\) and \(B\) is \((5, 3)\). Find the length of \(AB\) as a surd in its simplest form.

Across: \(5 – (-3) = 8\). Up: \(3 – (-1) = 4\).

\(AB = \sqrt{8^{2} + 4^{2}} = \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\)

(As a decimal, \(4\sqrt{5} = 8.94\) to 3 s.f. — but “exact” means leave \(4\sqrt{5}\).)

In the diagram, angle \(ABC = 90°\) and \(D\) is on \(BC\). \(AB = 9\) cm, \(AD = 15\) cm and \(DC = 7\) cm. Find \(AC\).

Step 1 — triangle \(ABD\): \(BD = \sqrt{15^{2} – 9^{2}} = \sqrt{144} = 12\) cm.

Step 2 — triangle \(ABC\): \(BC = 12 + 7 = 19\) cm, so

\(AC = \sqrt{9^{2} + 19^{2}} = \sqrt{442} = 21.02\ldots = 21.0\) cm (3 s.f.)

You cannot reach \(AC\) in one step. Find the side the two triangles share first.

Which of these triangles is right-angled? (a) 20 cm, 21 cm, 29 cm   (b) 7 cm, 9 cm, 12 cm

(a) \(20^{2} + 21^{2} = 400 + 441 = 841\) and \(29^{2} = 841\). Equal, so it is right-angled.

(b) \(7^{2} + 9^{2} = 49 + 81 = 130\) but \(12^{2} = 144\). Not equal, so it is not right-angled.

Always square the two shorter sides and compare with the square of the longest side.

A square has a diagonal of 10 cm. Find the exact length of its side and its area.

Call the side \(s\). The diagonal is the hypotenuse of a right-angled triangle with two sides \(s\):

\(s^{2} + s^{2} = 10^{2} \Rightarrow 2s^{2} = 100 \Rightarrow s^{2} = 50 \Rightarrow s = \sqrt{50} = 5\sqrt{2}\) cm

Area \(= s^{2} = 50\) cm². You do not even need \(s\) itself for the area — \(s^{2}\) is already there.