Pythagoras' Theorem in Plane Figures

Solution

(a) In $\triangle ABC$, $\angle B = 90^{\circ}$, so $AC$ is the hypotenuse.

$d^2 = 80^2 + 45^2 = 6400 + 2025 = 8425\text{,}$ so $d = \sqrt{8425} = 91.78\ldots = 91.8$ (3 s.f.).

(b) The diagonal of a square cuts it into two right-angled triangles whose other two sides are both $6$ cm.

$d^2 = 6^2 + 6^2 = 36 + 36 = 72\text{,}$ so $d = \sqrt{72} = 8.485\ldots = 8.49$ (3 s.f.).

Solution

In $\triangle ABC$, $\angle B = 90^{\circ}$ and the diagonal $AC$ is the hypotenuse.

$BC^2 = 17^2 - 15^2 = 289 - 225 = 64\text{,}$ so $BC = 8$ cm.

Area $= 15 \times 8 = 120$ cm².

Solution

The board is wider than the doorway and taller than it whichever way it is turned: $210$ cm and $250$ cm are both more than $80$ cm and $200$ cm. Carried upright, it does not go through.

But a thin board can be tilted, so that one of its edges lies along the diagonal of the doorway while its other edge points straight through the doorway, at right angles to the wall. The longest line across the doorway is its diagonal $PR$.

$PR^2 = 80^2 + 200^2 = 6400 + 40\,000 = 46\,400\text{,}$ so $PR = \sqrt{46\,400} = 215.4\ldots$ cm.

The board's shorter edge, $210$ cm, is less than $215.4\ldots$ cm. So, tilted with its $210$ cm edge along the diagonal and its $250$ cm edge going through the doorway, the board fits.

Solution

$AM$ is the line of symmetry, so $\angle AMB = 90^{\circ}$ and $BM = 10 \div 2 = 5$ cm.

In $\triangle ABM$, $AB$ is the hypotenuse: $AM^2 = 13^2 - 5^2 = 169 - 25 = 144\text{,}$ so $AM = 12$ cm.

Area $= \frac{1}{2} \times 10 \times 12 = 60$ cm².

Solution

The height $PN$ meets $QR$ at right angles at its midpoint $N$, so $QN = 4$ cm.

In $\triangle PQN$: $h^2 = 8^2 - 4^2 = 64 - 16 = 48\text{,}$ so $h = \sqrt{48} = 6.928\ldots = 6.93$ cm (3 s.f.).

Area $= \frac{1}{2} \times 8 \times 6.928\ldots = 27.71\ldots = 27.7$ cm² (3 s.f.). Use the full value of $h$, not $6.93$.

Solution

The diagonals cut each other in half at right angles at $O$: $AO = 16 \div 2 = 8$ cm, $BO = 12 \div 2 = 6$ cm, and $\angle AOB = 90^{\circ}$.

In $\triangle AOB$: $AB^2 = 8^2 + 6^2 = 64 + 36 = 100\text{,}$ so $AB = 10$ cm.

All four sides of a rhombus are equal: perimeter $= 4 \times 10 = 40$ cm.

Solution

$DCFE$ is a rectangle, so $EF = DC = 10$ cm. The two right-angled triangles at the ends are the same, so $AE = FB = (22 - 10) \div 2 = 6$ cm.

In $\triangle ADE$: $h^2 = 10^2 - 6^2 = 100 - 36 = 64\text{,}$ so $h = 8$ cm.

Area $= \frac{1}{2} \times (22 + 10) \times 8 = 128$ cm².

Solution

(a) In $\triangle APC$, $\angle C = 90^{\circ}$ and $AP$ is the hypotenuse: $AC^2 = 25^2 - 15^2 = 625 - 225 = 400\text{,}$ so $AC = 20$ cm.

(b) $BC = BP + PC = 21 + 15 = 36$ cm. In $\triangle ABC$, $AB$ is the hypotenuse: $AB^2 = 36^2 + 20^2 = 1296 + 400 = 1696\text{.}$

$AB = \sqrt{1696} = 41.18\ldots = 41.2$ cm (3 s.f.).

Solution

(a) In $\triangle XYZ$, $\angle Y = 90^{\circ}$: $XZ^2 = 12^2 + 16^2 = 144 + 256 = 400\text{,}$ so $XZ = 20$ cm.

(b) In $\triangle XYM$, $\angle XMY = 90^{\circ}$ and $XY$ is the hypotenuse: $XM^2 = 12^2 - 9.6^2 = 144 - 92.16 = 51.84\text{,}$ so $XM = \sqrt{51.84} = 7.2$ cm.