Geometry Basics: Chapter Test

A complement is whatever brings the angle up to a right angle, and a supplement is whatever brings it up to a straight angle, so both are subtractions.

$90^\circ - 41^\circ = 49^\circ, \qquad 180^\circ - 41^\circ = 139^\circ$

The two checks are $41^\circ + 49^\circ = 90^\circ$ and $41^\circ + 139^\circ = 180^\circ$. The supplement is the larger of the two, because $180$ is the larger total.

The reflex angle is the opening measured the other way round the vertex, so the two together sweep one complete rotation.

$360^\circ - 41^\circ = 319^\circ$

It is reflex because $180^\circ < 319^\circ < 360^\circ$, and the check is that $41^\circ + 319^\circ = 360^\circ$.

What each completion asks for. Each of the three is the amount left over after the opening is taken from a fixed total, so each exists only while that total is larger than the opening itself.

$90^\circ - 153^\circ < 0, \qquad 180^\circ - 153^\circ = 27^\circ, \qquad 360^\circ - 153^\circ = 207^\circ$

Reading the three results. The first subtraction runs past zero, and an angle cannot have a negative measure, so a $153^\circ$ opening has no complement at all. The other two land on genuine angles: a supplement of $27^\circ$ and a reflex partner of $207^\circ$.

The general rule. The test is simply which total the opening is smaller than. An opening below $90^\circ$ is below all three totals and has all three completions; one between $90^\circ$ and $180^\circ$ has a supplement and a reflex partner but no complement; and once an opening reaches $180^\circ$ only the full turn is still larger than it, though what the full turn leaves over is then $180^\circ$ or less and so no longer a reflex angle.

A sideways move changes only the first coordinate and an up-or-down move only the second, so handle the two separately. Right is positive and down is negative.

$(-4 + 11, \; 9) = (7, 9), \qquad (7, \; 9 - 13) = (7, -4)$

Each quadrant follows from the pair of signs: $(-, +)$ is the upper left, $(+, +)$ the upper right, and $(+, -)$ the lower right.

The distance from an axis is how far the point sits to one side of it, so it is the size of the coordinate measured across that axis, with the sign discarded.

$|7| = 7 \text{ units from the } y\text{-axis}, \qquad |-4| = 4 \text{ units from the } x\text{-axis}$

The first coordinate records position left and right of the y-axis, so it is the one that measures distance from that axis; the second does the same job for the x-axis.

Why the finish is the same. The two moves act on different coordinates. Whatever order they are made in, the first coordinate ends up as $-4 + 11$ and the second as $9 - 13$, so both routes finish on one square.

$(-4 + 11, \; 9 - 13) = (7, -4)$

Why the route is not. Reversing the order changes which coordinate is altered first, and the intermediate square records only that first change.

$(-4, \; 9 - 13) = (-4, -4)$

That square has both coordinates negative, which puts it in the lower left, Quadrant III, while the original route's intermediate square $(7, 9)$ has both positive and sits in Quadrant I. So the claim is right about the destination and wrong about the journey.

A triangle covers half of the parallelogram built from two copies of it, and that parallelogram has the same base and the same perpendicular height, so the rule is half the base times the height. It makes no difference that the perpendicular lands outside the triangle: the height is still the straight-across distance from $F$ to the line the base lies on.

$A = \tfrac{1}{2} \times b \times h = \tfrac{1}{2} \times 24 \times 13 = \tfrac{1}{2} \times 312 = 156 \text{ cm}^2$

Taking half of the $24$ first gives the same figure: $12 \times 13 = 156$.

The triangle has one area, whichever side is called the base, so the same area rule must hold with the new base in it. Undo the rule in the order it was built: multiply by $2$ to remove the one half, then divide by the new base.

$\tfrac{1}{2} \times 26 \times h = 156 \quad\Longrightarrow\quad 26h = 312 \quad\Longrightarrow\quad h = \frac{312}{26} = 12 \text{ cm}$

A check forward: $\tfrac{1}{2} \times 26 \times 12 = 156$, which matches part A.

Why the product is fixed. A triangle covers one definite amount of space, and the area rule says that amount is half of the base times the perpendicular height to that base. Rearranging the rule shows the product directly.

$A = \tfrac{1}{2} b h \quad\Longrightarrow\quad b h = 2A$

The right-hand side does not mention which side was chosen, so every choice of base gives the same product. Here both choices give $2 \times 156 = 312$, once as $24 \times 13$ and once as $26 \times 12$. A longer base is therefore always paired with a shorter height, which is exactly what happened between parts A and B.

Why the height falls outside. The perpendicular height to $DE$ is measured from $F$ straight down to the line that $DE$ lies on. Its foot lands between $D$ and $E$ only when $F$ sits above the stretch between them. Here the corner at $E$ is obtuse, which tips $F$ out past $E$, so the line has to be extended before the perpendicular can meet it. The height is still the straight-across distance from $F$ to that line, so the area rule is unaffected.

The measurement straight across through the centre is the diameter, so the rim is $\pi$ times it, and the strip across is that diameter itself.

$C = \pi d \approx \frac{22}{7} \times 84 = 22 \times 12 = 264 \text{ cm}$

Adding the two pieces gives the metal.

$264 + 84 = 348 \text{ cm}$

Both pieces are lengths, so the total is in centimetres. The fraction $\frac{22}{7}$ suits these numbers because $84$ divides by $7$.

The area rule needs the radius, and the stem gives the measurement across, so halve it first.

$r = \frac{84}{2} = 42 \text{ cm}$

Now square the radius and multiply by $\pi$.

$A = \pi r^2 \approx \frac{22}{7} \times 42^2 = \frac{22 \times 1764}{7} = 5544 \text{ cm}^2$

The radius was squared, so the unit is square centimetres.

What the glazier computed. The area rule squares the RADIUS. Putting $84$ into that slot works out the area of a circle whose radius is $84$ cm, which is a circle twice as wide as the window.

$\frac{22}{7} \times 84^2 = \frac{22 \times 7056}{7} = 22176 \text{ cm}^2$

How far out it is. Comparing that with the window's own area shows the size of the error.

$\frac{22176}{5544} = 4$

Why it is always four. The diameter is $2r$, so putting the diameter where the radius belongs replaces $r^2$ by $(2r)^2 = 4r^2$. The factor of $2$ is squared along with the radius, so this slip is four times too large for every circle, not just this one. That is also why the answer looks plausible: nothing in the arithmetic goes wrong, only the length that was fed into it.

Every slice taken across the post is the same square, so the post is a prism and its volume is that square's area multiplied by the length. Put both in one unit first: $5 \text{ m} = 500$ cm.

$B = 30 \times 30 = 900 \text{ cm}^2, \qquad V = B \times 500 = 450000 \text{ cm}^3$

A cubic metre is a cube $100$ cm along every edge, so it holds $100 \times 100 \times 100$ cubic centimetres.

$\frac{450000}{1{,}000{,}000} = 0.45 \text{ m}^3$

Unfold the post. Two of its faces are the square ends and the other four are long rectangles, each running the whole $500$ cm and $30$ cm across.

$2 \times 900 = 1800 \text{ cm}^2, \qquad 4 \times (30 \times 500) = 4 \times 15000 = 60000 \text{ cm}^2$

Adding the two groups gives the area to coat.

$S = 1800 + 60000 = 61800 \text{ cm}^2$

A square metre is a square $100$ cm on each side, so it holds $10{,}000$ square centimetres.

$\frac{61800}{10{,}000} = 6.18 \text{ m}^2$

Checking the label. A litre is a fixed number of cubic centimetres, so the conversion is a single division.

$\frac{450000}{1000} = 450 \text{ litres}$

The label claims $45$, ten times smaller, so the two do not agree. That figure is what comes of dividing by $10{,}000$ instead, which is the number of square centimetres in a square metre and has no business converting a volume.

Why the two measurements carry different units. Count the lengths each calculation multiplies. The volume took a side, a side and the length, three lengths in all, and three factors of centimetres make cubic centimetres. Each face of the net took only two lengths: the two sides of a square end, or the width and the length of a long face. Two factors make square centimetres, and adding flat pieces together cannot turn them into anything else. This is why the post can be described both as holding $450$ litres of timber and as taking $6.18$ square metres of preservative without the two figures being in competition: one counts space and the other counts material.

Two copies of a trapezoid fit together into a parallelogram whose base is the sum of the two parallel edges, so one trapezoid covers half of that, which is the average of the parallel edges times the height.

$\frac{23 + 31}{2} = \frac{54}{2} = 27 \text{ m}, \qquad A = 27 \times 16 = 432 \text{ m}^2$

The terrace therefore covers as much ground as a rectangle $27$ m wide and $16$ m deep.

The pond sits wholly inside the terrace, so the paving is the terrace with one rectangle taken out of it. Area is additive, so subtract.

$A_{\text{pond}} = 6 \times 5 = 30 \text{ m}^2, \qquad 432 - 30 = 402 \text{ m}^2$

Each slab covers exactly one square metre, so the count of slabs matches the number of square metres.

What the contractor computed. Adding the two parallel edges and multiplying by the height is the parallelogram step, the one taken before the halving.

$(23 + 31) \times 16 = 54 \times 16 = 864 \text{ m}^2$

That is the parallelogram built from the terrace and an upside-down copy of it, so it covers twice what the terrace does: $2 \times 432 = 864$.

What the quote becomes. The pond is subtracted from the inflated figure rather than from the terrace.

$864 - 30 = 834 \text{ m}^2$

Why the error does not simply double the answer. The pond is taken off only once, so the quoted figure is not twice the true $402 \text{ m}^2$ either; it is $432 \text{ m}^2$ too large, the extra copy of the terrace. An error made early and then built on does not stay a clean multiple of the right answer.

The side $PQ$ is a straight line meeting both $PS$ and $QR$, and those two are parallel, so $PQ$ is a transversal of them. The angles at $P$ and at $Q$ lie between the two parallel sides and on the same side of $PQ$, which makes them co-interior.

$\angle P + \angle Q = 180^\circ \quad\Longrightarrow\quad \angle Q = 180^\circ - 119^\circ = 61^\circ$

The relationship needs $PS$ and $QR$ to be parallel, and the definition of a parallelogram supplies exactly that.

Work round the frame, taking each side in turn as the transversal of the pair it crosses.

$\angle R = 180^\circ - \angle Q = 180^\circ - 61^\circ = 119^\circ, \qquad \angle S = 180^\circ - \angle R = 61^\circ$

The first uses $QR$ cutting the parallel pair $PQ$ and $SR$; the second uses $SR$ cutting the parallel pair $PS$ and $QR$.

Extending $SR$ past $R$ puts a new angle beside $\angle R$ with their outer sides in one straight line, which is a linear pair.

$180^\circ - 119^\circ = 61^\circ$

A check on the whole frame: the four corner angles come to $119 + 61 + 119 + 61 = 360$ degrees.

Adjacent corners. Label the parallelogram $ABCD$ with the corners in order, so that $AB$ is parallel to $DC$ and $AD$ is parallel to $BC$. Take the side $AB$. It is a straight line meeting $AD$ at $A$ and $BC$ at $B$, and $AD$ and $BC$ are parallel, so $AB$ is a transversal of them. The angles at $A$ and $B$ both lie between those parallel sides and both on the same side of $AB$, so they are co-interior.