4 Geometry and trigonometry

A small square marks a right angle. An arc normally marks the intended angle; for a reflex angle, follow the larger turn around the vertex.

Write one equation at a time and name the fact used. Transfer an angle through equalities first, then use a straight-line or point sum where needed.

If a triangle has angles $30^\circ$, $(4x+10)^\circ$ and $(x+20)^\circ$, then $30+4x+10+x+20=180$, so $x=24$.

If the equal base angles of an isosceles triangle are each $(x+52)^\circ$ and another expression for one is $(3x+10)^\circ$, equate them first, then use the triangle sum to find the apex angle.

Equal angles also face equal sides. This converse can prove that a triangle is isosceles when side equality is not given.

Parallelogram, rectangle, square, rhombus, trapezium and kite are all quadrilaterals distinguished by side, angle and parallel-line properties.

If the angles are $90^\circ$, $(x+15)^\circ$, $(x+25)^\circ$ and $(x+35)^\circ$, then $90+x+15+x+25+x+35=360$, giving $x=65$.

A rectangle has equal diagonals that bisect each other; a rhombus has perpendicular diagonals that bisect each other; a square has both sets of properties.

If each exterior angle is $24^\circ$, then $n=360/24=15$. If each interior angle is $162^\circ$, the exterior angle is $18^\circ$, so $n=20$.

Subtract all known interior angles from the total to find a missing angle. For algebraic angles, form one equation equal to the total.

Drawing diagonals from one vertex divides an $n$-gon into $n-2$ triangles, which explains the formula.

A congruent copy may face a different direction or be reflected. Orientation and position do not change length or angle measurements.

Match vertices in order and compare every corresponding side and angle. A single mismatch proves the figures are not congruent.

A congruence statement must list corresponding vertices in matching order. If $ABCD$ matches $PQRS$, then $AB$ corresponds to $PQ$ and angle $B$ to angle $Q$.

A regular hexagon has rotational order 6. A non-square rhombus has two diagonal lines of symmetry and rotational order 2.

If 300 and 400 are separated by five equal spaces, each space represents 20. A pointer two spaces after 300 reads $300+2(20)=340$.

From 1635 to 2015: 25 minutes to 1700, then 3 hours 15 minutes to 2015, so the interval is 3 hours 40 minutes.

Choose the measure type, select a plausible unit, compare with a known benchmark, then reject values that are ten or a hundred times too large or small.

Reverse bearings differ by $180^\circ$: add $180^\circ$ if the bearing is below $180^\circ$; subtract $180^\circ$ if it is at least $180^\circ$. Thus the reverse of $073^\circ$ is $253^\circ$.

If the ray falls between degree marks, read the closest mark; for example $109.6^\circ$ rounds to $110^\circ$.

For 40 km in 2 hours 15 minutes, convert time to $2.25$ hours. Then $v=40/2.25=17.77\ldots$, so the average speed is $18 ext{ km/h}$ to the nearest whole number.

A $12 ext{ cm} imes8 ext{ cm} imes5 ext{ cm}$ block has volume $480 ext{ cm}^3$. At density $0.7 ext{ g/cm}^3$, its mass is $0.7(480)=336 ext{ g}$.

Convert before substituting: $1 ext{ kg}=1000 ext{ g}$ and $1 ext{ m/s}=3.6 ext{ km/h}$. The numerator and denominator units determine the compound unit.

A reading of $5.37 ext{ cm}$ is $53.7 ext{ mm}$, which rounds to $54 ext{ mm}$ or $5.4 ext{ cm}$ to the nearest millimetre.

For sides $AB=8$ cm, $AC=6$ cm and $BC=9$ cm: draw $AB$; draw an arc radius 6 cm centred at $A$ and an arc radius 9 cm centred at $B$; their intersection is $C$; join $AC$ and $BC$.

At scale $1 ext{ cm}:5 ext{ km}$, a drawing distance of $3.6$ cm represents $18$ km. A real distance of $12$ km is drawn as $12/5=2.4$ cm.

Points on a perpendicular bisector are equidistant from the segment endpoints. Points on an angle bisector are equidistant from the two angle arms.

A diameter is always a chord, but a chord is a diameter only when it passes through the centre.

Add the radius to a tangent diagram to create a right angle. Join the centre to a chord midpoint to create two congruent right triangles when useful.

The sides of a cyclic quadrilateral are chords of the circle; its diagonals are also chords.

If chord $AC$ subtends $38^\circ$ at point $B$, then angle $AOC=76^\circ$. Since $OA=OC$, each base angle in triangle $AOC$ is $(180-76)/2=52^\circ$.

Mark the chord or arc each angle stands on before selecting a theorem, then combine with triangle, straight-line or point-angle facts.

Mark known values, find one angle at a time, state the property used, then substitute that result into the next shape. Finish by checking every local angle sum.

If $ABCD$ is cyclic and angle $A=112^\circ$, then angle $C=68^\circ$ because opposite angles in a cyclic quadrilateral sum to $180^\circ$.

Mark the right angle, identify the hypotenuse, substitute lengths with units, then take the positive square root. In a compound shape, form one right triangle at a time and carry the unrounded result forward.

For a length, rearrange the chosen ratio. For an angle, use the matching inverse function, such as $\theta=\tan^{-1}(O/A)$. Keep the calculator in degree mode and round only the final answer.

A bearing such as $142^\circ$ describes a direction from north; an interior triangle angle such as $38^\circ$ is not automatically the final bearing.

Sketch the angle in the second quadrant, find its acute reference angle, apply the correct sign, and check the calculator is in degree mode.

Draw a horizontal through the observer, label the vertical difference and horizontal distance, then use the right-triangle ratio that connects the known sides to the required angle or length.

Label opposite pairs, choose a form with one unknown, substitute without premature rounding, and test the result against the largest-side/largest-angle relationship. For the sine-rule ambiguous case, check whether the supplementary angle also satisfies the data and angle sum.

Identify the two endpoints, draw or name a helpful face projection, prove the relevant angle is $90^\circ$, then apply Pythagoras once or twice. Keep surds exact when requested.

Split compound shapes into triangles, calculate each contribution, and add or subtract as the geometry requires. Preserve full precision until the final stated accuracy.

A useful check is that the projection is shorter than the sloping line and the chosen angle lies between $0^\circ$ and $90^\circ$.

Write the unit relationship first, raise its numerical factor to the power shown by the unit, then multiply toward smaller units or divide toward larger units. Include the converted unit in the answer.

For shapes made from identical rectangles or triangles, shared edges disappear from the perimeter. A cost per metre is applied only after the perimeter has been found.

Draw decomposition lines, label each base and perpendicular height, calculate in one unit, and preserve full precision before any coverage or cost decision. Round a required number of whole tins or tiles upward.

Mark the parallel sides, identify or derive their perpendicular separation, substitute with consistent units, and rearrange the same formula if a missing length is required.

For a composite region, find each radius from the given diameters, then add or subtract the relevant circular and polygonal parts. Use exact $\pi$ during working and round only at the end.

Identify the centre, radius and intended minor or major angle, calculate the circle fraction in degrees, then include only the requested boundary or area. Radian measure is outside this syllabus objective.

Identify any repeated parallel cross-section, then check the number and shape of plane faces and whether a curved surface or single apex is present.

For a prism with an $n$-sided end face: faces $=n+2$, edges $=3n$, vertices $=2n$. Trace systematically so hidden dashed edges are included.

For a right prism, lateral area equals perimeter of cross-section $\times$ prism length; then add the two end areas when both are exposed.

Confirm whether the given circular measure is radius or diameter, find any missing height from other data if needed, and include exactly the exposed circular ends.

For packing, volume alone gives an upper bound; whole boxes must also fit by their dimensions. For filling, divide the required volume by a rate or container capacity and round according to context.

Write the one-dimensional relationship, cube its factor for cubic units, then multiply toward smaller units or divide toward larger units. Use the litre bridge only after units are compatible.

For joined solids, add volumes but count only external surfaces. For a hollow or removed part, subtract its volume or exposed area. Similar cones scale lengths by $k$, areas by $k^2$ and volumes by $k^3$.

A valid scale factor gives the same ratio for every corresponding side. Angles remain unchanged and are never multiplied by $k$.

Measure between the specified points with the same ruler convention used by the scale, keep units explicit, and allow for stated measurement tolerance before comparing routes or distances.

Keep the direction consistent—new divided by original—then apply the squared factor to every corresponding area, including curved surface area.

Match corresponding solids, write the factor direction, cube only the linear factor, and attach cubic units to the result.

Choose one direction and convert the given ratio back to the linear factor before moving to the required dimension. Apply totals or differences only after corresponding measures have been expressed consistently.