4. Geometry

A semicircle is half a circle, formed when a diameter splits the circle. For example, if the diameter is $16.8$ cm, the radius is $16.8\div2=8.4$ cm.

Perpendicular lines meet at $90^\circ$. When a task asks for a perpendicular line, use the grid or a protractor/set square to fix the right angle, then rule the line.

Every point on the first arc is the second length from its centre, and every point on the second arc is the third length from its centre. Their intersection therefore satisfies both required distances.

For sides $6.5$ cm, $6.5$ cm and $8$ cm, draw an $8$ cm base. Two arcs of radius $6.5$ cm—one from each endpoint—meet at the third vertex. Joining that point to both endpoints completes the triangle.

Start with one face, attach neighbouring faces along edges of equal length, and account for every face once. Use a ruler for all straight edges. Then mentally fold around each shared edge: faces must meet to close the solid, not cover the same position.

Dimensions transfer across matching edges. A cuboid net for dimensions $3$ cm by $2$ cm by $1$ cm must contain two $3\times2$, two $3\times1$ and two $2\times1$ rectangles. Their total area is the cuboid's surface area, $2(3\times2+3\times1+2\times1)=22$ cm$^2$, and the enclosed volume is $3\times2\times1=6$ cm$^3$.

Put both lengths in the same unit before using a ratio. For a scale of $1:50\,000$, $1$ cm on the drawing represents $50\,000$ cm $=0.5$ km in reality. A measured map length of $8.5$ cm therefore represents $8.5\times0.5=4.25$ km.

To draw a distance, reverse the conversion: drawing length $=$ actual length $\div n$, after matching units. Rule straight edges. If a point is fixed by its distances from two known points, convert both distances and use two arcs; their intersection locates the point.

For the bearing of $B$ from $A$, place north at $A$, centre the protractor at $A$, and measure clockwise from the north line to $AB$. To draw a bearing, mark that clockwise angle from north, rule the ray, then use the stated scale to mark the distance.

If the bearing of $B$ from $A$ is $025^\circ$, the bearing of $A$ from $B$ is $025^\circ+180^\circ=205^\circ$. If the original bearing is $322^\circ$, the reverse is $322^\circ-180^\circ=142^\circ$.

Choose one complete pair of corresponding lengths and divide in the direction you need to find $k$. Apply that same multiplier to the matching unknown side. Keep every ratio in the same shape order; an equivalent proportion is also valid.

Two similar plant pots have corresponding diameters $27$ cm and $33$ cm. If the smaller height is $21.6$ cm, the enlargement factor is $33/27$. The larger height is $21.6\times33/27=26.4$ cm.

An enlargement factor greater than $1$ must produce a longer matching side; a reduction factor between $0$ and $1$ must produce a shorter one. This size check often catches an inverted ratio.

A line of symmetry must pair every point, edge and shaded region with a mirror image the same perpendicular distance on the other side. Test vertical, horizontal and diagonal candidates, but draw only those that reflect the entire figure.

Turn the shape through equal angles around its centre and count each exact match before reaching $360^\circ$. If the smallest matching turn is $\theta$, the order is $360^\circ/\theta$. A half-turn match gives order $2$.

In an isosceles triangle with equal base angles $41^\circ$, the third angle is $180^\circ-41^\circ-41^\circ=98^\circ$. If its exterior angle lies on a straight line, that exterior angle is $180^\circ-98^\circ=82^\circ$.

In $\angle ABC$, the middle letter $B$ is the vertex. This three-letter notation distinguishes the intended angle when several rays meet at one point.

Start from the given angle and move to the unknown in one justified step when possible. If the required pair is not immediate, first use a straight-line or vertically opposite fact, then apply the named parallel-line relationship.

If an interior angle is $112^\circ$, its alternate angle is also $112^\circ$. The co-interior angle on the same side of the transversal is $180^\circ-112^\circ=68^\circ$.

Equal corresponding or alternate angles can also establish that two lines are parallel; so can co-interior angles whose sum is $180^\circ$. State the pair used.

For a regular 15-sided polygon, the exterior angle is $360^\circ/15=24^\circ$, so each interior angle is $180^\circ-24^\circ=156^\circ$. The interior-angle sum is $13\times180^\circ=2340^\circ$.

If each exterior angle is $20^\circ$, then $n=360^\circ/20^\circ=18$. A non-integer result means identical turns of that size cannot close to form a regular polygon.

If $DF$ is a diameter and $E$ lies on the circle, $\angle DEF=90^\circ$. When $\angle DFE=49^\circ$, $\angle EDF=180^\circ-90^\circ-49^\circ=41^\circ$.

If $OB$ is a radius and $AB$ is tangent at $B$, then $\angle OBA=90^\circ$. With $\angle OAB=36^\circ$, triangle $OAB$ gives $\angle AOB=180^\circ-90^\circ-36^\circ=54^\circ$.