Areas and Perimeters Worked ExamplesIGCSE Maths 0580

Identify what the formula needs: the two parallel sides ($8\text{ cm}$ and $12\text{ cm}$) and the perpendicular height between them ($4\text{ cm}$, the dashed line). The $5\text{ cm}$ sloping side is not needed.

Substitute.

Evaluate.

(a) A parallelogram is base times perpendicular height.

(b) A triangle is half of the corresponding parallelogram.

Write the area formula with the height unknown.

Simplify the right-hand side.

Divide.

(a) Find the area of the full rectangle.

Subtract the removed rectangle.

(b) For the perimeter, notice the two cut edges replace exactly the two edges they removed — the total distance around is unchanged.

Area of the rectangle.

Area of the triangle.

Add, since the shapes sit side by side without overlapping.

Let the length be $l$ and the width $w$, then write both conditions.

The area gives a second equation.

Substitute $w = 23 - l$ into the area equation.

Rearrange into a quadratic.

Factorise — two numbers multiplying to $120$ and adding to $23$.

The two roots are the length and the width — they are the same rectangle either way.

(a) No side lies along a grid line, so there is no base and height to read off. Instead, surround the triangle with the smallest rectangle whose sides are grid lines: $6\text{ cm}$ across (from $A$ to $B$) and $5\text{ cm}$ up (from $A$ to $C$).

Three right-angled triangles fill the corners of the rectangle outside $ABC$. Read their legs along the grid: $A$ to $B$ is $6$ across and $2$ up; $B$ to $C$ is $4$ across and $3$ up; $C$ to $A$ is $2$ across and $5$ down.

Subtract the corners from the rectangle.

(b) Take $AB$ as the base. Its length comes from Pythagoras on the grid steps.

The perpendicular distance from $C$ to $AB$ is the height that goes with this base, so use $\tfrac{1}{2} \times AB \times h = 13$.

(a) The shaded square is the big square minus four identical right-angled triangles in the corners. Each has legs $3\text{ cm}$ and $12 - 3 = 9\text{ cm}$.

Subtract all four from the big square.

(b) Each side of $PQRS$ is the hypotenuse of a corner triangle.

A square has four equal sides.

(c) Now each corner triangle has legs $x$ and $12 - x$. Form the same subtraction with the new area.

Expand and rearrange into a quadratic, then divide through by $2$.

Factorise. Both roots lie between $0$ and $12$, so both are valid — they are mirror images of each other ($AP = 4$, $PB = 8$ or the other way round).

Let $AB = DC = b$ and let $h$ be the perpendicular distance between $AB$ and $DC$, so $bh = 60$. Triangle $AEF$ is the parallelogram with three unshaded triangles removed.

$\triangle ADE$ has base $DE = \tfrac{1}{3}b$ on $DC$, and $A$ is a full height $h$ from $DC$.

$F$ is the midpoint of $BC$, so it is exactly halfway between the parallel sides: its distance from $AB$ is $\tfrac{1}{2}h$.

$\triangle ECF$ has base $EC = \tfrac{2}{3}b$ on $DC$, and $F$ is also $\tfrac{1}{2}h$ from $DC$.

Subtract.

Nothing depended on how far the parallelogram leans, so triangle $AEF$ is always $\tfrac{5}{12}$ of the parallelogram for these positions of $E$ and $F$.

Each trapezium has parallel sides $AE = x$ and $DM = 10.2$ (both vertical), with perpendicular distance $AM = 6.5$ between them. The two trapeziums are congruent, so each has half the area of the wall.

Form an equation with the trapezium formula, keeping the bracket round the parallel sides.

Undo each operation in turn.

Check the whole wall.

(a) The sloping side is not given. Drop a vertical from the top-right corner: it makes a right-angled triangle with height $8\text{ cm}$ and horizontal side $15 - 9 = 6\text{ cm}$.

Use Pythagoras for the sloping side.

Add all four sides.

(b) Area in $\text{cm}^2$ first, with the $8\text{ cm}$ side as the perpendicular height.

$1\text{ m} = 100\text{ cm}$, so $1\text{ m}^2 = 100 \times 100 = 10\,000\text{ cm}^2$. Divide by $10\,000$.