Exercises — Area — triangle, parallelogram, trapezium, composite shapes

The two sides meet at a right angle, so one is the base and the other is already the perpendicular height. No extra work needed. $A=\frac{1}{2}bh=\frac{1}{2}\times 9\times 6=27{\text{cm}}^2$

The height given is already perpendicular (the problem says so), so plug straight in — no halving, because a parallelogram is two triangles. $A=bh=12\times 5=60{\text{m}}^2$

Add the two parallel sides, then halve their sum times the height. $A=\frac{1}{2}(a+b)h=\frac{1}{2}(7+11)\times 4=\frac{1}{2}\times 18\times 4=36{\text{cm}}^2$

Start from $A=\frac{1}{2}bh$ and rearrange for $h$. We divide both sides by $\frac{1}{2}b$, i.e. multiply by $\frac{2}{b}$: $h=\frac{2A}{b}=\frac{2\times 40}{16}=\frac{80}{16}=5\text{cm}$

Guided look at the figure. The blue outline is the leaning parallelogram. The orange line is the $8\text{m}$ slant side. Drop straight down from its top end: that dashed green line is the true height $h$, and the dotted gray line along the bottom is its horizontal shadow. Slant (orange), shadow (gray) and height (green) form a right triangle, with the red arc marking the $30^{\circ }$ lean at the base corner. The slant side is not the height. The height is the vertical leg — the side opposite the $30^{\circ }$ angle — so we use sine (opposite over hypotenuse): $h=8\sin 30^{\circ }=8\times 0.5=4\text{m}$ Now the perpendicular height is known: $A=bh=10\times 4=40{\text{m}}^2$

Guided look at the figure. The blue trapezium has its short top ($a=6$) centred over its long bottom ($b=14$). The red segment on the left is the overhang — the bit of the bottom that pokes out past the top. The orange line is the $5\text{cm}$ slant, and the dashed green line is the height $h$ we want. Overhang (red), height (green) and slant (orange) form a right triangle. The height isn't given, so build it. Because the trapezium is symmetric, the bottom sticks out equally on both sides: total overhang $=14-6=8\text{cm}$, split into two, so each side overhangs $4\text{cm}$. Each slant side is now the hypotenuse of a right triangle with base $4\text{cm}$ and vertical leg $h$. By Pythagoras: $h=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3\text{cm}$ Now apply the trapezium formula: $A=\frac{1}{2}(6+14)\times 3=\frac{1}{2}\times 20\times 3=30{\text{cm}}^2$

Key idea: the area of the triangle does not change just because we choose a different base. So both base–height pairs must give the same $12{\text{cm}}^2$. $\frac{1}{2}\times 8\times h=12\Rightarrow h=\frac{2\times 12}{8}=3\text{cm}$ Notice: a longer base ($8>6$) forces a shorter height ($3<4$) — the product $\frac{1}{2}bh$ stays pinned at $12$.

Guided look at the figure. The gray outline is the whole L. The blue shaded block is the full-width bottom strip; the orange shaded block is the top-left piece. The red dashed rectangle in the top-right corner is the bite that was removed. Route (a) adds the blue and orange fills; route (b) takes the big rectangle and cuts away the red dashed corner — both must land on the same number. (a) Add two rectangles. Split the L along the inner corner. One piece is the full-width bottom strip $10\times 5=50{\text{m}}^2$ (height $8-3=5$), the other is the top-left block $(10-4)\times 3=6\times 3=18{\text{m}}^2$. $A=50+18=68{\text{m}}^2$ (b) Subtract the bite. Full rectangle minus the cut-out: $A=(10\times 8)-(4\times 3)=80-12=68{\text{m}}^2$ Both routes agree — always a good check.

The two pieces sit on top of each other and don't overlap, so add. $A_{\text{rect}}=12\times 5=60{\text{m}}^2$ $A_{△}=\frac{1}{2}\times 12\times 4=24{\text{m}}^2$ $A=60+24=84{\text{m}}^2$

First fix the geometry — the two clues must agree. Only the right side leans, so the whole horizontal difference between the parallel sides is the slant's shadow along the bottom: $\text{run}=b-a=15-9=6\text{cm}$ The run is the side touching (adjacent to) the $37^{\circ }$ angle, so it fixes the slant length through cosine: $s=\frac{\text{run}}{\cos 37^{\circ }}=\frac{6}{0.7986}=7.513\text{cm}$ Why we do it this way: the run is a hard geometric fact — it is forced by $b-a$ — whereas a stated slant length could easily be inconsistent with it. So we trust the run, use it to pin the slant, and only then read off the height. This avoids the contradiction of "taking" a height from a number that fights the base geometry. The height is the side opposite the angle, so: $h=\text{run}\times \tan 37^{\circ }=6\times 0.7536=4.521\text{cm}\approx 4.52\text{cm}$ (Equivalently $h=s\sin 37^{\circ }=7.513\times 0.6018=4.52$ — the two now agree, because we built $s$ from the run first.) Finally the area: $A=\frac{1}{2}(9+15)\times 4.52=\frac{1}{2}\times 24\times 4.52=12\times 4.52=54.25{\text{cm}}^2$

Total minus the removed triangle. $A_{\text{rect}}=20\times 12=240{\text{cm}}^2$ $A_{△}=\frac{1}{2}\times 6\times 4=12{\text{cm}}^2$ $A_{\text{remaining}}=240-12=228{\text{cm}}^2$

A semicircle of radius $r$ has area $\frac{1}{2}\pi r^2$ — half a full circle (see circle area). Here $r=7$, and its flat diameter $14\text{m}$ matches the rectangle's end exactly, so it fits cleanly. $A_{\text{rect}}=20\times 14=280{\text{m}}^2$ $A_{\text{semi}}=\frac{1}{2}\pi (7{)}^2=\frac{1}{2}\times 3.1416\times 49=76.97{\text{m}}^2$ $A_{\text{lawn}}=280-76.97=203.03{\text{m}}^2$ Sanity check: the bed is removed, so the answer must be less than $280$ — it is.

Trapezium first, then subtract the pond triangle. $A_{\text{trap}}=\frac{1}{2}(30+50)\times 20=\frac{1}{2}\times 80\times 20=800{\text{m}}^2$ $A_{\text{pond}}=\frac{1}{2}\times 12\times 9=54{\text{m}}^2$ $A_{\text{grass}}=800-54=746{\text{m}}^2$

Area is invariant to which side is the base, so still $12{\text{m}}^2$. Solve for the new height on the $5\text{m}$ base: $h=\frac{2A}{b}=\frac{2\times 12}{5}=\frac{24}{5}=4.8\text{m}$ The covered area is identical ($12{\text{m}}^2$) — only the height changes: shorter base forces a taller height. The triangle occupies the same amount of flat space no matter which edge it sits on.

A triangle and a parallelogram share the same base and same height. How do their areas compare? ::: The triangle is exactly half the parallelogram ($\frac{1}{2}bh$ vs $bh$). Why can't you use a slant side as the height in an area formula? ::: The slant is longer than the true vertical gap; area only counts the perpendicular squeeze between base and top. A shape has area $12$ on a base of $6$. On a base of $8$, what is the height? ::: $h=\frac{2\times 12}{8}=3$; area is unchanged by the base choice.