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An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Answer: Let sides be $a=15$, $b=15$, $c$. Perimeter $= 15+15+c = 40 \Rightarrow c = 10$ cm.
Base $c = 10$. Drop a perpendicular height $h$ to bisect the base into two 5 cm segments.
By Pythagoras: $h^2 + 5^2 = 15^2 \Rightarrow h^2 + 25 = 225 \Rightarrow h^2 = 200 \Rightarrow h = 10\sqrt{2}$.
Area $= \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times 10 \times 10\sqrt{2} =$ $50\sqrt{2}$ cm².
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Answer: Area = $\frac{1}{2} \times \text{base} \times h = 60 \Rightarrow \frac{1}{2} \times 10 \times h = 60 \Rightarrow 5h = 60 \Rightarrow h = 12$ cm.
The height bisects the base into 5 cm segments. Let equal sides be $a$.
By Pythagoras: $a^2 = h^2 + 5^2 = 12^2 + 5^2 = 144 + 25 = 169$.
$a = \sqrt{169} =$ 13 cm.
If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.
Answer: Distance in 1 rotation = Circumference $= \pi \times d = \frac{22}{7} \times 60 \approx 188.57$ cm.
Distance in 100 rotations = $188.57 \times 100 = 18857$ cm.
Distance in meters = $18857 / 100 =$ 188.57 m (or exactly $\frac{1320}{7}$ m).
Find the area of a quadrant of a circle whose circumference is 66 cm.
Answer: Circumference $= 2\pi r = 66 \Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow \frac{44}{7}r = 66 \Rightarrow r = \frac{66 \times 7}{44} = \frac{3 \times 7}{2} = 10.5$ cm.
Area of quadrant = $\frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{441}{4} = \frac{1}{4} \times 22 \times \frac{63}{4} = \frac{1386}{16} =$ 86.625 cm².
Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.
Answer:
(i) Algebra: A kite's diagonals are perpendicular. Let diagonals be $d_1$ and $d_2$. The diagonals split the kite into four right triangles. Let $d_1$ be split into segments $x$ and $y$ ($x+y=d_1$), and $d_2$ is bisected into $d_2/2$ and $d_2/2$.
Area = $\frac{1}{2}(x)(d_2/2) + \frac{1}{2}(x)(d_2/2) + \frac{1}{2}(y)(d_2/2) + \frac{1}{2}(y)(d_2/2)$
Area = $\frac{1}{2}(d_2/2)[x + x + y + y] = \frac{1}{4}d_2[2(x+y)] = \frac{1}{2}d_2(d_1)$.
(ii) Geometry: A kite can be divided by its main diagonal into two congruent triangles. If main diagonal is $d_1$, the height of each triangle is half the other diagonal ($d_2/2$).
Area of one triangle = $\frac{1}{2} \times d_1 \times (d_2/2)$.
Total Area = $2 \times \left[ \frac{1}{2} \times d_1 \times (d_2/2) \right] = \frac{1}{2} \times d_1 \times d_2$.
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