Notes
Polygon angles
- Interior angle sum: $(n - 2) \cdot 180^\circ$. Exterior angle sum: $360^\circ$.
- Regular $n$-gon: each interior angle $\displaystyle \frac{(n - 2) \cdot 180^\circ}{n}$, each exterior angle $\displaystyle \frac{360^\circ}{n}$.
Parallelograms
- Opposite sides congruent and parallel, opposite angles congruent, consecutive angles supplementary, diagonals bisect each other.
- To prove one: show any of those side, angle or diagonal facts, or one pair of opposite sides both congruent and parallel.
Special quadrilaterals
| Shape | Extra properties |
|---|---|
| Rectangle | 4 right angles, congruent diagonals |
| Rhombus | 4 congruent sides, perpendicular diagonals that bisect the angles |
| Square | All rectangle and rhombus properties |
| Isosceles trapezoid | Congruent legs, congruent base angle pairs, congruent diagonals |
| Kite | Perpendicular diagonals, one pair of congruent opposite angles |
- Trapezoid midsegment $\displaystyle = \frac{b_1 + b_2}{2}$.
- On a grid, use slope (parallel, perpendicular), distance (congruent) and midpoint (bisecting diagonals).
Topics in this unit
Each topic has its own short notes and practice.
Practice
$(n - 2) \cdot 180 = 1260$, so $n - 2 = 7$.
$5x - 3 = 3x + 9$ gives $x = 6$, so $AB = CD = 27$ and $BC = AD = 12$. Perimeter $= 2(27 + 12) = 78$.
Bisecting diagonals make a parallelogram; congruent diagonals make it a rectangle; not perpendicular rules out a square.
$\displaystyle \frac{12 + b}{2} = 17$, so $12 + b = 34$.
Diagonals bisect each other at right angles, making right triangles with legs 5 and 12: $\displaystyle \sqrt{25 + 144} = 13$.
Each exterior angle is $30^\circ$, and $360 \div 30 = 12$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A garden designer marks the corners of a flower bed at $A(1, 1)$, $B(5, 3)$, $C(7, 7)$ and $D(3, 5)$ on a grid measured in feet.
(a) Show that $ABCD$ is a parallelogram.
Show the answer and scoring guide
Slopes: $\displaystyle AB = \frac{2}{4} = \frac{1}{2}$, $\displaystyle DC = \frac{2}{4} = \frac{1}{2}$, $\displaystyle BC = \frac{4}{2} = 2$, $\displaystyle AD = \frac{4}{2} = 2$. Both pairs of opposite sides are parallel. (Or: diagonals $\overline{AC}$ and $\overline{BD}$ both have midpoint $(4, 4)$.)
Check off the points you earned:
(b) Show that $ABCD$ is a rhombus.
Show the answer and scoring guide
$\displaystyle AB = \sqrt{4^2 + 2^2} = \sqrt{20}$ and $\displaystyle BC = \sqrt{2^2 + 4^2} = \sqrt{20}$. A parallelogram with consecutive sides congruent has four congruent sides, so it is a rhombus. (Or: diagonal slopes are $1$ and $-1$, so the diagonals are perpendicular.)
Check off the points you earned:
(c) Is $ABCD$ a square? Explain, and find the length of fencing needed to go around it, to the nearest tenth of a foot.
Show the answer and scoring guide
Not a square: slopes of $\overline{AB}$ and $\overline{BC}$ are $\displaystyle \frac{1}{2}$ and 2, whose product is $1$, not $-1$, so there is no right angle. Perimeter $\displaystyle = 4\sqrt{20} \approx 17.9$ feet.
Check off the points you earned: