Special quadrilaterals
Kite (formed by joining two isosceles triangle together):
Properties
- Adjacent sides have the same length, i.e. AB = AD and BC = DC
- M is the midpoint of BD (but not AC)
- Diagonals are perpendicular
Trapezium:
Properties
- One pair of parallel sides
Parallelogram:
Properties
- Opposite sides are parallel
- Opposite sides have the same length
- Diagonals have the same midpoint
Rectangle:
Properties
- Opposite sides are parallel
- Opposite sides have the same length
- Diagonals have the same midpoint
- Diagonals have the same length
- Adjacent sides are perpendicular
(Thus, all rectangles are parallelograms!)
Rhombus:
Properties
- Opposite sides are parallel
- All sides have the same length
- Diagonals have the same midpoint
- Diagonals have the same length
- Diagonals are perpendicular
(Thus, all rhombuses are parallelograms!)
Square:
Properties
- Opposite sides are parallel
- All sides have the same length
- Diagonals have the same midpoint
- Diagonals have the same length
- Diagonals are perpendicular
- Adjacent sides are perpendicular
(Thus, all squares are rhombuses!)
Questions
Q1. Three of the vertices of a parallelogram $PQRS$ are $P(2, 6)$, $Q(7, 8)$ and $R(5, 3)$.
(from think! A Maths Workbook Worksheet 7A)
(i) Find the coordinates of $S$.
Answer: $ S(0, 1) $
Solutions
\begin{align}
x \text{-coordinate of } S & = 5 - 5 = 0 \\
\\
y \text{-coordinate of } S & = 3 = 2 = 1 \\
\\
\therefore & \phantom{.} S(0, 1)
\end{align}
(ii) Show that $PQRS$ is a rhombus.
Solutions
\begin{align}
\text{Gradient of } PR & = {3 - 6 \over 5 - 2} = -1 \\
\\
\text{Gradient of } QS & = {1 - 8 \over 0 - 7} = 1 \\
\\
-1 \times 1 & = -1 \\
\\ \\
\therefore \text{Diagonals } PR \text{ and } QS & \text{ are perpendicular and } PQRS
\text{ is a rhombus}
\end{align}
Alternate solutions
\begin{align*}
PQ & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\
& = \sqrt{ (7 - 2)^2 + (8 - 6)^2 } \\
& = \sqrt{29} \text{ units} \\
\\
QR & = \sqrt{ (5 - 7)^2 + (8 - 3)^2 } \\
& = \sqrt{29} \text{ units} \\
\\
RS & = \sqrt{ (0 - 5)^2 + (1 - 3)^2 } \\
& = \sqrt{29} \text{ units} \\
\\
SP & = \sqrt{ (2 - 0)^2 + (6 - 1)^2} \\
& = \sqrt{29} \text{ units} \\
\\
\text{Since all sides } & \text{ have the same length, } PQRS \text{ is a rhombus}
\end{align*}
Q2. The diagram shows a rhombus $ABCD$ in which the coordinates of the points $A$ and $C$ are $(-10, 8)$ and $(2, 2)$ respectively, while point $B$ lies on the $y$-axis.
(from think! A Maths Workbook Worksheet 7B)
(i) Find the coordinates of $B$ and of $D$.
Answer: $ B (0,13), D(-8, -3) $
Solutions
\begin{align}
\text{Let } M \text{ denote } & \text{the common midpoint of } AC \text{ and } BD \\
\\
M & = \left( {-10 + 2 \over 2}, {8 + 2 \over 2} \right) \\
& = (-4, 5) \\
\\
\text{Gradient of } AC & = {2 - 8 \over 2 - (-10)} \\
& = -{1 \over 2} \\
\\
\text{Gradient of } BD & = {-1 \over -{1 \over 2}}
\phantom{000000} [m_1 \times m_2 = - 1] \\
& = 2 \\
\\
y & = mx + c \\
y & = 2x + c \\
\\
\text{Using } & M(-4, 5), \\
5 & = 2(-4) + c \\
5 & = -8 + c \\
13 & = c \\
\\
\text{Eqn of } BD: & \phantom{0} y = 2x + 13 \\
\\
\text{Let } & x = 0,
\phantom{000000} [B \text{ lies on the } y \text{-axis}] \\
y & = 2(0) + 13 \\
y & = 13 \\
\\
\therefore & \phantom{.} B(0, 13) \\
\\ \\
\text{Let coordinates} & \text{ of } D \text{ be } (a, b) \\
\\
M & = \text{Midpoint of } BD \\
(-4, 5) & = \left( {0 + a \over 2}, {13 + b \over 2} \right)
\end{align}
\begin{align}
{0 + a \over 2} & = -4 &&& {13 + b \over 2} & = 5 \\
a & = 2(-4) &&& 13 + b & = 2(5) \\
a & = -8 &&& b & = 10 - 13 \\
& &&& b & = -3
\end{align}
$$ \therefore B (0,13), D(-8, -3) $$
(ii) Find the acute angle that the line $CD$ makes with the $x$-axis.
Answer: $ 26.6^\circ $
Solutions
\begin{align*}
\text{Gradient of } CD & = {-3 - 2 \over -8 - 2} \\
& = {1 \over 2} \\
\\
y & = mx + c \\
y & = {1 \over 2}x + c \\
\\
\text{Using } & C(2, 2), \\
2 & = {1 \over 2}(2) + c \\
2 & = 1 + c \\
1 & = c \\
\\
\text{Eqn of } CD: & \phantom{0} y = {1 \over 2}x + 1 \\
\\
\implies & y \text{-intercept is } 1 \\
\\
\text{Let } & y = 0, \\
0 & = {1 \over 2}x + 1 \\
-{1 \over 2}x & = 1 \\
{1 \over 2}x & = -1 \\
x & = - 2 \phantom{000000} [x \text{-intercept}]
\end{align*}
\begin{align*}
\tan \theta & = {2 \over 4} \phantom{000000} \left[ {Opp \over Adj} \right] \\
\theta & = \tan^{-1} \left(2 \over 4\right) \\
& = 26.565^\circ \\
& \approx 26.6^\circ
\end{align*}
Q3. The diagram shows a kite $OPQR$ in which $OP = PQ$ and $OR = QR$. The coordinates of $P$ and $R$ are $(4, 0)$ and $(0, 8)$ respectively.
(from think! A Maths Workbook Worksheet 7B)
(i) Find the equation of $OQ$.
Answer: $ y = {1 \over 2}x $
Solutions
\begin{align}
\text{Gradient of } PR & = {8 - 0 \over 0 - 4} \\
& = -2 \\
\\
\text{Gradient of } OQ & = {-1 \over -2}
\phantom{000000} [m_1 \times m_2 = -1, \text{ since diagonals are perpendicular}] \\
& = {1 \over 2} \\
\\
y & = mx + c \\
y & = {1 \over 2}x + c \\
\\
\text{Using } & O(0, 0), \\
0 & = {1 \over 2}(0) + c \\
0 & = 0 + c \\
0 & = c \\
\\
\text{Eqn: } & y = {1 \over 2}x
\end{align}
(ii) Find the coordinates of $Q$.
Answer: $ Q \left(6{2 \over 5}, 3{1 \over 5} \right) $
Solutions
\begin{align}
\text{Gradient of } PR & = -2 \\
\\
y & = mx + c \\
y & = -2x + c \\
\\
\text{Using } & P(4, 0), \\
0 & = -2(4) + c \\
0 & = -8 + c \\
8 & = c \\
\\
\text{Eqn of } PR: & \phantom{0} y = -2x + 8 \phantom{00} \text{--- (1)} \\
\\
\text{Eqn of } OQ: & \phantom{0} y = {1 \over 2}x \phantom{00} \text{--- (2)} \\
\\
\text{Let } M \text{ denote } & \text{the midpoint of } OQ \\
\\
\text{Substitute } & \text{(1) into (2),} \\
-2x + 8 & = {1 \over 2}x \\
-2x - {1 \over 2}x & = -8 \\
-{5 \over 2}x & = -8 \\
x & = {-8 \over -{5 \over 2}} \\
x & = 3{1 \over 5} \\
\\
\text{Substitute } & \text{into (1),} \\
y & = -2 \left(3{1 \over 5}\right) + 8 \\
y & = 1{3 \over 5} \\
\\
\therefore & \phantom{.} M \left(3{1 \over 5}, 1{3 \over 5}\right) \\
\\
\text{Let coordinates} & \text{ of } Q \text{ be } (a, b) \\
\\
M & = \text{Midpoint of } OQ \\
\left(3{1 \over 5}, 1{3 \over 5}\right) & = \left( {0 + a \over 2}, {0 + b \over 2} \right) \\
& = \left( {a \over 2}, {b \over 2} \right)
\end{align}
\begin{align}
{a \over 2} & = 3{1 \over 5} &&& {b \over 2} & = 1{3 \over 5} \\
a & = 2 \left(3{1 \over 5}\right) &&& b & = 2 \left(1{3 \over 5}\right) \\
a & = 6{2 \over 5} &&& b & = 3{1 \over 5}
\end{align}
$$ \therefore Q \left(6{2 \over 5}, 3{1 \over 5} \right) $$
Q4. The diagram shows a trapezium $ABCD$ in which $AB$ is parallel to $DC$ and angle $BAD = 90^\circ$. The point $A$ is $(0, 5)$ and the point $D$ is $(2, -3)$.
(from think! A Maths Workbook Worksheet 7C)
(i) Find the equation of $AB$.
Answer: $ y = {1 \over 4}x + 5 $
Solutions
\begin{align}
\text{Gradient of } AD & = {-3 - 5 \over 2 - 0} \\
& = -4 \\
\\
\text{Gradient of } AB & = {-1 \over -4}
\phantom{000000} [m_1 \times m_2 = -1] \\
& = {1 \over 4} \\
\\
y & = mx + c \\
y & = {1 \over 4}x + c \\
\\
\text{Using } & A(0, 5) \\
5 & = {1 \over 4}(0) + c \\
5 & = c \\
\\
\text{Eqn: } & y = {1 \over 4}x + 5
\end{align}
(ii) Given that point $B$ lies on the line $16y = 9x$, find the coordinates of $B$.
Answer: $ B(16, 9) $
Solutions
\begin{align}
\text{Eqn of } AB: & \phantom{0} y = {1 \over 4}x + 5 \phantom{00} \text{--- (1)} \\
\\
16y & = 9x \phantom{00} \text{--- (2)} \\
\\
\text{Substitute } & \text{(1) into (2),} \\
16 \left({1 \over 4}x + 5\right) & = 9x \\
4x + 80 & = 9x \\
4x - 9x & = -80 \\
-5x & = -80 \\
x & = {-80 \over -5} \\
x & = 16 \\
\\
\text{Substitute } & \text{into (1),} \\
y & = {1 \over 4}(16) + 5 \\
y & = 9 \\
\\
\therefore & \phantom{.} B(16, 9)
\end{align}
(iii) Given that the length of $AB$ is twice the length of $DC$, find the coordinates of $C$.
Answer: $ C(10, -1) $
Solutions
\begin{align}
x \text{-coordinate of } C & = 2 + \left(16 \over 2\right) = 10 \\
\\
y \text{-coordinate of } C & = -3 + \left(4 \over 2\right) = -1 \\
\\
\therefore & \phantom{.} C(10, -1)
\end{align}
(iv) Find the area of the trapezium $ABCD$.
Answer: $ 102 \text{ units}^2 $
Solutions
\begin{align}
\text{Area} & = {1 \over 2} \left| \begin{matrix}
16 & 0 & 2 & 10 & 16 \\ 9 & 5 & -3 & -1 & 9
\end{matrix}\right|
\phantom{000000} [B, A, D, C, B] \\
& = {1 \over 2} [(16)(5)+0+(2)(-1)+(10)(9)] - {1 \over 2}[0+(5)(2)+(-3)(10)+(-1)(16)] \\
& = 102 \text{ units}^2
\end{align}
Past year O level questions
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