Shapes that split into right-angled triangles

Geometry, grade 10 (4 units) Lesson 13 of 20Learning

Shapes that split into right-angled triangles

An isosceles trapezoid

The two altitudes from the ends of the short base split an isosceles trapezoid into a rectangle and two congruent right-angled triangles.

Example: in the isosceles trapezoid $\displaystyle ABCD$ the bases are $\displaystyle {BC}={4}$ and $\displaystyle {AD}={10}$ , and the legs are $\displaystyle 5$ long.

$\displaystyle BCFE$ is a rectangle, so $\displaystyle {EF}={BC}={4}$ ; the triangles $\displaystyle ABE$ and $\displaystyle DCF$ are congruent, so $\displaystyle {AE}={FD}={\frac{10-4}{2}}={3}$ ;
the altitude, by Pythagoras’ theorem in the triangle $\displaystyle ABE$ : $\displaystyle {BE}={\sqrt{5^{2}-3^{2}}}={4}$ ;
the base angle: $\displaystyle {\cos\alpha}={\frac{3}{5}}$ , so $\displaystyle {\alpha}={\cos^{-1}\frac{3}{5}\approx53.13^{\circ}}$ ;
the area: $\displaystyle {\frac{4+10}{2}\cdot4}={28}$ .

A rhombus

A rhombus’s diagonals are perpendicular to each other, bisect each other and bisect its angles. So they split it into four congruent right-angled triangles.

Example: the side of the rhombus $\displaystyle ABCD$ is $\displaystyle 6$ long, and $\displaystyle {\angle DAB}={60^{\circ}}$ .

in the triangle $\displaystyle AOB$ the angle $\displaystyle O$ is right, and $\displaystyle {\angle OAB}={30^{\circ}}$ – the diagonal bisects the angle;
$\displaystyle {BO}={6\sin30^{\circ}}={3}$ , so $\displaystyle {BD}={6}$ ;
$\displaystyle {AO}={6\cos30^{\circ}}={3\sqrt{3}}$ , so $\displaystyle {AC}={6\sqrt{3}}$ ;
the area, half the product of the diagonals: $\displaystyle {\frac{1}{2}\cdot6\cdot6\sqrt{3}}={18\sqrt{3}}$ .

Exercise 1

$\displaystyle ABCD$ is a rectangle, and $\displaystyle F$ is the midpoint of the side $\displaystyle AB$ .
The line $\displaystyle DF$ crosses the extension of the side $\displaystyle CB$ at the point $\displaystyle E$ .
Prove: $\displaystyle \triangle AFD\cong\triangle BFE$ .

Given also: $\displaystyle {EC}={9}$ and $\displaystyle {\angle AFD}={40^{\circ}}$ .
Find $\displaystyle BC$ and the rectangle’s area (rounded to two decimal places, only at the end).

\displaystyle BC

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The rectangle’s area

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What is the ratio of the area of the triangle $\displaystyle EBF$ to the area of the rectangle $\displaystyle ABCD$ ?
Explain without finding areas.

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Exercise 2

In the trapezoid $\displaystyle ABCD$ the bases are $\displaystyle BC$ and $\displaystyle AD$ , and $\displaystyle BC<AD$ .
$\displaystyle {AB}={BC}$ , and the diagonal $\displaystyle AC$ is equal to the leg $\displaystyle CD$ .
Prove: the diagonal $\displaystyle AC$ bisects the angle $\displaystyle BAD$ .

Given also: $\displaystyle {\angle CAD}={30^{\circ}}$ and $\displaystyle {AB}={5}$ .
Calculate exactly.

The area of the triangle

\displaystyle ABC

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\displaystyle AC

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The area of the trapezoid

\displaystyle ABCD

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Exercise 3

In the isosceles trapezoid $\displaystyle ABCD$ ( $\displaystyle BC\parallel AD$ ), $\displaystyle EF$ is the midsegment.
The diagonal $\displaystyle AC$ crosses the midsegment at the point $\displaystyle G$ , so that $\displaystyle {EG:GF}={1:3}$ .
$\displaystyle {EG}={3}$ and $\displaystyle {\angle CDA}={50^{\circ}}$ .
Calculate (rounded to two decimal places, only at the end).

a

\displaystyle AD

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b

The trapezoid’s altitude

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c

The trapezoid’s area

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Formulas: geometry, grade 10 (4 units)

Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$