The ratio of perimeters, areas and special lines

Geometry, grade 10 (4 units) Lesson 7 of 20Learning

The ratio of perimeters, areas and special lines

The perimeters and the areas

In triangles similar in the ratio $\displaystyle k$ , every side of the first triangle is $\displaystyle k$ times the corresponding side of the second. So:

the perimeters – a sum of sides, each $\displaystyle k$ times as long, so the perimeter is $\displaystyle k$ times as long too:

\displaystyle \frac{P_1}{P_2}

\displaystyle =

\displaystyle k

the areas – a side times an altitude: both $\displaystyle k$ times as long, so the area is $\displaystyle k^{2}$ times as large:

\displaystyle \frac{S_1}{S_2}

\displaystyle =

\displaystyle k^{2}

The altitudes, medians and angle bisectors

Theorem: in similar triangles, the ratio of corresponding altitudes, of corresponding medians and of corresponding angle bisectors is the similarity ratio.

\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k

Given: $\displaystyle \triangle ABC\sim\triangle KLM$ ; $\displaystyle CD$ and $\displaystyle MN$ are corresponding altitudes.
Prove: $\displaystyle {\frac{CD}{MN}}={\frac{AC}{KM}}$ – that is, the ratio of the altitudes is the similarity ratio.

The proof below finds a pair of similar right-angled triangles – $\displaystyle ADC$ and $\displaystyle KNM$ – in which the altitudes are corresponding sides.

	Claim	Reason
1	$\displaystyle {\angle CAB}={\angle MKL}$	Given ( $\displaystyle \triangle ABC\sim\triangle KLM$ )
2	$\displaystyle {\angle CDA}={\angle MNK}$	$\displaystyle CD$ and $\displaystyle MN$ are altitudes – both angles are right
3	$\displaystyle \triangle ADC\sim\triangle KNM$	(1), (2); the AA similarity theorem
4	$\displaystyle {\frac{CD}{MN}}={\frac{AC}{KM}}$	(3); corresponding sides of similar triangles

So with medians (Exercise 4) and with angle bisectors: find a pair of similar triangles in which the special line is a side.

Example: from the ratio of the areas to the similarity ratio

Two similar triangles. The area of one is $\displaystyle 12$ , the area of the other $\displaystyle 48$ . The altitude of the small triangle to a certain side is $\displaystyle 3$ .

$\displaystyle {k^{2}}={\frac{48}{12}}={4}$ , so $\displaystyle {k}={2}$ ;
the corresponding altitude in the large triangle: $\displaystyle {2\cdot3}={6}$ ;
the large triangle’s perimeter is $\displaystyle 2$ times the small one’s, and its area $\displaystyle 4$ times.

Exercise 1

The perimeters of two similar triangles are $\displaystyle 24$ and $\displaystyle 36$ .
The area of the large triangle is $\displaystyle 81$ .

a

The similarity ratio, small to large

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

The area of the small triangle

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

The ratio of the areas of two similar triangles is $\displaystyle 9:25$ .
A median of the small triangle is $\displaystyle 6$ long.
How long is the corresponding median of the large triangle?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The rectangle $\displaystyle EFGH$ is inscribed in the triangle $\displaystyle ABC$ : $\displaystyle E$ on the side $\displaystyle AB$ , $\displaystyle F$ on the side $\displaystyle AC$ , and $\displaystyle G$ , $\displaystyle H$ on the side $\displaystyle BC$ .
$\displaystyle AD$ is the altitude to the side $\displaystyle BC$ , and it crosses $\displaystyle EF$ at the point $\displaystyle M$ .
$\displaystyle {AD}={15}$ , $\displaystyle {BC}={27}$ , and $\displaystyle {AF:FC}={1:2}$ .

Prove: $\displaystyle \triangle AEF\sim\triangle ABC$ .

Find $\displaystyle EF$ , $\displaystyle AM$ and the area of the rectangle $\displaystyle EFGH$ .
(A hint: $\displaystyle AM$ is the altitude of the triangle $\displaystyle AEF$ .)

\displaystyle EF

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle AM

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The rectangle’s area

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

$\displaystyle \triangle ABC\sim\triangle KLM$ .
$\displaystyle AP$ is the median to the side $\displaystyle BC$ , and $\displaystyle KQ$ the corresponding median to the side $\displaystyle LM$ .
Prove: $\displaystyle {\frac{AP}{KQ}}={\frac{AB}{KL}}$ .
(A hint: prove first that $\displaystyle \triangle ABP\sim\triangle KLQ$ .)

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 $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$