Practice: similar triangles and midsegments

Geometry, grade 10 (4 units) Lesson 8 of 20Practice

Practice: similar triangles and midsegments

1

The triangle $\displaystyle ABC$ is isosceles: $\displaystyle {AC}={BC}={15}$ .
The point $\displaystyle D$ is on $\displaystyle AC$ and the point $\displaystyle E$ on $\displaystyle BC$ , so that $\displaystyle {CD}={\frac{1}{3}AC}$ and $\displaystyle {CE}={\frac{1}{3}BC}$ .
$\displaystyle {DE}={6}$ .
Find the area of the trapezoid $\displaystyle ADEB$ , step by step.

a

\displaystyle AB

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b

The altitude from

\displaystyle C

to the base

\displaystyle AB

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c

The area of the trapezoid

\displaystyle ADEB

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2

The vertices of the parallelogram $\displaystyle ABCD$ are $\displaystyle A(4,-2)$ , $\displaystyle B(10,0)$ , $\displaystyle C(9,3)$ and $\displaystyle D(3,1)$ .
The side $\displaystyle BC$ is extended by its own length, to the point $\displaystyle E(8,6)$ : $\displaystyle {BC}={CE}$ .
The segment $\displaystyle AE$ crosses the side $\displaystyle CD$ at the point $\displaystyle F$ .
Prove: $\displaystyle \triangle EFC\sim\triangle EAB$ .

What is the similarity ratio?
Find the length of the segment $\displaystyle FC$ .

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Find the coordinates of the point $\displaystyle F$ .
Explain.

3

$\displaystyle ABCD$ is a trapezoid, $\displaystyle BC\parallel AD$ .
Its diagonals cross at the point $\displaystyle O$ .
Prove: $\displaystyle \triangle AOD\sim\triangle COB$ .

$\displaystyle {AD}={12}$ , $\displaystyle {BC}={4}$ and $\displaystyle {AO}={9}$ .
Find $\displaystyle OC$ .

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4

Two similar triangles.
Corresponding angle bisectors in them are $\displaystyle 4$ and $\displaystyle 10$ long.
The area of the small triangle is $\displaystyle 12$ .
What is the area of the large triangle?

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5

The area of the triangle $\displaystyle ABC$ is $\displaystyle 40$ .
$\displaystyle DE$ is a midsegment: $\displaystyle D$ is the midpoint of $\displaystyle AB$ , and $\displaystyle E$ the midpoint of $\displaystyle AC$ .
What is the area of the trapezoid $\displaystyle DBCE$ ?

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6

In the triangle $\displaystyle ABC$ the angle $\displaystyle C$ is right, and $\displaystyle CD$ is the altitude to the hypotenuse.
$\displaystyle {AD}={4}$ and $\displaystyle {DB}={9}$ .
(A hint: $\displaystyle \triangle ADC\sim\triangle CDB$ , and $\displaystyle \triangle ADC\sim\triangle ACB$ .)

a

\displaystyle CD

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b

\displaystyle AC

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