How much bigger?

SAT-style · warm-up vote: 30 sec in chat, then GO

The radius of circle B is 3 times the radius of circle A. The area of circle B is how many times the area of circle A?

A3 times

B6 times

C9 times

D27 times

!

Hold on to your vote. We will come back to this in Section 3.

01

Section 1 · about 10 minutes

Angles.

Three facts and one big unlock: parallel lines.

Three facts to keep in your pocket.

Straight Line = 180°

The angles along a line add to 180.

Triangle = 180°

The three inside angles add to 180.

Vertical Angles are Equal

Where two lines cross, the angles across from each other match.

On the SAT, the trick is knowing when to use each of these properties.

Two facts, one problem.

SAT-style · your turn: 45 sec, then GO

In the triangle shown, side \(BC\) is extended to point \(D\). What is the value of \(x\)?

A47180 − 98 − 35

B63✓

C133added the angles

D145180 − 35

Step 1 · Straight Line = 180°
\(BCD\) is a straight line, so the inside angle at \(C\) is \(180-98=82^\circ\).

Step 2 · Triangle = 180°
The three angles of \(ABC\) add to 180: \(x=180-35-82=63\).

Parallel lines: only two angles.

YOU DO NOT NEED:Alternate InteriorAlternate ExteriorConsecutive InteriorCorresponding

8 angles: 4 small and 4 big.

• All the smalls are equal.
• All the bigs are equal.

Small + Big = 180°

One line crossing two parallel lines: know one of the 8 angles and you know all 8. (If it crosses at 90°, all eight are 90°.) On the SAT, this unlocks most of the angles in a figure.

Small or big?

SAT-style · your turn: 45 sec, then GO

In the figure, lines \(p\) and \(q\) are parallel. What is the value of \(x\)?

A30✓

B50set them equal

C70that is the small angle

D110that is the big angle

Step 1 · Equal, or add to 180°?
One angle is small and one is big, so they are not equal. They add to 180°:
\((3x-20)+(x+80)=180\)

Step 2 · Solve
\(4x+60=180\), so \(4x=120\) and \(x=30\). (Check: \(70^\circ+110^\circ=180^\circ\).)

02

Section 2 · about 12 minutes

Similar Triangles.

One reminder, then two SAT problems that tie it all together.

Similar triangles.

Same Angles

All corresponding angles are equal.

Proportional Sides

All sides scale by the same factor.

\(\dfrac{10}{5}=\dfrac{8}{4}=\dfrac{12}{6}=2\)

An Example Problem

SAT-style · your turn: 45 sec, then GO

In the figure, \(AB \parallel DE\), and segments \(AE\) and \(BD\) meet at \(C\). What is the length of \(DE\)?

A48flipped

B27✓

C24wrong match

D30guess

Step 1 · Parallel lines

\(AE\) crosses the parallel lines. \(\angle A\) and \(\angle E\) are both small, so \(\angle A=\angle E\).

Step 2 · Parallel lines again

\(BD\) crosses them too. \(\angle B\) and \(\angle D\) are both small, so \(\angle B=\angle D\).

Step 3 · Vertical angles

The two angles at \(C\) are vertical angles, so they are equal.

Step 4 · Similar

All angles match, so \(\triangle ABC\sim\triangle EDC\).

Step 5 · Proportion

\(\dfrac{DE}{36}=\dfrac{18}{24}\ \Rightarrow\ DE=27\)

Another Example Problem

SAT-style · your turn: 45 sec, then GO

In right triangle \(ABC\), angle \(C\) is a right angle. Point \(D\) lies on \(AB\) so that \(CD\perp AB\). If \(AD=4\) and \(DB=9\), what is the length of \(CD\)?

A6✓

B6.5averaged

C13added

D36no √

Step 1 · Right angles

\(CD\perp AB\), so both angles at \(D\) are 90°.

Step 2 · Triangle = 180°

In \(\triangle ABC\): \(\angle A+\angle B=90^\circ\). In \(\triangle ADC\): \(\angle A+\angle ACD=90^\circ\). So \(\angle ACD=\angle B\).

Step 3 · Same idea

In \(\triangle CDB\): \(\angle B+\angle BCD=90^\circ\), so \(\angle BCD=\angle A\).

Step 4 · Similar

All angles match, so \(\triangle ADC\sim\triangle CDB\).

Step 5 · Proportion

\(\dfrac{AD}{CD}=\dfrac{CD}{DB}\ \Rightarrow\ \dfrac{4}{CD}=\dfrac{CD}{9}\)

\(CD^2=36\ \Rightarrow\ CD=6\)

03

Section 3 · about 8 minutes

Scaling laws.

Multiply every length by k. What happens to area? To volume?

Triple every length. What happens?

Works for every shape, as long as every length scales by the same \(k\):
lengths \(\times k\), areas \(\times k^2\), volumes \(\times k^3\).

A model car and a real car.

An exact scale model, inside and out: model lengths are \(\tfrac{1}{10}\) of the real car’s.

model

real car

1D Measurements (×10)

Height
Width
Steering-wheel radius
Width of the trunk

2D Measurements (×100)

Paint for the outside
Windex for the windows
Tire shine for the tires

3D Measurements (×1000)

Air in the tires
Gas in the tank
Trunk and glove-box space

Lengths: \(\times10\). Areas you cover: \(\times10^2=100\). Volumes you fill: \(\times10^3=1000\).

Back to the circles.

SAT-style · your turn: 20 sec, then GO

The radius of circle B is 3 times the radius of circle A. The area of circle B is how many times the area of circle A?

A3 timesthat is the radius (1D)

B6 timesdoubled the 3

C9 times✓

D27 timesthat is volume (3D)

Since the radius (1D) scales by a factor of 3, the area (2D) must scale by a factor of \(3^2=9\).

Check: say circle A has radius 1, so its area is \(\pi(1)^2=\pi\).
Then circle B has radius 3, so its area is \(\pi(3)^2=9\pi\): 9 times as big.

04

Section 4 · about 15 minutes

Right Triangles.

A quick SOH-CAH-TOA reminder, then the one identity classes skip.

SOH-CAH-TOA and Pythagoras.

SOH \(\sin A=\dfrac{\text{opp}}{\text{hyp}}\)

CAH \(\cos A=\dfrac{\text{adj}}{\text{hyp}}\)

TOA \(\tan A=\dfrac{\text{opp}}{\text{adj}}\)

Pythagorean Theorem

\(a^2+b^2=c^2\)

Pythagoras, then SOH-CAH-TOA.

SAT-style · your turn: 45 sec, then GO

In right triangle \(ABC\), angle \(C\) is a right angle, \(BC=8\), and \(AB=17\). What is the value of \(\cos A\)?

A\(\dfrac{8}{17}\)sin

B\(\dfrac{15}{17}\)✓

C\(\dfrac{8}{15}\)tan

D\(\dfrac{17}{15}\)flipped

Step 1 · Name the sides from \(A\)

\(BC\) is opposite \(A\), \(AB\) is the hypotenuse, \(AC\) is adjacent.

\(\cos A=\dfrac{\text{ADJ}}{\text{HYP}}\), but \(AC\) is not given.

Step 2 · Pythagorean Theorem

\(8^2+AC^2=17^2\Rightarrow AC^2=225\Rightarrow AC=15\)

Step 3 · SOH-CAH-TOA

\(\cos A=\dfrac{AC}{AB}=\dfrac{15}{17}\)

A rectangle inside a circle.

SAT-style · your turn: 60 sec, then GO

Rectangle \(ABCD\) is inscribed in a circle. \(BC=6\), and diagonal \(AC\) makes a \(30^\circ\) angle with side \(AB\). What is the area of the circle?

A\(144\pi\)12 as radius

B\(36\pi\)✓

C\(12\pi\)circumference

D\(9\pi\)radius 3

Step 1 · Find the right triangle

Corner \(B\) is 90°, so \(\triangle ABC\) is a right triangle. Diagonal \(AC\) is its hypotenuse, and \(BC=6\) is opposite the 30°. So find \(AC\).

Step 2 · SOH

\(\sin30^\circ=\dfrac{BC}{AC}\Rightarrow\dfrac12=\dfrac{6}{AC}\Rightarrow AC=12\)

Step 3 · Diagonal = diameter

\(AC=12\) is the diameter of the circle,
so the radius is 6.

Step 4 · Area

\(A=\pi r^2=\pi(6)^2=36\pi\)

Sine and cosine trade places.

The two small angles add to 90°: \(A+B=90^\circ\).

\(\sin A=\dfrac{\textcolor{#e06666}{a}}{\textcolor{#a8c63e}{c}}\) and \(\cos B=\dfrac{\textcolor{#e06666}{a}}{\textcolor{#a8c63e}{c}}\)

\(\sin B=\dfrac{\textcolor{#7fb2e5}{b}}{\textcolor{#a8c63e}{c}}\) and \(\cos A=\dfrac{\textcolor{#7fb2e5}{b}}{\textcolor{#a8c63e}{c}}\)

Complementary angles swap sin and cos. For acute angles, \(\sin A=\cos B\) means \(A+B=90^\circ\).

Example: \(70^\circ+20^\circ=90^\circ\), so \(\sin70^\circ=\cos20^\circ\) and \(\cos70^\circ=\sin20^\circ\).

No triangle needed.

SAT-style · your turn: 30 sec, then GO

In right triangle \(RST\), angle \(T\) is a right angle. If \(\sin R=\dfrac{\sqrt{15}}{4}\), what is \(\cos S\)?

A\(\dfrac{1}{4}\)that is cos R

B\(\dfrac{1}{4}-\dfrac{\sqrt{15}}{4}\)subtracted

C\(\dfrac{\sqrt{15}}{4}\)✓

D\(\dfrac{4}{\sqrt{15}}\)flipped the fraction

Step 1 · Triangle = 180°
\(T=90^\circ\), so the other two angles share the remaining 90°: \(R+S=90^\circ\).

Step 2 · Sine and cosine trade places
\(R\) and \(S\) add to 90°, so \(\cos S=\sin R=\dfrac{\sqrt{15}}{4}\).

A sin equal to a cos.

SAT-style · your turn: 45 sec, then GO

If \(\cos(x+17)=\sin(2x+10)\) and both angles are acute, what is the value of \(x\)?

All angles are measured in degrees.

A7set the angles equal

B21✓

C30dropped the 17 and 10

D38that is x + 17

Step 1 · Sine and cosine trade places
A cos equals a sin, so the two angles add to 90°: \((x+17)+(2x+10)=90\)

Step 2 · Solve
\(3x+27=90\Rightarrow 3x=63\Rightarrow x=21\)

05

Section 5 · about 7 minutes

Circles.

One equation. Read the center and the radius straight off it.

Equation ↔ graph.

\((x-\textcolor{#e7c24a}{h})^2+(y-\textcolor{#e7c24a}{k})^2=\textcolor{#a8c63e}{r}^2\)

Center \((\textcolor{#e7c24a}{h},\textcolor{#e7c24a}{k})\) Radius \(\textcolor{#a8c63e}{r}\)

\((x-\textcolor{#e7c24a}{3})^2+(y+\textcolor{#e7c24a}{2})^2=\textcolor{#a8c63e}{16}\)

Center \((\textcolor{#e7c24a}{3},\textcolor{#e7c24a}{-2})\): flip the signs. Radius \(\sqrt{16}=\textcolor{#a8c63e}{4}\)

Is the point \((3,2)\) on the circle? Plug it in:

\((\textcolor{#e06666}{3}-\textcolor{#e7c24a}{3})^2+(\textcolor{#e06666}{2}+\textcolor{#e7c24a}{2})^2=\textcolor{#a8c63e}{16}\)

\(0^2+4^2=16\)

\(16=16\ \checkmark\)

Build the equation.

SAT-style · your turn: 45 sec, then GO

A circle in the \(xy\)-plane has its center at \((4,-7)\) and passes through the point \((4,8)\). Which equation represents the circle?

A\((x-4)^2+(y+7)^2=225\)✓

B\((x-4)^2+(y+7)^2=15\)r, not r²

C\((x+4)^2+(y-7)^2=225\)wrong center

D\((x+4)^2+(y-7)^2=15\)wrong center

Step 1 · Center

Center \((4,-7)\) means \((x-4)\) and \((y+7)\): flip the signs. C and D have the wrong center.

Step 2 · Radius

Same \(x\), so measure up: from \(y=-7\) to \(y=8\) is \(8-(-7)=15\). So \(r=15\).

Step 3 · Right side is \(r^2\)

\(r^2=15^2=225\), so the answer is A. (B used \(r\), not \(r^2\).)

Tonight in one slide

Five takeaways.

Angles

Straight line \(=180^\circ\) · Triangle \(=180^\circ\) · Vertical angles are equal
Parallel lines: Small + Big \(=180^\circ\)

Similar Triangles

Same angles · Proportional sides: every side \(\times k\)

Scaling Laws

Length \(\times k\) · Area \(\times k^2\) · Volume \(\times k^3\)

Right Triangles

\(a^2+b^2=c^2\) · SOH-CAH-TOA · \(\sin A=\cos B\) when \(A+B=90^\circ\)

Circles

\((x-h)^2+(y-k)^2=r^2\): center \((h,k)\), radius \(r\)

Practice this week.

Q

Practice questions from tonight, with worked answers.

Try → check → explain one mistake out loud.

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NEXT WEEK

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Geometry& Trig.

How much bigger?

Angles.

Three facts to keep in your pocket.

Two facts, one problem.

Parallel lines: only two angles.

Small or big?

Similar Triangles.

Similar triangles.

An Example Problem

Another Example Problem

Scaling laws.

Triple every length. What happens?

A model car and a real car.

Back to the circles.

Right Triangles.

SOH-CAH-TOA and Pythagoras.

Pythagoras, then SOH-CAH-TOA.

A rectangle inside a circle.

Sine and cosine trade places.

No triangle needed.

A sin equal to a cos.

Circles.

Equation ↔ graph.

Build the equation.

Five takeaways.

Practice this week.

One questionbefore you go?

Geometry
& Trig.

One question
before you go?