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Angles & Trianglesmedium

77k

Similarity

Figure for Similarity

In the figure shown, $\overline{JM}$ and $\overline{KL}$ intersect at point $P$ , $JP = 45$ , $LP = 18$ , $JK = 40$ , and $KP = 60$ . What is the length of $\overline{LM}$ ?

Note: Figure not drawn to scale.

Solution

Step 1. The figure marks angle $J$ and angle $L$ with the same $a^\circ$ . $\spotgreen{\angle JPK}$ and $\spotgreen{\angle LPM}$ are vertical angles, so they're equal too. Two matching angles are enough for similarity, so triangle $JKP$ is similar to triangle $LMP$ , with $J$ corresponding to $L$ , $K$ to $M$ , and $P$ to $P$ .

Segments JM and KL crossing at P, with the a-degree marks at J and L and the vertical angles at P arced in green

Step 2. Similar triangles are scaled copies, so every side is multiplied by the same number. Pair the sides by their endpoints. $J$ and $P$ match $L$ and $P$ , so $\spotred{\overline{JP}}$ corresponds to $\spotred{\overline{LP}}$ . $J$ and $K$ match $L$ and $M$ , so $\spotblue{\overline{JK}}$ corresponds to $\spotblue{\overline{LM}}$ .

The same figure with JP and LP in red labeled 45 and 18, and JK and LM in blue labeled 40 and a question mark

Step 3. The problem gives us $\spotred{JP} = 45$ and $\spotred{LP} = 18$ , so triangle $LMP$ is triangle $JKP$ scaled by $\dfrac{18}{45}$ . Call that number $k$ .

Step 4. Every side scales by that same $k$ , and the problem gives us $\spotblue{JK} = 40$ :

\spotblue{LM} = 40k

\displaystyle k = \frac{18}{45}

\displaystyle 40k

$\overline{LM}$ has a length of $16$ . Choice B.

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Angles & Trianglesmedium

77k

Similarity

Figure for Similarity

In the figure shown, $\overline{JM}$ and $\overline{KL}$ intersect at point $P$ , $JP = 45$ , $LP = 18$ , $JK = 40$ , and $KP = 60$ . What is the length of $\overline{LM}$ ?

Note: Figure not drawn to scale.

Solution

Step 1. The figure marks angle $J$ and angle $L$ with the same $a^\circ$ . $\spotgreen{\angle JPK}$ and $\spotgreen{\angle LPM}$ are vertical angles, so they're equal too. Two matching angles are enough for similarity, so triangle $JKP$ is similar to triangle $LMP$ , with $J$ corresponding to $L$ , $K$ to $M$ , and $P$ to $P$ .

Segments JM and KL crossing at P, with the a-degree marks at J and L and the vertical angles at P arced in green

Step 2. Similar triangles are scaled copies, so every side is multiplied by the same number. Pair the sides by their endpoints. $J$ and $P$ match $L$ and $P$ , so $\spotred{\overline{JP}}$ corresponds to $\spotred{\overline{LP}}$ . $J$ and $K$ match $L$ and $M$ , so $\spotblue{\overline{JK}}$ corresponds to $\spotblue{\overline{LM}}$ .

The same figure with JP and LP in red labeled 45 and 18, and JK and LM in blue labeled 40 and a question mark

Step 3. The problem gives us $\spotred{JP} = 45$ and $\spotred{LP} = 18$ , so triangle $LMP$ is triangle $JKP$ scaled by $\dfrac{18}{45}$ . Call that number $k$ .

Step 4. Every side scales by that same $k$ , and the problem gives us $\spotblue{JK} = 40$ :

\spotblue{LM} = 40k

\displaystyle k = \frac{18}{45}

\displaystyle 40k

$\overline{LM}$ has a length of $16$ . Choice B.

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Preview of the Bedrock Pro video explanation for Systems of Inequalities

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The complete SAT Math curriculum, built for your score level
Level 6 - Ruby
650–700
60 min
Problem D15: Custom Regressions, hard
Problem 7b: Custom Regressions, medium, requires Bedrock Pro
Problem 7c: Custom Regressions, medium, requires Bedrock Pro
Problem 7d: Custom Regressions, hard, requires Bedrock Pro
Problem 7f: Custom Regressions, hard, requires Bedrock Pro
Problem 8: Radical Models, hard
Problem 8b: Radical Models, hard, requires Bedrock Pro
Problem 8c: Radical Models, hard, requires Bedrock Pro
Problem 44: Points on Exponential Function, medium
Problem 44e: Points on Exponential Function, medium, requires Bedrock Pro
Problem 7: Custom Regressions, hard, free video solution
Problem 7e: Custom Regressions, hard, requires Bedrock Pro
Problem 8d: Radical Models, hard, requires Bedrock Pro
Problem 44h: Points on Exponential Function, hard, requires Bedrock Pro
Problem 44i: Points on Exponential Function, hard, requires Bedrock Pro
30 min
Problem D22: Percentages, hard
Problem 37: Hard Percentages, hard
Problem 37b: Hard Percentages, hard, requires Bedrock Pro
Problem 37c: Hard Percentages, hard, requires Bedrock Pro
Problem 37h: Hard Percentages, hard, requires Bedrock Pro
Problem 37d: Hard Percentages, hard, requires Bedrock Pro
Problem 37e: Hard Percentages, hard, requires Bedrock Pro
Problem 37f: Hard Percentages, hard, requires Bedrock Pro
Problem 37g: Hard Percentages, hard, requires Bedrock Pro
65 min
Problem 62b: Vertex Form, medium, requires Bedrock Pro
Problem 62d: Vertex Form, medium, requires Bedrock Pro
Problem 60c: Quadratic Applications, medium, requires Bedrock Pro
Problem 60d: Quadratic Applications, medium, requires Bedrock Pro
Problem 60h: Quadratic Applications, hard, requires Bedrock Pro
Problem 60: Quadratic Applications, hard
Problem 62: Vertex Form, hard
Problem 62e: Vertex Form, hard, requires Bedrock Pro
Problem 63c: Lowest Possible Sum, hard, requires Bedrock Pro
Problem 63: Lowest Possible Sum, hard
Problem 60b: Quadratic Applications, medium, requires Bedrock Pro
Problem 60e: Quadratic Applications, medium, requires Bedrock Pro
Problem 60f: Quadratic Applications, hard, requires Bedrock Pro
Problem 60g: Quadratic Applications, hard, requires Bedrock Pro
Problem 62c: Vertex Form, medium, requires Bedrock Pro
Problem 62f: Vertex Form, hard, requires Bedrock Pro
Problem 62g: Vertex Form, hard, requires Bedrock Pro
Problem 63b: Lowest Possible Sum, hard, requires Bedrock Pro
Problem 63d: Lowest Possible Sum, hard, requires Bedrock Pro
45 min
Problem 85b: Line Tangent to Circle, medium, requires Bedrock Pro
Problem 85c: Line Tangent to Circle, medium, requires Bedrock Pro
Problem 85: Line Tangent to Circle, medium
Problem 86b: Quadrilateral from Tangent Lines, hard, requires Bedrock Pro
Problem 86c: Quadrilateral from Tangent Lines, hard, requires Bedrock Pro
Problem 86: Quadrilateral from Tangent Lines, hard, free video solution
Problem 85d: Line Tangent to Circle, medium, requires Bedrock Pro
Problem 85e: Line Tangent to Circle, hard, requires Bedrock Pro
Problem 86d: Quadrilateral from Tangent Lines, hard, requires Bedrock Pro
Problem 87h: Standard Form of a Circle, hard, requires Bedrock Pro
15 min
Problem 125b: Margin of Error, easy, requires Bedrock Pro
Problem 125c: Margin of Error, medium, requires Bedrock Pro
Problem 125: Margin of Error, medium
Problem 125e: Margin of Error, medium, requires Bedrock Pro
Problem 125d: Margin of Error, medium, requires Bedrock Pro
Video lessons for 127 core problems
Concept guides and drills to build fundamentals
The Inscribed Angle Theorem
3 min
Two angles that face the same arc

Here's a circle with center $O$ and three points $P$ , $Q$ , and $R$ on it.

The $\spotblue{\text{blue}}$ angle has its vertex at the center. That makes it a central angle. A central angle and the arc it cuts off are the same measure: a $70^\circ$ central angle cuts off a $70^\circ$ arc. Think of the angle and the $\spotorange{\text{arc}}$ as one measurement, just drawn at different distances from the center.

The $\spotred{\text{red}}$ angle has its vertex on the circle. That makes it an inscribed angle. Its two sides are chords, and it opens onto the same orange arc the central angle does.

The inscribed angle is half

Here's the same figure with the numbers on.

The $\spotred{35^\circ}$ at $R$ is half of the $\spotblue{70^\circ}$ at $O$ . That is always true, for any inscribed angle:
$\spotred{\text{inscribed angle}} = \frac{1}{2} \cdot \spotblue{\text{central angle}}$ $\spotblue{\text{central angle}} = 2 \cdot \spotred{\text{inscribed angle}}$
This is the inscribed angle theorem. It doesn't matter where $R$ sits on the circle. Slide it anywhere along the far side and the red angle stays $\spotred{35^\circ}$ , because it still faces the same $70^\circ$ arc.

You should memorize this for the SAT. It is not on the formula sheet.

When the arc is half the circle

Suppose $\overline{PQ}$ passes through the center, so it's a $\spotblue{\text{diameter}}$ .

The central angle from $P$ to $Q$ is a straight line, so it's $180^\circ$ . Half of that is $90^\circ$ , so the inscribed angle at $R$ is a $\spotred{\text{right angle}}$ .

It runs the other way too. If a triangle sits inside a circle with one side a diameter, the angle across from that side is $90^\circ$ . Recognize this figure on sight: it's a right triangle, and everything you know about right triangles applies.
Correct
960+ problems with written explanations
Angles & Trianglesmedium
Problem 77k
Similarity
In the figure shown, $\overline{JM}$ and $\overline{KL}$ intersect at point $P$ , $JP = 45$ , $LP = 18$ , $JK = 40$ , and $KP = 60$ . What is the length of $\overline{LM}$ ?

Note: Figure not drawn to scale.
Solution
Solution approach
Step 1. The figure marks angle $J$ and angle $L$ with the same $a^\circ$ . $\spotgreen{\angle JPK}$ and $\spotgreen{\angle LPM}$ are vertical angles, so they're equal too. Two matching angles are enough for similarity, so triangle $JKP$ is similar to triangle $LMP$ , with $J$ corresponding to $L$ , $K$ to $M$ , and $P$ to $P$ .

Step 2. Similar triangles are scaled copies, so every side is multiplied by the same number. Pair the sides by their endpoints. $J$ and $P$ match $L$ and $P$ , so $\spotred{\overline{JP}}$ corresponds to $\spotred{\overline{LP}}$ . $J$ and $K$ match $L$ and $M$ , so $\spotblue{\overline{JK}}$ corresponds to $\spotblue{\overline{LM}}$ .

Step 3. The problem gives us $\spotred{JP} = 45$ and $\spotred{LP} = 18$ , so triangle $LMP$ is triangle $JKP$ scaled by $\dfrac{18}{45}$ . Call that number $k$ .

Step 4. Every side scales by that same $k$ , and the problem gives us $\spotblue{JK} = 40$ :
$\spotblue{LM} = 40k$
$\displaystyle k = \frac{18}{45}$
$\displaystyle 40k$

$\overline{LM}$ has a length of $16$ . Choice B.
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