Home / Digital SAT Math Practice Questions -Advanced : Lines, angles, and triangles

Digital SAT Math Practice Questions -Advanced : Lines, angles, and triangles

SAT MAth Practice questions – all topics

  • Geometry and Trigonometry Weightage: 15% Questions: 5-7
    • Area and volume
    • Lines, angles, and triangles
    • Right triangles and trigonometry
    • Circles

SAT MAth and English  – full syllabus practice tests

  Question   Hard

 

Line \(m\) is shown in the \(x y\)-plane, and the point with coordinates \((0.25, r)\) is on line \(m\). What is the value of \(r\) ?

▶️Answer/Explanation

Ans:5/3

To find the equation of line \(m\), we can first determine its slope using the given points \((0,2)\) and \((1.5,0)\). Then, we can use the slope-intercept form of a line (\(y = mx + b\)), where \(m\) is the slope and \(b\) is the y-intercept.

Calculate the slope (\(m\)):
\[m = \frac{{y_2 – y_1}}{{x_2 – x_1}}\]
\[m = \frac{{0 – 2}}{{1.5 – 0}}\]
\[m = \frac{{-2}}{{1.5}}\]
\[m = -\frac{4}{3}\]

Use the slope-intercept form to find the equation of the line:
Given that the y-intercept is 2, we can substitute \(m\) and \(b\) into the equation:
\[y = -\frac{4}{3}x + 2\]

So, the equation of line \(m\) is \(y = -\frac{4}{3}x + 2\).

Now, we’re given that the point \((0.25, r)\) is on line \(m\). We can substitute \(x = 0.25\) into the equation of line \(m\) to find \(r\):
\[r = -\frac{4}{3}(0.25) + 2\]
\[r = -\frac{1}{3} + 2\]
\[r = \frac{5}{3}\]

Therefore, the coordinates of the point on line \(m\) are \((0.25, \frac{5}{3})\).

  Question    Hard

For the linear equation \(y=m x+b\), where \(m\) and \(b\) are positive constants, which of the following tables gives three values of \(x\) and their corresponding values of \(y\) ?

 

▶️Answer/Explanation

Ans:B

Given the linear equation \( y = mx + b \), we need to match the given values of \( x \) with their corresponding values of \( y \) from the options.

1. For \( x = -2 \):
\[
y = m(-2) + b = -2m + b
\]

2. For \( x = 1 \):
\[
y = m(1) + b = m + b
\]

3. For \( x = \frac{-b}{m} \):
\[
y = m \left( \frac{-b}{m} \right) + b = -b + b = 0
\]

Option A:
For \( x = -2 \): \( y = -2m \) (incorrect, should be \( -2m + b \))
For \( x = 1 \): \( y = m \) (incorrect, should be \( m + b \))
For \( x = \frac{b}{m} \): \( y = 0 \) (incorrect \( x \) value)

Option B:
For \( x = -2 \): \( y = -2m + b \) (correct)
For \( x = 1 \): \( y = m + b \) (correct)
For \( x = \frac{-b}{m} \): \( y = 0 \) (correct)

Option C:
For \( x = -2 \): \( y = -2m + b \) (correct)
For \( x = 1 \): \( y = m + b \) (correct)
For \( x = b \): \( y = 0 \) (incorrect \( x \) value)

Option D:
For \( x = -2 \): \( y = -2m \) (incorrect, should be \( -2m + b \))
For \( x = 1 \): \( y = m \) (incorrect, should be \( m + b \))
For \( x = b \): \( y = 0 \) (incorrect \( x \) value)

  Question   Hard

Triangle ABC and triangle DEF each have two angles measuring 35°, as shown. Which of the following additional pieces of information is sufficient to prove that triangle ABC is congruent to triangle DEF ?

A) the measures of LACB and ∠DFE are equal

B) The lengths of \(\overline{BC}\) and \(\overline{EF}\) are equal

C) The lengths of AC and DE are equal.

D)No additional information is necessary to prove that the two triangles are congruent.

▶️Answer/Explanation

B) The lengths of \(\overline{BC}\) and \(\overline{EF}\) are equal

To prove that two triangles are congruent, we need to satisfy one of the triangle congruence conditions. Given the information provided in the diagram, we can use the Angle-Side-Angle (ASA) congruence condition.

The ASA congruence condition states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

In this case, we have: ∠A = 35° and ∠B = 35° ∠D = 35° and ∠E = 35°

So we know that two angles of triangle ABC are congruent to two angles of triangle DEF.

The additional piece of information needed is to ensure that the included side between those congruent angles is also congruent.

The lengths of \(\overline{BC}\) and \(\overline{EF}\) are equal

Therefore, option B is sufficient to prove that triangle ABC is congruent to triangle DEF by the ASA congruence condition.

  Question   Hard

The measure of angle A is \(\frac{7}{12}\pi\) radians greater than the measure of angle B. How much greater is the measure of angle A than the measure of angle B, in degrees ? (Disregard the degree symbol when entering your answer.)

▶️Answer/Explanation

105

We need to find how much greater the measure of angle \(A\) is than the measure of angle \(B\), in degrees, given that \(\text{angle } A\) is \(\frac{7}{12}\pi\) radians greater than \(\text{angle } B\).

1. Convert \(\frac{7}{12}\pi\) radians to degrees:
Use the conversion factor \(\frac{180^\circ}{\pi}\):
\[
\frac{7}{12}\pi \times \frac{180^\circ}{\pi} = \frac{7}{12} \times 180
\]
\[
= \frac{7 \times 180}{12} = \frac{1260}{12} = 105^\circ
\]

Thus, the measure of angle \(A\) is \(105\) degrees greater than the measure of angle \(B\).

  Question   Hard

Triangle LMN and triangle PQR each have an angle measuring 60° and a given side length, as shown.

For triangles LMN and PQR, which additional piece of information is sufficient to prove that the triangles are similar?
I. The length of line segment PQ is \(\frac{2}{3}\) the length of line segment LM.
II. The length of line segment PR is \(\frac{2}{3}\) the length of line segment LN.
A) I is sufficient but II is not.
B) II is sufficient but I is not.

▶️Answer/Explanation

C) I is sufficient and II is sufficient.
D) Neither I nor II is sufficient.

C) I is sufficient and II is sufficient.

For these triangles to be similar
\[
\begin{aligned}
\frac{M N}{Q R} & =\frac{L N}{P R}=\frac{L M}{P Q} \\
\frac{9}{6} & =\frac{L N}{P R}=\frac{L M}{P Q} \\
\frac{3}{2} & =\frac{L N}{P R}=\frac{L M}{P Q} \\
\frac{P R}{L N} & =\frac{P Q}{L M}=\frac{2}{3} .
\end{aligned}
\]

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