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 k is tangent to the circle with center O at point P as shown. What is the slope of line k ?
▶️Answer/Explanation
Ans:3/4
Lets denotes slope of line $\rm{OP}$ is $m$
\[m = \frac{{y_2 – y_1}}{{x_2 – x_1}}\]
\[m = \frac{{0 – (-4)}}{{0- 3}}\]
\[m = \frac{4}{{-3}}\]
\[m = -\frac{4}{3}\]
If the slope of line \(OP\) is \(-\frac{4}{3}\), then the slope of line \(k\), denoted as \(m_k\), will be the negative reciprocal of \(-\frac{4}{3}\).
The negative reciprocal of \(-\frac{4}{3}\) is obtained by flipping the fraction and changing the sign:
\[m_k = \frac{3}{4}\]
So, the slope of line \(k\) is \(\frac{3}{4}\).
Question Hard
A circle in the \(x y\)-plane has its center at \((-3,4)\) and the point \((-2,1)\) lies on the circle. Which equation represents this circle?
A. \((x-3)^2+(y+4)^2=\sqrt{10}\)
B. \((x+3)^2+(y-4)^2=\sqrt{10}\)
C. \((x-3)^2+(y+4)^2=10\)
D. \((x+3)^2+(y-4)^2=10\)
▶️Answer/Explanation
Ans:D
To find the equation representing the circle, we’ll use the standard form of the equation for a circle:
\[(x – h)^2 + (y – k)^2 = r^2\]
where \((h, k)\) are the coordinates of the center of the circle, and \(r\) is the radius of the circle.
We can use the distance formula to find the radius of the circle, which is the distance from the center \((-3, 4)\) to the point \((-2, 1)\):
\[r = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\]
\[r = \sqrt{(-2 – (-3))^2 + (1 – 4)^2}\]
\[r = \sqrt{(1)^2 + (-3)^2}\]
\[r = \sqrt{1 + 9}\]
\[r = \sqrt{10}\]
So, the radius of the circle is \(\sqrt{10}\).
Now, we can plug the center and the radius into the standard form equation of the circle:
\[(x + 3)^2 + (y – 4)^2 = (\sqrt{10})^2\]
\[ (x + 3)^2 + (y – 4)^2 = 10\]
Therefore, the equation representing this circle is:
\[(x + 3)^2 + (y – 4)^2 = 10\]
Question Hard
In the figure shown, point \(B\) and the center of each circle lie on \(\overline{A C}\). The ratio of \(A B\) to \(B C\) is 4 to 1 . If the area of the small circle is 72 , what is the area of the shaded region?
▶️Answer/Explanation
Ans:40.5,81 / 2
$\rm{A B: B C=4: 1}$
Let’s assume that the length of \(A B\) is \(4 a\) and the length of \(B C\) is \( a\)
Area of the smaller circle
\[
\begin{aligned}
& =\pi(2 a)^2 \\
& 4 \pi a^2=72 \rightarrow \pi a^2=18
\end{aligned}
\]
The radius of the larger circle is 5a divided by 2.
Area of the larger circle
\[
=\pi\left(\frac{5}{2} a\right)^2
\]
$\text{Area of the shaded circular region = Area of the larger circle -Area of the small circle}$
\(\begin{aligned} & \text{Area of the shaded circular region}\\&=\pi \times \frac{25}{4} a^2-\pi \times 4 a^2 \\ & =\pi \frac{9}{4} a^2\end{aligned}\)
substituting \(\pi a^2=18\)
\(\text{Area of the shaded circular region}=\frac{81}{2}\)
Question Hard
\(x^{2}-10x+y^{2}-6y=30\)
In the xy-plane, the graph of the given equation is a circle. What is the area of this circle?
A. 8\(\pi \)
B. 15\(\pi \)
C. 46\(\pi \)
D. 64\(\pi \)
▶️Answer/Explanation
Ans: D
To find the area of the circle represented by the equation \(x^2-10 x+y^2-6 y=30\), we need to rewrite the equation in standard form by completing the square.
\[
(x-5)^2+(y-3)^2-34=30
\]
\[
(x-5)^2+(y-3)^2=64
\]
Standard Form of a Circle:
The standard form of a circle’s equation is \((x-h)^2+(y-k)^2=r^2\), where \((h, k)\) is the center and \(r\) is the radius.
From \((x-5)^2+(y-3)^2=64\), we identify:
Center: \((5,3)\)
Radius: \(r=\sqrt{64}=8\)
The area \(A\) of a circle is given by \(A=\pi r^2\).
Substitute \(r=8\) :
\[
A=\pi\left(8^2\right)=\pi \cdot 64=64 \pi
\]
Question Hard
Point P is the center of the circle in the figure shown. What is the value of x?
▶️Answer/Explanation
100
Angle subtended by the arc at circumference\(=\frac{1}{2}\) Angle subtended by the are at the circle
\(\begin{aligned} \angle BAC & =\frac{1}{2} \angle BPC \\ & =\frac{1}{2} x \\ \angle BAC & =x / 2\end{aligned}\)
In quadrilateral ABPC Sum of all angles is 360.,
$\angle A+ \angle B + \angle P + \angle C = 360^{\circ}$
$\angle P = 360 – x $
$\frac{1}{2} x+50+360-x=360$
$x=100$