DSAT 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
DSAT MAth and English – full syllabus practice tests
▶️ Answer/Explanation
Answer: \(\frac{3}{4}\)
Slope of \(OP\) = \(\frac{0 – (-4)}{0 – 3} = \frac{4}{-3} = -\frac{4}{3}\). The slope of tangent \(k\) is the negative reciprocal, so \(m_k = \frac{3}{4}\).
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:
\[(x + 3)^2 + (y – 4)^2 = 10\]
▶️ Answer/Explanation
Answer: 40.5 or \(\frac{81}{2}\)
With \(AB:BC = 4:1\), let \(AB = 4a\), \(BC = a\). Area of small circle \(4\pi a^2 = 72\), so \(\pi a^2 = 18\). Radius of large circle is \(\frac{5a}{2}\). Shaded area = \(\pi \cdot \frac{25}{4} a^2 – 4\pi a^2 = \frac{9}{4} \pi a^2 = \frac{9}{4} \cdot 18 = 40.5\).
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 \(r=8\) :
\[
A=\pi\left(8^2\right)=\pi \cdot 64=64 \pi
\]
▶️ Answer/Explanation
Answer: 100
Angle at circumference \(\angle BAC = \frac{1}{2} \angle BPC = \frac{x}{2}\). In quadrilateral \(ABPC\), \(\angle P = 360 – x\), so \(\frac{x}{2} + 50 + 360 – x = 360\), solving gives \(x = 100\).