Step 1: Understanding the Concept:
This problem requires finding the distance between a given point and the origin (0, 0) in a 2D Cartesian coordinate system.
Step 2: Key Formula or Approach:
The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
When one point is the origin (0, 0) and the other is (x, y), the formula simplifies to:
\[ d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]
Step 3: Detailed Explanation:
We need to find the distance of the point (3, 4) from the origin (0, 0).
Here, \(x = 3\) and \(y = 4\).
Using the simplified distance formula:
\[ d = \sqrt{3^2 + 4^2} \]
\[ d = \sqrt{9 + 16} \]
\[ d = \sqrt{25} \]
\[ d = 5 \]
Distance is a scalar quantity and is always non-negative, so option (D) -5 is incorrect.
Step 4: Final Answer:
The distance of the point (3, 4) from the origin is 5 units.
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.