Step 1: Understanding the Concept:
A Pythagorean triplet consists of three positive integers \(a\), \(b\), and \(c\) such that the sum of the squares of the two smaller integers equals the square of the largest integer.
Step 2: Key Formula or Approach:
The formula to verify a Pythagorean triplet is \(a^2 + b^2 = c^2\), where \(c\) is the greatest of the three numbers. We will test each option against this formula.
Step 3: Detailed Explanation:
Let's check each option:
(A) (1, 5, 10):
Here, \(a = 1\), \(b = 5\), and \(c = 10\).
\[ 1^2 + 5^2 = 1 + 25 = 26 \]
\[ 10^2 = 100 \]
Since \(26 \neq 100\), this is not a Pythagorean triplet.
(B) (3, 4, 5):
Here, \(a = 3\), \(b = 4\), and \(c = 5\).
\[ 3^2 + 4^2 = 9 + 16 = 25 \]
\[ 5^2 = 25 \]
Since \(25 = 25\), this is a Pythagorean triplet.
(C) (2, 2, 2):
All numbers are equal. This represents the sides of an equilateral triangle, not a right-angled triangle. Also, \(2^2 + 2^2 = 8\), which is not equal to \(2^2 = 4\).
(D) (5, 5, 2):
The two smaller numbers are 2 and 5. The largest is 5.
\[ 2^2 + 5^2 = 4 + 25 = 29 \]
\[ 5^2 = 25 \]
Since \(29 \neq 25\), this is not a Pythagorean triplet.
Step 4: Final Answer:
Based on the calculations, the only set of numbers that satisfies the Pythagorean theorem is (3, 4, 5).
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.