- Given that the vertices of the right-angled triangle lie on the circumference of a circle, the hypotenuse of the triangle is the diameter of the circle.
- The sides of the right-angled triangle are 14 cm and 48 cm.
- Using the Pythagorean theorem to find the hypotenuse:
\[ h = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50 \text{ cm} \]
- Thus, the diameter of the circle is 50 cm.
- The radius \(r\) of the circle is half of the diameter:
\[ r = \frac{50}{2} = 25 \text{ cm} \]
- The area \(A\) of the circle is given by the formula:
\[ A = \pi r^2 = \pi \times 25^2 = \pi \times 625 = 625\pi \text{ cm²} \]
Thus, the area of the circle is 625π cm².
Conclusion: The correct answer is 625π cm².
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.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$