Step 1: Use the Pythagorean theorem to find the height of the triangle:
\[ \text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2 \]
\[ 13^2 = 12^2 + \text{Height}^2 \]
\[ 169 = 144 + \text{Height}^2 \]
Step 2: Solving for the height:
\[ \text{Height}^2 = 169 - 144 = 25 \]
\[ \text{Height} = 5 \text{ cm} \]
Step 3: The area of the triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
\[ = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2 \]
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 $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$