The area A of a triangle is given by:
A = √(s (s - a) (s - b) (s - c))
where a, b, and c are the sides of the triangle, and s is the semi-perimeter:
s = (a + b + c) / 2
s = (32 + 27 + 11) / 2 = 35
A = √(35 × (35 - 32) × (35 - 27) × (35 - 11))
= √(35 × 3 × 8 × 24)
= √20160
≈ 142 m²
Thus, the correct answer is (B) 142 m².
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 $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$