The formula for the volume of a sphere is Upper V equals four thirds pi r cubedV=

1 answer:

8 0

$\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\quad 
\begin{cases}
r=radius=\frac{diameter}{2}\\
d=diameter\\
----------\\
d=10\\
r=5
\end{cases}\implies V=\cfrac{4\pi \cdot 5^3}{3}$

now, that rounded to 2 decimals is just 523.60

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1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m

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1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m².

a = 51 m, b = 37 m, c = 20 m

semiperimeter: p = (51+37+20):2 = 54 m

Area of triangle:

$A=\sqrt{p(p-a)(p-b)(p-c)}\\\\A=\sqrt{54(54-51)(54-37)(54-20)}\\\\A=\sqrt{54\cdot3\cdot17\cdot34}\\\\A=\sqrt{9\cdot2\cdot3\cdot3\cdot17\cdot17\cdot2}\\\\A=3\cdot2\cdot3\cdot17\\\\A=306\,m^2$

Rate: ₹ 3 per m².

Cost: ₹ 3•306 = ₹ 918

2. Find the area of the isosceles triangle whose perimeter is 11 cm and the base is 5 cm.

a = 5 cm

a+2b = 11 cm ⇒ 2b = 6 cm ⇒ b = 3 cm

p = 11:2 = 5.5

$A=\sqrt{5.5(5.5-3)^2(5.5-5)}\\\\ A=\sqrt{5.5\cdot(2.5)^2\cdot0.5}\\\\ A=\sqrt{11\cdot0.5\cdot(2.5)^2\cdot0.5}\\\\A=0.5\cdot2.5\cdot\sqrt{11}\\\\A=1.25\sqrt{11}\,cm^2\approx4.146\,cm^2$

3. Find the area of the equilateral triangle whose each side is 8 cm.

a = b = c = 8 cm

p = (8•3):2 = 12 cm

$A=\sqrt{12(12-8)^3}\\\\ A=\sqrt{12\cdot4^3}\\\\ A=\sqrt{3\cdot4\cdot4\cdot4^2}\\\\A=4\cdot4\cdot\sqrt{3}\\\\A=16\sqrt3\ cm^2\approx27.713\ cm^2$

4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

a = 2x

b = 3x

2x + 2•3x = 32 cm ⇒ 8x = 32 cm ⇒ x = 4 cm ⇒ a = 8 cm, b = 12 cm

p = 32:2 = 16 cm

$A=\sqrt{16(16-8)(16-12)^2}\\\\ A=\sqrt{16\cdot8\cdot4^2}\\\\ A=\sqrt{2\cdot8\cdot8\cdot4^2}\\\\ A=8\cdot4\cdot\sqrt2\\\\ A=32\sqrt2\ cm^2\approx45.2548\ cm^2$