![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=23.9\\ h=100 \end{cases}\implies V=\cfrac{\pi (23.9)^2(100)}{3} \\\\\\ V=\cfrac{57121\pi }{3}\implies V\approx 59816.97\implies \stackrel{\textit{rounded up}}{V=59817} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D23.9%5C%5C%20h%3D100%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%2823.9%29%5E2%28100%29%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20V%3D%5Ccfrac%7B57121%5Cpi%20%7D%7B3%7D%5Cimplies%20V%5Capprox%2059816.97%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BV%3D59817%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
now, for the second one, we know the diameter is 10, thus its radius is half that or 5.
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=5\\ V=225 \end{cases}\implies 225=\cfrac{\pi (5)^2 h}{3}\implies 225=\cfrac{25\pi h}{3} \\\\\\ \cfrac{225}{25\pi }=\cfrac{h}{3}\implies \cfrac{9}{\pi }=\cfrac{h}{3}\implies \cfrac{27}{\pi }=h\implies 8.59\approx h\implies \stackrel{\textit{rounded up}}{8.6=h}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D5%5C%5C%20V%3D225%20%5Cend%7Bcases%7D%5Cimplies%20225%3D%5Ccfrac%7B%5Cpi%20%285%29%5E2%20h%7D%7B3%7D%5Cimplies%20225%3D%5Ccfrac%7B25%5Cpi%20h%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B225%7D%7B25%5Cpi%20%7D%3D%5Ccfrac%7Bh%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B9%7D%7B%5Cpi%20%7D%3D%5Ccfrac%7Bh%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B27%7D%7B%5Cpi%20%7D%3Dh%5Cimplies%208.59%5Capprox%20h%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B8.6%3Dh%7D)
7) Area trapezoid = (B+b).H/2, but the Median is equal to (B+b)./ 2
84 =12 . H ==> H = 84/12 ==> H = 7
11) Area Equilateral triangle inscribe in a cercle with Radius R = 2√3
Area = (B.. Altitude) / 2. Calculate H, the altitude. The altitude in an equilateral triangle bisects the opposite side. Apply Pythagoras
(2√3)² = (√3)² + H² ==> 12 = 3 + H² ==> H² = 9 & H = 3
Hence Are = (2√3 x 3) /2 ==> 3√3 unit² or 5.2 unit²
12) Area of regular hexagone with perimeter =12
regular hexagone is formed with 6 equilateral triangles with
each side =12/6 = 2 units
Let's calculate the area of 1 equilateral triangle. Follow the same logic as in problem 11 & you will find that Altitude = √3, Area =(2.√3)/2 =√3 =1.73 Unit³
13) a) Circumference of a circle : 2πR==> 2π(30) =60π = 188.4
b) Area of a circle =πR³ ==> π(30)² = 900π = 2,826 unit²
Answer:
(3i+5)+(2-6i)
⇒ 3i+5+2-6i
⇒3i-6i+5+2
⇒-3i+7
<u>The simplified form of (3i+5)+(2-6i) will be -3i+7</u>
