Solved Examples on Mensuration

$ A = \pi \\[3ex] r = ? \\[3ex] r = \dfrac{\sqrt{A\pi}}{\pi} \\[5ex] = \dfrac{\sqrt{\pi * \pi}}{\pi} \\[5ex] = \dfrac{\sqrt{\pi^2}}{\pi} \\[5ex] = \dfrac{\pi}{\pi} \\[5ex] = 1\:\:inch $

$ A = 1815\:yd^2 \\[3ex] l = 50\:m \\[3ex] w =...to\:\:be\:\:in\:\:m \\[3ex] Convert\:\:1815\:\:yd^2\:\:to\:\:m^2 \\[3ex] 1815\:\:yd * yd * \dfrac{1\:m * 1\:m}{1.1\:yd * 1.1\:yd} \\[5ex] A = \dfrac{1815m^2}{1.21} \\[5ex] A = 1500\:m^2 \\[3ex] w = \dfrac{A}{l} \\[5ex] w = \dfrac{1500\:m^2}{50\:m} \\[5ex] w = 30\:m $

The diameter of the semicircle is the width of the rectangle

$ (i) \\[3ex] diameter\:\:of\:\:semicircle = 3.5\:m \\[3ex] $ The perimeter of the swimming pool is the sum of:
Circumference of the semicircle $AFE$ and
Length of the rectangle $ED$ and
Circumference of the semicircle $DCB$ and
Length of the rectangle $BA$

$ (ii) \\[3ex] Circumference\:\:of\:\:semicircle\:\:AFE = \dfrac{\pi d}{2} \\[5ex] Circumference\:\:of\:\:semicircle\:\:DCB = \dfrac{\pi d}{2} \\[5ex] Length\:\:of\:\:rectangle\:\:ED = 8\:m \\[3ex] Length\:\:of\:\:rectangle\:\:BA = 8\:m \\[3ex] Perimeter\:\:of\:\:the\:\:swimming\:\:pool \\[3ex] = \dfrac{\pi d}{2} + \dfrac{\pi d}{2} + 8 + 8 \\[5ex] = \pi d + 16 \\[3ex] = \dfrac{22}{7} * 3.5 + 16 \\[5ex] = \dfrac{22}{7} * \dfrac{7}{2} + 16 \\[5ex] = 11 + 16 \\[3ex] = 27\:m $

$ Area\:\:of\:\:rectangle \\[3ex] = (x + 5)(x + 7) \\[3ex] = x^2 + 7x + 5x + 35 \\[3ex] = x^2 + 12x + 35 \\[3ex] Area\:\:of\:\:square \\[3ex] = x^2 \\[3ex] Remaining\:\:area\:\:of\:\:rectangle \\[3ex] = x^2 + 12x + 35 - x^2 \\[3ex] = 12x + 35 $

Because it is a right triangle, the hypotenuse is $17$ meters (longest side)
This implies that the other two sides are the base and the height.
Any of those two sides could be either the base or the height. No information in the question suggets otherwise.

$ \underline{Short\:\:Method:\:\:Recommended\:\:for\:\:the\:\:ACT} \\[3ex] A = \dfrac{1}{2} * b * h \\[5ex] A = \dfrac{1}{2} * 8 * 15 \\[5ex] A = 4 * 15 \\[3ex] A = 60\:m^2 \\[3ex] \underline{Long\:\:Method:\:\:Hero's\:\:Formula} \\[3ex] Semiperimeter = \dfrac{height + base + hypotenuse}{2} \\[5ex] Semiperimeter = \dfrac{15 + 8 + 17}{2} = \dfrac{40}{2} = 20\:cm \\[5ex] first\:\:difference = Semiperimeter - height = 20 - 15 = 5\:cm \\[3ex] second\:\:difference = Semiperimeter - base = 20 - 8 = 12\:cm \\[3ex] third\:\:difference = Semiperimeter - hypotenuse = 20 - 17 = 3\:cm \\[3ex] Area = \sqrt{Semiperimeter * first\:\:difference * second\:\:difference * third\:\:difference} \\[3ex] Area = \sqrt{20 * 5 * 12 * 3} \\[3ex] Area = \sqrt{3600\:m^4} \\[3ex] Area = 60\:m^2 $

The vertex of one square is at the center of the other square.
Do not let the titled square confuse you.
If you stand it directly, it has the same area as when it is titled because it's vertex is at the center of the other square.
This implies that half of the length of the titled square is at the center of the other square.
The $2$ squares in the figure below have the same dimensions.
This implies that $\dfrac{1}{2} * 8 = 4\:inches$ of the titled square is at the center of the other square.

$ Area\:\:of\:\:the\:\:shaded\:\:region \\[3ex] = 4(4) \\[3ex] = 16\:in^2 \\[3ex] $ Alternatively, one can say that the area of the shaded region (region of the titled square) is one-quarter of the area of the first square.

$ Area\:\:of\:\:the\:\:first\:\:square = 8(8) = 64\:in^2 \\[3ex] Area\:\:of\:\:the\:\:shaded\:\:region = \dfrac{1}{4} * 64 \\[3ex] = 16\:in^2 $

$ \underline{First\:\:Cone} \\[3ex] r = 5\:cm \\[3ex] l = 12\:cm \\[3ex] LSA = \pi rl = \pi (5)(12) = 60\pi cm^2 \\[3ex] \underline{Second\:\:Cone} \\[3ex] r = 6\:cm \\[3ex] LSA = 60\pi cm^2...from\:\:the\:\:question:\:\:curved\:\:surface\:\:areas\:\:are\:\:equal \\[3ex] h = ? \\[3ex] h = \dfrac{\sqrt{LSA^2 - \pi^2 r^4}}{\pi r} \\[5ex] = \dfrac{\sqrt{(60\pi)^2 - \pi^2 * 6^4}}{\pi * 6} \\[5ex] = \dfrac{\sqrt{3600\pi^2 - 1296\pi^2}}{6\pi} \\[5ex] = \dfrac{\sqrt{2304\pi^2}}{6\pi} \\[5ex] = \dfrac{48\pi}{6\pi} \\[5ex] = 8 \\[3ex] h = 8\:cm $

The vertex of one square is at the center of the other square.
Do not let the titled square confuse you.
If you stand it directly, it has the same area as when it is titled because it's vertex is at the center of the other square.
This implies that half of the length of the titled square is at the center of the other square.
The $2$ squares in the figure below have the same dimensions.
This implies that $\dfrac{1}{2} * 8 = 4\:inches$ of the titled square is at the center of the other square.

$ Area\:\:of\:\:the\:\:shaded\:\:region \\[3ex] = 4(4) \\[3ex] = 16\:in^2 \\[3ex] $ Alternatively, one can say that the area of the shaded region (region of the titled square) is one-quarter of the area of the first square.

$ Area\:\:of\:\:the\:\:first\:\:square = 8(8) = 64\:in^2 \\[3ex] Area\:\:of\:\:the\:\:shaded\:\:region = \dfrac{1}{4} * 64 \\[3ex] = 16\:in^2 $

(a)
(i)
The diagram can be represented as:

$ (ii) \\[3ex] r = 21\:cm \\[3ex] V = 4.158\:litres = 4.158\:dm^3 \\[3ex] Convert\:\:to\:\:cm^3 \\[3ex] 4.158\:dm^3 \\[3ex] = 4.158\:dm * dm * dm * \dfrac{0.1\:m * 0.1\:m * 0.1\:m}{1\:dm * 1\:dm * 1\:dm} * \dfrac{1\:cm * 1\:cm * 1\:cm}{0.01\:m * 0.01\:m * 0.01\:m} \\[5ex] = \dfrac{4.158 * 0.1 * 0.1 * 0.1 * cm^3}{0.01 * 0.01 * 0.01} \\[5ex] = \dfrac{0.001458\:cm^3}{0.000001} \\[5ex] = 4158\:cm^3 \\[3ex] h = \dfrac{3V}{\pi r^2} \\[5ex] = 3V \div \pi r^2 \\[3ex] = 3(4158) \div \left(\dfrac{22}{7} * 21 * 21\right) \\[5ex] = 3(4158) \div (22 * 3 * 21) \\[3ex] = \dfrac{3 * 4158}{22 * 3 * 21} \\[5ex] = \dfrac{4158}{462} \\[5ex] = 9\:cm \\[3ex] (b) \\[3ex] Volume\:\:of\:\:cone = 4158\:cm^3 \\[3ex] Water\:\:was\:\:drawn\:\:out \\[3ex] \underline{Remaining\:\:volume\:\:of\:\:water} \\[3ex] d_{remain} = 28\:cm \\[3ex] r_{remain} = \dfrac{d_{remain}}{2} \\[5ex] r_{remain} = \dfrac{28}{2} = 14\:cm \\[5ex] \dfrac{h_{remain}}{h} = \dfrac{r_{remain}}{r}...Similar\:\:\triangle s \\[5ex] \dfrac{h_{remain}}{9} = \dfrac{14}{21} \\[5ex] h_{remain} = \dfrac{9 * 14}{21} = 6\:cm \\[5ex] V_{remain} = \dfrac{\pi * r_{remain}^2 * h_{remain}}{3} \\[5ex] V_{remain} = \dfrac{1}{3} * \dfrac{22}{7} * 14 * 14 * 6 \\[5ex] V_{remain}= 22 * 2 * 14 * 2 \\[3ex] V_{remain} = 1232\:cm^3 \\[3ex] \underline{Volume\:\:of\:\:water\:\:drawn\:\:out} \\[3ex] V_{drawn} + V_{remain} = V \\[3ex] V_{drawn} = V - V_{remain} \\[3ex] V_{drawn} = 4158 - 1232 \\[3ex] V_{drawn} = 2926\:cm^3 $

The square feet of carpeting needed to cover the entire floor is the area of the composite shape.
The area of the composite shape is the sum of the area of section $1$ and the area of section $2$

$ Area\:\:of\:\:section\:1 = 12(12) = 144\:ft^2 \\[3ex] Area\:\:of\:\:section\:2 = 6(21) = 126\:ft^2 \\[3ex] Area\:\:of\:\:composite\:\:shape = 144 + 126 = 270\:ft^2 \\[3ex] $ $270$ square feet of carpeting are needed to cover the entire floor.

$ \underline{Square} \\[3ex] Let\;\;side = x \\[3ex] Perimeter = 4 * side \\[3ex] P = 4x \\[3ex] \underline{Circle} \\[3ex] radius = 2\;cm \\[3ex] Circumference = 2 * \pi * radius \\[3ex] C = 2 * \pi * 2 \\[3ex] C = 4\pi \\[3ex] \underline{Question} \\[3ex] P = C \\[3ex] 4x = 4\pi \\[3ex] x = \dfrac{4\pi}{4} \\[5ex] x = \pi \\[3ex] $ The length of a side of the square is $\pi$ centimeters

$ \underline{Square\;ABCD} \\[3ex] Side = 15 \\[3ex] Area = Side^2 \\[3ex] A_{square} = 15^2 = 225 \\[3ex] \underline{Rectangle} \\[3ex] width = 10\;meters \\[3ex] length = p \\[3ex] Area = length * width \\[3ex] A_{rectangle} = p * 10 = 10p \\[3ex] \underline{Question} \\[3ex] A_{rectangle} = A_{square} \\[3ex] 10p = 225 \\[3ex] p = \dfrac{225}{10} \\[5ex] p = 22.5 \\[3ex] $ The length of the rectangle is $22.5$ meters

$ Area\:\;in\:\:square\:\:feet = 90\:feet * 30\:feet = 2700\:square\:feet \\[3ex] Convert\:\: 2700\:feet^2 \:\:to\:\: yard^2 \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 2700\:ft^2\:\:to\:\:yd^2 \\[3ex] 2700\:ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 2700\:ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{2700 * 1 * 1}{3 * 3} \\[5ex] = \dfrac{2700}{9} \\[5ex] = 300\:yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method

Let $c$ = converted unit from $ft^2$ to $yd^2$

$ 3\:ft = 1\:yd \\[3ex] Square\:\:both\:\:sides \\[3ex] (3\:ft)^2 = (1\:yd)^2 \\[3ex] 3^2 * ft^2 = 1^2 * yd^2 \\[3ex] 9\:ft^2 = 1\:yd^2 \\[3ex] $

$ft^2$	$yd^2$
$9$	$1$
$2700$	$c$

$ \dfrac{c}{1} = \dfrac{2700}{9} \\[5ex] c = 300\:yd^2 \\[3ex] \therefore 2700\:ft^2 = 300\:yd^2 \\[3ex] $ The area of the rectangular stage is $300$ square yards

$ d = 2r \\[3ex] C = 2\pi r \\[3ex] \dfrac{d}{C} = \dfrac{2r}{2\pi r} \\[5ex] \dfrac{d}{C} = \dfrac{1}{\pi} $

Figure 8 pattern around the entire track exactly 1 time is the circumference of the 2 circles

$ C = \pi d \\[3ex] \underline{Small\;\;circle} \\[3ex] d = 30\;feet \\[3ex] C = 30\pi \\[3ex] \underline{Large\;\;circle} \\[3ex] d = 50\;feet \\[3ex] C = 50\pi \\[3ex] \underline{Figure\;\;8\;\;pattern} \\[3ex] C = 30\pi + 50\pi = 80\pi \\[3ex] $

Proportional Reasooning Method to calculate speed

Distance (feet)	Time (minute)
$75$	$1$
$80\pi$	$p$

$ \dfrac{x}{1} = \dfrac{80\pi}{75} \\[5ex] x = 80 * \dfrac{22}{7} * \dfrac{1}{75} \\[5ex] x = \dfrac{1760}{525} \\[5ex] x = 3.352380952\; feet \\[3ex] x \approx 3\;minutes $