Mensuration
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These are the solutions to WASSCE Mathematics questions on Mensuration.
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Formulas
Right Triangle
$ perpendicular\:\:height = height \\[3ex] Area = \dfrac{1}{2} * base * height \\[5ex] height = \dfrac{2 * Area}{base} \\[5ex] base = \dfrac{2 * Area}{height} \\[5ex] hypotenuse^2 = height^2 + base^2...Pythagorean\:\:Theorem \\[3ex] hypotenuse = \sqrt{height^2 + base^2} \\[3ex] height = \sqrt{hypotenuse^2 - base^2} \\[3ex] base = \sqrt{hypotenuse^2 - height^2} \\[3ex] Perimeter = hypotenuse + height + base \\[3ex] Area = \dfrac{1}{2} * height * base * \sin (hypotenuseAngle) \\[5ex] Area = \dfrac{1}{2} * height * hypotenuse * \sin (baseAngle) \\[5ex] Area = \dfrac{1}{2} * base * hypotenuse * \sin (heightAngle) \\[5ex] Semiperimeter = \dfrac{height + base + hypotenuse}{2} \\[5ex] Semiperimeter - height = firstdifference \\[3ex] Semiperimeter - base = seconddifference \\[3ex] Semiperimeter - hypotenuse = thirddifference \\[3ex] Area = \sqrt{Semiperimeter * firstdifference * seconddifference * thirddifference}...Hero's\:\:Formula\:\:or\:\:Heron's\:\:Formula \\[5ex] hypotenuse = {Perimeter^2 - 4 * Area}{2 * Perimeter} \\[5ex] base = \dfrac{(Perimeter - hypotenuse) \pm Math.sqrt((hypotenuse - Perimeter)^2 - 8 * Area)}{2} \\[5ex] height = \dfrac{2 * Area}{base} $
Triangle
$ Perimeter = firstside + secondside + thirdside \\[5ex] Area = \dfrac{1}{2} * firstside * secondside * \sin (thirdAngle) \\[5ex] Area = \dfrac{1}{2} * firstside * thirdside * \sin (secondAngle) \\[5ex] Area = \dfrac{1}{2} * secondside * thirdside * \sin (firstAngle) \\[5ex] Semiperimeter = \dfrac{firstside + secondside + thirdside}{2} \\[5ex] Semiperimeter - firstside = firstdifference \\[3ex] Semiperimeter - secondside = seconddifference \\[3ex] Semiperimeter - thirdside = thirddifference \\[3ex] Area = \sqrt{Semiperimeter * firstdifference * seconddifference * thirddifference}...Hero's\:\:Formula\:\:or\:\:Heron's\:\:Formula \\[5ex] \underline{Cosine\:\:Law} \\[3ex] firstside^2 = secondside^2 + thirdside^2 - 2 * secondside * thirdside * \cos (firstAngle) \\[3ex] secondside^2 = firstside^2 + thirdside^2 - 2 * firstside * thirdside * \cos (secondAngle) \\[3ex] thirdside^2 = firstside^2 + secondside^2 - 2 * firstside * secondside * \cos (thirdAngle) \\[5ex] firstAngle = \cos^{-1} \left(\dfrac{secondside^2 + thirdside^2 - firstside^2}{2 * secondside * thirdside}\right) \\[5ex] secondAngle = \cos^{-1} \left(\dfrac{firstside^2 + thirdside^2 - secondside^2}{2 * firstside * thirdside}\right) \\[5ex] thirdAngle = \cos^{-1} \left(\dfrac{firstside^2 + secondside^2 - thirdside^2}{2 * firstside * secondside}\right) \\[7ex] \underline{\text{Area of a Triangle given the vertices}} \\[3ex] \text{Let the vertices be:} \\[3ex] Vertex\;1:\;\;(x_1, y_1) \\[4ex] Vertex\;2:\;\;(x_2, y_2) \\[4ex] Vertex\;3:\;\;(x_3, y_3) \\[4ex] Area = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $
Insribed Triangle (Circumscribed Circle)
$ Circumradius = \dfrac{firstside \cdot secondside \cdot thirdside}{4 \cdot Area} $
Square
$ side = length = width = height \\[3ex] Area = side^2 \\[3ex] side = \sqrt{Area} \\[3ex] Perimeter = 4 * side \\[3ex] side = \dfrac{Perimeter}{4} \\[5ex] diagonal = side * \sqrt{2} \\[3ex] side = \dfrac{diagonal * \sqrt{2}}{2} \\[5ex] Area = \dfrac{Perimeter^2}{16} \\[5ex] Perimeter = 4 * \sqrt{Area} \\[3ex] Area = \dfrac{diagonal^2}{2} \\[5ex] diagonal = \sqrt{2 * Area} \\[3ex] Perimeter = 2 * diagonal * \sqrt{2} \\[3ex] diagonal = \dfrac{Perimeter * \sqrt{2}}{4} $
Rectangle
$ Area = length * width \\[3ex] length = \dfrac{Area}{width} \\[5ex] width = \dfrac{Area}{length} \\[5ex] Area = \dfrac{(length * Perimeter) - (2 * length^2)}{2} \\[5ex] Area = \dfrac{(width * Perimeter) - (2 * width^2)}{2} \\[5ex] Perimeter = 2(length + width) \\[3ex] length = \dfrac{Perimeter - 2 * width}{2} \\[5ex] width = \dfrac{Perimeter - 2 * length}{2} \\[5ex] Perimeter = \dfrac{2(length^2 + Area)}{length} \\[5ex] Perimeter = \dfrac{2(width^2 + Area)}{width} \\[5ex] diagonal = \sqrt{length^2 + width^2} \\[4ex] length = \sqrt{diagonal^2 - width^2} \\[4ex] width = \sqrt{diagonal^2 - length^2} \\[4ex] diagonal = \dfrac{\sqrt{length^4 + Area^2}}{length} \\[5ex] diagonal = \dfrac{\sqrt{width^4 + Area^2}}{width} \\[5ex] diagonal = \dfrac{\sqrt{(Perimeter^2) + (5 * length^2) - (4 * Perimeter * length)}}{2} \\[5ex] diagonal = \dfrac{\sqrt{(Perimeter^2) + (5 * width^2) - (4 * Perimeter * width)}}{2} $
Circle
$ Area = A \\[3ex] Circumference = C \\[3ex] Radius = r \\[3ex] Diameter = d \\[3ex] d = 2r \\[3ex] r = \dfrac{d}{2} \\[5ex] A = \pi r^2 \\[3ex] A = \dfrac{\pi d^2}{4} \\[5ex] C = 2\pi r \\[3ex] C = \pi d \\[3ex] r = \dfrac{\sqrt{A\pi}}{\pi} \\[5ex] r = \dfrac{C}{2\pi} \\[5ex] d = \dfrac{2\sqrt{A\pi}}{\pi} \\[5ex] r = \dfrac{C}{\pi} \\[5ex] A = \dfrac{C^2}{4\pi} \\[5ex] C = 2\sqrt{A\pi} $
Cube
6 square faces
12 edges
$
edge = side = length = width = height \\[3ex]
Surface\:\:Area = 6 * edge^2 \\[3ex]
edge = \sqrt{\dfrac{Surface\:\:Area}{6}} \\[5ex]
Volume = edge^3 \\[3ex]
edge = \sqrt[3]{Volume} \\[3ex]
Volume = \dfrac{edge * Surface\:\: Area}{6} \\[5ex]
edge = \dfrac{6 * Volume}{Surface\:\:Area} \\[5ex]
Surface\:\:Area = \dfrac{6 * Volume}{edge} \\[5ex]
Volume = \dfrac{Surface\:\:Area * \sqrt{6 * Surface\:\:Area}}{36} \\[5ex]
edge = \dfrac{diagonal * \sqrt{3}}{3} \\[5ex]
diagonal = \sqrt{3} * edge \\[3ex]
Surface\:\:Area = 2 * diagonal^2 \\[3ex]
diagonal = \dfrac{\sqrt{2 * Surface\:\:Area}}{2} \\[5ex]
Volume = \dfrac{diagonal^3 * \sqrt{3}}{9} \\[5ex]
diagonal = \sqrt{3} * \sqrt[3]{Volume}
$
Cuboid (Right Rectangular Prism)
$ Volume = Length \cdot Width \cdot Height \\[3ex] $
Right Cone
Curved Surface Area = Lateral Surface Area
Height = Perpendicular Height
$
Volume\:\:of\:\:Cone = \dfrac{1}{3} * Volume\:\:of\:\:Cylinder \\[5ex]
Lateral\:\:Surface\:\:Area = LSA \\[3ex]
Base\:\:Area = BA \\[3ex]
Total\:\:Surface\:\:Area = TSA \\[3ex]
Volume = V \\[3ex]
Diameter = d \\[3ex]
Radius = r \\[3ex]
Height = h \\[3ex]
Slant Height = l \\[3ex]
r = \dfrac{d}{2} \\[5ex]
d = 2r \\[3ex]
l = \sqrt{h^2 + r^2} \\[3ex]
l = \dfrac{\sqrt{4h^2 + d^2}}{2} \\[5ex]
h = \sqrt{l^2 - r^2} \\[3ex]
h = \dfrac{\sqrt{4l^2 - d^2}}{2} \\[5ex]
r = \sqrt{l^2 - h^2} \\[3ex]
d = 2 * \sqrt{l^2 - h^2} \\[3ex]
BA = \pi r^2 \\[3ex]
r = \dfrac{\sqrt{BA * \pi}}{\pi} \\[5ex]
BA = \dfrac{\pi d^2}{4} \\[5ex]
d = \dfrac{2\sqrt{BA * \pi}}{\pi} \\[5ex]
LSA = \pi rl \\[3ex]
LSA = \dfrac{\pi dl}{2} \\[5ex]
l = \dfrac{LSA}{\pi r} \\[5ex]
LSA = \pi r\sqrt{h^2 + r^2} \\[3ex]
h = \dfrac{\sqrt{LSA^2 - \pi^2 r^4}}{\pi r} \\[5ex]
TSA = BA + LSA \\[3ex]
TSA = \pi r(r + l) \\[3ex]
l = \dfrac{TSA - \pi r^2}{\pi r} \\[5ex]
TSA = \dfrac{\pi d(d + 2l)}{4} \\[5ex]
l = \dfrac{4 * TSA - \pi d^2}{2\pi d} \\[5ex]
r = \dfrac{-\pi l \pm \sqrt{\pi^2 l^2 + 4\pi * TSA}}{2\pi} \\[5ex]
TSA = \pi r(r + \sqrt{h^2 + r^2}) \\[3ex]
h = \dfrac{\sqrt{TSA(TSA - 2\pi r^2)}}{\pi r} \\[5ex]
V = \dfrac{BA * h}{3} \\[5ex]
V = \dfrac{\pi r^2h}{3} \\[5ex]
V = \dfrac{\pi hd^2}{12} \\[5ex]
V = \dfrac{\pi h(l^2 - h^2)}{3} \\[5ex]
h = \dfrac{3V}{\pi r^2} \\[5ex]
r = \dfrac{\sqrt{3V\pi h}}{\pi h}
$
Right Cylinder
Curved Surface Area = Lateral Surface Area
Height = Perpendicular Height
$
Volume\:\:of\:\:Cylinder = 3 * Volume\:\:of\:\:Cone \\[3ex]
Lateral\:\:Surface\:\:Area = LSA \\[3ex]
Base\:\:Area = BA \\[3ex]
Total\:\:Surface\:\:Area = TSA \\[3ex]
Volume = V \\[3ex]
Diameter = d \\[3ex]
Radius = r \\[3ex]
Height = h \\[3ex]
r = \dfrac{d}{2} \\[5ex]
d = 2r \\[3ex]
LSA = 2\pi rh \\[3ex]
r = \dfrac{LSA}{2\pi h} \\[5ex]
h = \dfrac{LSA}{2\pi r} \\[5ex]
LSA = \pi dh \\[3ex]
h = \dfrac{LSA}{\pi d} \\[5ex]
d = \dfrac{LSA}{\pi h} \\[5ex]
BA = \pi r^2 \\[3ex]
r = \dfrac{\sqrt{\pi BA}}{\pi} \\[5ex]
r = \dfrac{1}{\pi} * \sqrt{\dfrac{\pi(TSA - 2 * LSA)}{2}} \\[5ex]
BA = \dfrac{\pi d^2}{4} \\[5ex]
d = \dfrac{2\sqrt{\pi BA}}{\pi} \\[5ex]
d = \dfrac{\sqrt{2\pi (TSA - LSA)}}{\pi} \\[5ex]
TSA = 2\pi r(r + h) \\[3ex]
h = \dfrac{TSA - 2\pi r^2}{2\pi r} \\[5ex]
r = \dfrac{-\pi h \pm \sqrt{\pi(\pi h^2 + 2 * TSA)}}{2\pi} \\[5ex]
TSA = 2BA + LSA \\[3ex]
BA = \dfrac{TSA - LSA}{2} \\[5ex]
LSA = TSA - 2BA \\[3ex]
TSA = \pi d\left(\dfrac{d + 2h}{2}\right) \\[5ex]
h = \dfrac{2 * TSA - \pi d^2}{2\pi d} \\[5ex]
d = \dfrac{-\pi h \pm \sqrt{\pi(h^2 + 2 * TSA)}}{\pi} \\[5ex]
h = \dfrac{LSA * \sqrt{\pi * BA}}{\pi * BA} \\[5ex]
h = \dfrac{LSA}{\sqrt{2\pi(TSA - LSA)}} \\[5ex]
BA = \dfrac{LSA^2}{\pi h^2} \\[5ex]
BA = \dfrac{(4 * TSA + \pi h^2) \pm h\sqrt{\pi(\pi h^2 - 8 * TSA)}}{8} \\[5ex]
LSA = h\sqrt{BA * \pi} \\[3ex]
LSA = \dfrac{-\pi h^2 \pm h\sqrt{\pi(\pi h^2 + 8 * TSA)}}{4} \\[5ex]
TSA = 2 * BA \pm h\sqrt{\pi * BA} \\[3ex]
TSA = \dfrac{LSA(2 * LSA + \pi h^2)}{\pi h^2} \\[5ex]
V = \pi r^2h \\[3ex]
r = \dfrac{2V}{LSA} \\[5ex]
d = \dfrac{4V}{LSA} \\[5ex]
r = \dfrac{\sqrt{Vh\pi}}{h\pi} \\[5ex]
V = BA * h \\[3ex]
BA = \dfrac{V}{h} \\[5ex]
h = \dfrac{V}{BA} \\[5ex]
h = \dfrac{V}{\pi r^2} \\[5ex]
h = \dfrac{4V}{\pi d^2} \\[5ex]
V = \dfrac{\pi d^2h}{4} \\[5ex]
d = \dfrac{\sqrt{Vh\pi}}2{h\pi} \\[5ex]
V = \dfrac{LSA^2}{h\pi} \\[5ex]
LSA = \sqrt{Vh\pi} \\[3ex]
h = \dfrac{LSA^2}{4V\pi} \\[5ex]
V = \dfrac{(h^3\pi + 4 * TSA * h) \pm h^2\sqrt{\pi(h^2\pi + 8 * TSA)}}{8} \\[5ex]
TSA = \dfrac{2V + h\sqrt{Vh\pi}}{h} \\[5ex]
TSA = \dfrac{2V + 2\pi rh^2}{h} \\[5ex]
r = \dfrac{TSA * h - 2V}{2\pi h^2} \\[5ex]
d = \dfrac{TSA * h - 2V}{\pi h^2} \\[5ex]
h = \dfrac{TSA \pm \sqrt{TSA^2 - 16\pi rV}}{4\pi r}
$
Trapezoid
Trapezoid's Midpoint Segment Theorem states that the line segment connecting the nonparallel sides of a
trapezoid is parallel to the bases, and it's length is the average of the lengths of the bases.
$
Midline = \dfrac{short\;\;base + long\;base}{2}
$
The diagram shows the net of a rectangular pyramid.
Calculate, correct to two decimal places, the:
(a.) slant height;
(b.) perpendicular height;
(c.) total surface area of the pyramid.
The net of the rectangular-based pyramid has 4 triangular faces (2 triangles of base 4 cm and 2 triangles of base 10 cm) and a rectangle.
When folded into the pyramid, the perpendicular height of the triangle is the slant height of the pyramid.
This implies that there are two slant heights of the pyramid:
the slant height as a result of the perpendicular heights of 2 triangles of base 4 cm
and
the slant height as a result of the perpendicular heights of 3 triangles of base 10 cm
Let us indicate points on the diagram to help us analyze it.
$ \underline{\triangle EJD} \\[3ex] \perp height = |JD| = 6\;cm \quad\dots\text{1st value: Given} \\[3ex] base = |EJ| = \dfrac{4}{2} = 2\;cm \\[5ex] hypotenuse = |ED| = ? \\[3ex] |ED|^2 = |EJ|^2 + |JD|^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] |ED|^2 = 2^2 + 6^2 \\[3ex] |ED| = \sqrt{4 + 36} \\[3ex] = \sqrt{40} \\[3ex] = 6.32455532\;cm \\[3ex] \approx 6.32\;cm \quad\dots\text{to two decimal places} \\[3ex] $ The slant edges (the hypotenuse of the right triangles) are the same length because, in a right rectangular pyramid, the apex is positioned symmetrically above the center of the rectangular base, hence the distance from the apex of the pyramid to each vertex of the rectangular base is the same.
$ \underline{\text{Slant Edge}} \\[3ex] |ED| = |DF| = |GE| = |GB| = |BA| = |AC| = |CH| = |HF| = \sqrt{40} = 6.32455532\;cm \\[5ex] \underline{\triangle GKE} \\[3ex] \perp height = |GK| = ? \\[3ex] base = |KE| = \dfrac{10}{2} = 5\;cm \\[5ex] hypotenuse = |GE| = \sqrt{40}\;cm \\[3ex] |GE|^2 = |GK|^2 + |KE|^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] |GK|^2 = |GE|^2 - |KE|^2 \\[3ex] |GK|^2 = (\sqrt{40})^2 - 5^2 \\[3ex] |GK|^2 = 40 - 25 \\[3ex] |GK| = \sqrt{15} \\[3ex] |GK| = 3.872983346\;cm \quad\dots\text{2nd value: Calculated} \\[3ex] $ For the 6cm:
The base of the right triangle formed inside the pyramid will be half the length of the other base dimension (10 cm)
For the \sqrt{15}\;cm:
The base of the right triangle formed inside the pyramid will be half the length of the other base dimension (4 cm)
$ (b.) \\[3ex] \text{Perpendicular Height, H} \\[3ex] \text{1st Approach}: \\[3ex] hyp = 6\;cm\quad\dots\text{1st value: Given} \\[3ex] leg = base = \dfrac{10}{2} = 5\;cm \\[5ex] leg = \perp height = H \\[3ex] hyp^2 = leg^2 + leg^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] 6^2 = H^2 + 5^2 \\[3ex] H^2 = 6^2 - 5^2 \\[3ex] H = \sqrt{36 - 25} \\[3ex] H = \sqrt{11} \\[3ex] H = 3.31662479 \\[3ex] H \approx 3.32\;cm \quad\dots\text{to two decimal places} \\[3ex] \text{2nd Approach}: \\[3ex] hyp = \sqrt{15}\;cm\quad\dots\text{1st value: Given} \\[3ex] leg = base = \dfrac{4}{2} = 2\;cm \\[5ex] leg = \perp height = H \\[3ex] hyp^2 = leg^2 + leg^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] (\sqrt{15})^2 = H^2 + 2^2 \\[3ex] H^2 = (\sqrt{15})^2 - 2^2 \\[3ex] H = \sqrt{15 - 4} \\[3ex] H = \sqrt{11} \\[3ex] H = 3.31662479 \\[3ex] H \approx 3.32\;cm \quad\dots\text{to two decimal places} \\[5ex] (c.) \\[3ex] \text{Area of a Triangle} = \dfrac{1}{2} * base * height \\[5ex] \text{Area of 2 triangles of base, 4cm and } \perp \text{height, 6cm} \\[3ex] = 2 * \dfrac{1}{2} * 4 * 6 \\[5ex] = 24\;cm^2 \\[5ex] \text{Area of 2 triangles of base, 10cm and } \perp \text{height, 3.872983346cm} \\[3ex] = 2 * \dfrac{1}{2} * 10 * 3.872983346 \\[5ex] = 38.72983346\;cm^2 \\[5ex] \text{Area of a Rectangle} = Length * Width \\[3ex] \text{Area of a Rectangle of Length, 10cm and Width, 4cm} \\[3ex] = 10 * 4 \\[3ex] = 40\;cm^2 \\[5ex] \text{Total Surface Area of the Pyramid} \\[3ex] = \text{Area of 2 triangles of base 4cm} + \text{Area of 2 triangles of base 10cm} + \text{Area of the rectangle} \\[3ex] = 24 + 38.72983346 + 40 \\[3ex] = 102.7298335 \\[3ex] \approx 102.73\;cm^2 $
If $|\overline{PQ}| = |\overline{QR}| = 17\;cm \;\;|\overline{PR}| = 16\;cm$ and M is the midpoint of $|\overline{PR}|$, calculate:
(a.) $|\overline{QM}|$
(b.) correct to the nearest whole number, the radius of the circle.
Let us represent the information on a diagram
Please note the colors
$ (a.) \\[3ex] Let\;\;|\overline{QM}| = h \\[3ex] \underline{\triangle QMR} \\[3ex] h^2 + 8^2 = 17^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] h^2 = 17^2 - 8^2 \\[3ex] h^2 = 289 - 64 \\[3ex] h^2 = 225 \\[3ex] h = \sqrt{225} \\[3ex] h = 15\;cm \\[5ex] $ We can solve (b.) using at least two approaches.
Use any approach you prefer.
Let us update our diagram
Let O be the centre of the circle
Let r be the radius of the circle
Please note the colors
$ (b.) \\[3ex] \underline{\text{1st Approach: Pythagorean Theorem}} \\[3ex] \underline{\triangle OMR} \\[3ex] r^2 = (15 - r)^2 + 8^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] r^2 = (15 - r)(15 - r) + 64 \\[3ex] r^2 = 225 - 15r - 15r + r^2 + 64 \\[3ex] r^2 = 225 - 30r + r^2 + 64 \\[3ex] r^2 - r^2 + 30r = 225 + 64 \\[3ex] 30r = 289 \\[3ex] r = \dfrac{289}{30} \\[5ex] r = 9.633333333 \\[3ex] r \approx 10\;cm \quad\dots\text{to the nearest whole number} \\[3ex] $ Let:
a, b, c be the sides of the triangle
s be the perimeter A be the area of the triangle
$ \underline{\text{2nd Approach: Heron's Formula and Circumradius Formula}} \\[3ex] a = 16\;cm \\[3ex] b = 17\;cm \\[3ex] c = 17\;cm \\[3ex] s = \dfrac{a + b + c}{3} = \dfrac{16 + 17 + 17}{2} = 25 \\[5ex] s - a = 25 - 16 = 9 \\[3ex] s - b = 25 - 17 = 8 \\[3ex] s - c = 25 - 17 = 8 \\[3ex] s(s - a)(s - b)(s - c) = 25(9)(8)(8) = 14400 \\[3ex] A = \sqrt{s(s - a)(s - b)(s - c)}\quad\dots\text{Heron's Formula} \\[3ex] A = \sqrt{14400} \\[3ex] A = 120\;cm^2 \\[5ex] r = \dfrac{abc}{4A} \\[5ex] r = \dfrac{16 \cdot 17 \cdot 17}{4 \cdot 120} \\[5ex] r = 9.633333333 \\[3ex] r \approx 10\;cm \quad\dots\text{to the nearest whole number} $
