Solid Mensuration Circle = It is a closed figure with set of points that is equidistant to a fixed point called center. π
π
π π¨π = π
ππ = π πͺπΆ = ππ
π Example: Solve for the area and circumference of a circle whose radius is 10cm. π¨π = ππ 2 = π(10)2 = πππ. πππππ πΆπ = 2ππ = 2π(10) = ππ. ππππ Arc Length = It is a set of definite points that are equidistant to a fixed point called center. πΊπππ =
π
π½π πππ
Sector = A closed figure bounded by an arc and two radii. π¨π = π·=
π
π½ππ πππ
π
π½π π
π½ + ππ = π (π + ) πππ πππ
Example: Solve for the area and perimeter of a sector whose radius is 10cm and its central angle is 36o. π¨π =
πππ 2 π(36)(10)2 = = ππ. πππππ 360 360
π = π (2 +
ππ 36π ) = 10 (2 + ) = ππ. ππππ 180 180
Segment = A closed figure bounded by an arc and a chord. π
π½ππ ππ πΊπππ½ π
π½ πΊπππ½ π¨π = β = ππ ( β ) πππ π πππ π Example: Solve for the area of a segment whose radius is 10cm and its central angle is 36o. π¨π = ππ (
π
π½ πΊπππ½ πππ
πΊππππ β ) = πππ ( β ) = π. πππππ πππ π πππ π
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Area of Triangle Area = It is the surface included within the set of lines. Perimeter = it is the boundary of a closed plane figure.
h a
c
a
h
c
a
c
b
b b ππ π¨π = π π·= π+π+π Example: Solve the area and perimeter of a right triangle whose base is 15cm and its altitude is 5cm. ππ π(ππ) π¨= = = ππ. ππππ π π π = β52 + 152 = β25 + 225 = β250 = ππ. ππππ π = π + π + π = 5 + 15 + 15.81 = ππ. ππππ C
b
a A B c 1. Heronβs Formula: π
π¨π = βπ(π β π)(π β π)(π β π) Where: AΞ = Area of the triangle a, b, c = sides of the oblique triangle π+π+π π Example: The coordinates of an oblique triangle are (4, 5), (-2, 1) and (6, -4), what is the area and perimeter of the triangle? A(4, 5) B(-2, 1) C(6, -4) π = π + π + π = 7.21 + 9.43 + 9.22 = ππ. ππππ 2 2 2 2 π΄π΅ = β(4 β β2) + (5 β 1) = β6 + 4 = β52 = π. ππ π 25.86 π = = = ππ. ππ π΅πΆ = β(β2 β 6)2 + (1 β β4)2 = β82 + 52 2 2 = β89 = π. ππ 2 πΆπ΄ = β(6 β 4)2 + (β4 β 5)2 = β22 + 92 π¨π = βπ (π β π)(π β π)(π β π) = β85 = π. ππ = β12.93(5.72)(3.5)(3.71) = ππ. ππππ π=
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2. Coordinate Method: π¨π =
π ππ [ π ππ
ππ ππ ππ ππ
ππ ππ ]
Example: Solve the previous example by coordinate method. π¨π =
π π βπ π π π [ ] = ((π + π + ππ)β (βππ + π β ππ)) π π βπ π π π =
π ((ππ)β (βππ)) = ππππ π
3. Two sides and the included angle known π¨π =
πππΊππ π© π
Example: In triangle ABC, a = 40m, c = 50m and angle A = 53o. Find the area of the triangle? ππππΆ πππ53 = 50 40 50πππ53 πΆ1 = πππβ1 ( ) = 86.65 40 πΆ2 = 180 β 86.65 = 93.35 π΅1 = 180 β 86.65 β 53 = 40.35 π΅2 = 180 β 93.35 β 53 = 33.65 π¨π =
πππππ π΅ 40(50)πππ40.35 = = πππ. ππππ 2 2
π¨π =
πππππ π΅ 40(50)πππ33.65 = = πππ. ππππ 2 2
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4. One side and two adjacent angles known ππ πΊπππ¨ πΊππ πͺ π¨π = ππΊππ π© Example: The area of a triangle is 8346m2 and two of its interior angles are 37o25β and 56o17β. What is the length of the longest side? π΅ = 180 β 37Β°25β² β 56Β°17β² = ππ. π π¨π =
ππ πΊπππ¨ πΊππ πͺ ππΊππ π©
8346 = π2 =
π 2 πππ37Β°25β² πππ 56Β°17β² 2πππ 86.3
8346(2)πππ 86.3 πππ37Β°25β² πππ 56Β°17β²
8346(2)πππ 86.3 π=β = 181.54π πππ37Β°25β² πππ 56Β°17β² 5. Triangle inscribe in a circle π¨π =
πππ ππ
Example: The sides of a triangle are 8cm, 10cm and 14cm respectively. Determine the radius of the circle if the triangle is inscribed in a circle. π =
8+10+14 2
=
32 2
= ππ
2
π΄π₯ = βπ (π β π)(π β π)(π β π) 2
β16(16 β 8)(16 β 10)(16 β 14)
= β16(8)(6)(2) = ππ. πππππ π¨π =
πππ ππ
39.19 =
π=
8(10)(14) 4π
8(10)(14) = 7.14 4(39.19)
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6. The circle is inscribed in a triangle π¨ = ππ Example: The sides of the triangle are 8, 10 and 14 cm. Find the radius of the circle inscribed in the triangle. π =
8 + 10 + 14 32 = = ππ 2 2
π΄π₯ = 39.19ππ2 π΄π₯ = ππ 39.19 = π(16) π=
39.19 = 16
2.45cm
7. Triangle with an escribed circle π¨ = π(πΊ β π) Example: The radius of the circle escribed outside a triangle is 12cm and is tangent to the 18cm side. The other side of the triangle is 24cm long. If the area of the triangle is 216 sq.cm., find the third side of the triangle. π¨ = π(πΊ β π) πππ = ππ(πΊ β ππ) π =
216 + 18 = ππ 12
π =
π+π+π 2
36 =
18 + 24 + π 2
π = 36(2) β 18 β 24 = ππ
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8. Length of the median of a triangle ππ =
ππ βπππ + πππ β ππ π
Example: If the sides of the triangle ABC are AB = 15cm, BC = 18cm and CA = 24cm, compute the length of the median from C to the side AB βπ =
12 β2π2 + 2π 2 β π 2 2
βπ =
12 β2(18)2 + 2(24)2 β (15)2 = 19.84ππ 2
9. Length of bisector of an angle of a triangle to opposite side πβπππ(π β π) ππ = π+π Example: Triangle ABC has sides AB = 14cm, BC = 18cm and CA = 24cm respectively. Find the length of the bisector of angle A to the side BC. π =
π+π+π 2
π =
14 + 24 + 18 = 28πππ 2
βπ =
2βπππ (π β π) π+π
βπ =
2β24(14)(28)(28 β 18) = 16.14ππ 24 + 14
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10. Polygon Inscribed in a circle πππ π Where: Ο = central angle of an isosceles triangle formed from a polygon n = number of sides of a polygon β
=
πππ β β
π Where: π½ = interior angle of an isosceles triangle formed from a polygon π·=
π¨ = πππ πΊπππ·πͺπππ· π· = ππππͺπππ· Example: Compute the area and perimeter of a pentagon inscribed in a circle, r =10cm. β
=
360 360 = = ππ π 5
π½=
180 β β
180 β 72 = = ππ 2 2
π΄ = ππ 2 ππππ½πΆππ π½ = 5(10)2 πππ54πΆππ 54 = πππ. πππππ π = 2πππΆππ π½ = 2(5)(10)πΆππ 54 = ππ. ππππ 11. Polygon Circumscribing a Circle πππ π»πππ· πππ π·= π»πππ· π¨=
Example: Compute the area and perimeter of a pentagon circumscribing a circle, r = 10cm. β
=
360 360 = = ππ π 5
π½=
180 β β
180 β 72 = = ππ 2 2
π=
2ππ 2(5)(10) = = ππ. ππππ ππππ½ πππ54
π΄=
ππ 2 5(10)2 = = πππ. πππππ ππππ½ πππ54
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Name: Course/Year & Section: Compute the area and perimeter of the following if r = 15cm. 1.a. Inscribed Octagon β
=
360 360 = = ππ π 8
π½=
180 β β
180 β 72 = = ππ. π 2 2
π΄ = ππ 2 ππππ½πΆππ π½ = 8(15)2 πππ67.5πΆππ 67.5 = πππ. πππππ π = 2πππΆππ π½ = 2(8)(15)πΆππ 67.5 = ππ. ππππ 1.b. Circumscribing Octagon π=
2ππ 2(8)(15) = = ππ. ππππ ππππ½ πππ67.5
π΄=
ππ 2 8(15)2 = = πππ. πππππ ππππ½ πππ67.5
2.a. Inscribed Decagon β
=
360 360 = = ππ π 10
π½=
180 β β
180 β 36 = = ππ 2 2
π΄ = ππ 2 ππππ½πΆππ π½ = 10(15)2 πππ72πΆππ 72 = πππ. πππππ π = 2πππΆππ π½ = 2(10)(15)πΆππ 72 = ππ. ππππ 2.b. Circumscribing Decagon π=
2ππ 2(10)(15) = = ππ. ππππ ππππ½ πππ72
π΄=
ππ 2 10(15)2 = = πππ. πππππ ππππ½ πππ72
Page 8 of 25
Ellipse π¨ = π
ππ (ππ + ππ ) π· = ππ
β π Example: a = 5cm, b = 4cm π΄ = πππ = π(5)(4) = 20π = ππ. πππππ (π2 + π 2 ) (52 + 42 ) 25 + 16 41 π = 2πβ = 2πβ = 2πβ = 2πβ = ππ. ππππ 2 2 2 2 Spandrel = it is sometimes referred as the ornamented space between the left or right exterior curve of a line and the enclosing right triangle. Line Horizontal Diagonal Curve π¨=
n 0 1 2 or >
h
ππ π+π
n = degree of the curve
b
Example: Determine the area of a second degree curve whose base is 10cm and height 5cm. πβ 10(5) 50 π΄= = = = ππ. πππππ π+1 2+1 3
Area of Quadrilaterals Square = A quadrilateral whose sides are equal and consecutive sides are perpendicular to each other π¨ = ππ s P = 4s s Example: Solve the area and perimeter of a square whose side is 5cm each. π¨ = ππ = ππ = πππππ P = 4s = 4(5) = 20cm Rectangle = A quadrilateral whose opposite side are equal in length and consecutive sides are perpendicular to each other.
h
A = bh P = 2(b + h)
b Example: Solve the area and perimeter of a rectangle whose height is 5cm and base is 10cm. A = bh = 5(10) = 50cm2 Page 9 of 25
P = 2(b + h) = 2(5+10) = 30cm Rhombus = A quadrilateral whose sides are all equal but the consecutive sides are not perpendicular to each other. π΄ = π β = π 2 πππβ
P =4s
s h Ο
s Example: 1. Solve the area and perimeter of a rhomboid whose sides are 10cm and is inclined at 30o. π΄ = π β = π 2 πππβ
= 102 πππ30 = 100(0.5) = πππππ P =4s = 4(10) = 40cm 2. The area of a rhombus is 168sq.m. If one of its diagonal is 12m, find the length of the sides of a rhombus.
π
π π
π π 12π2 168 = 2 π2 = 28ππ π¨=
π₯ = β62 + 142 = ππ. ππππ Rhomboid= A quadrilateral whose opposite sides are equal in length but the consecutive sides are not perpendicular to each other.
s
h
π¨ = ππ = πππΊππβ
P = 2(s + b)
Ο b Example: 1. Solve the area and perimeter of a rhomboid whose base is 10cm and slant height is 5cm which is inclined at 30o. π΄ = ππ πππβ
= 10(5)πππ30 = πππππ P = 2(s + b) = 2(5 + 10) = 30cm
Page 10 of 25
2. The diagonals of a parallelogram are 18cm and 30cm respectively. One side of the parallelogram is 12cm. Find the area of the parallelogram. 15 9
ΞΈ
12 π = πΆππ β1 (
122 β 152 β 92 ) = 53.13 β2(15)(9)
π
π π
π πΊπππ½ π 18(30)πππ53.13 π΄= = ππππππ 2 Trapezoid = A quadrilateral whose opposite bases are parallel. a π¨=
π¨=π c
h
d
(π+π) π
= ππ =
ππ β ππ π(πͺπππ½ + πͺπππ·)
=
(ππ β ππ )π»πππ½π»πππ· π(π»πππ½+π»πππ·)
P=a+b+c+d
m ΞΈ
Ξ²
b Example: 1. Find the area and perimeter of trapezoid having length of median equal to 32m and an altitude of 6m. π΄ = πβ = 32(6) = πππππ 2. If a=100m, b=250m, ΞΈ=75o, Ξ²=60o, find the area and perimeter of the trapezoidal lot shown. 100 x
y
75 150 60
100
(π 2 β π2 )ππππππππ½ (2502 β 1002 )πππ75πππ60 = = ππ, πππ. ππππ 2(ππππ + ππππ½) 2(πππ75 + πππ60) π₯ 150 π¦ 150 = = πππ60 πππ45 πππ75 πππ45 π΄=
x= 183.71
y= 204.90
P = a + b + c + d = 100+250+183.71+204.90 = 738.61m Four sides not parallel having sides AB = 10cm, BC = 5cm, CD = 14.14cm and DA = 15cm, if the sum of the opposite angles is equal to 225o. π¨ = β(π β π)(π β π)(π β π)(π β π
) β ππππ
πͺπππ π½
Page 11 of 25
π=
π+π+π+π
π
Where: ΞΈ = average of opposite angles Examples: Find the area of a quadrilateral having sides AB=10cm, BC=5cm, CD=14.14cm and DA=15cm, if the sum of the opposite angles is equal to 225o. π+π+π+π 10 + 5 + 14.14 + 15 π = = = 22.07 2 2 π=
225 = 112.5 2
π΄ = β(π β π)(π β π)(π β π)(π β π) β πππππΆππ 2 π π΄ = β(22.07 β 10)(22.07 β 5)(22.07 β 14.14)(22.07 β 15) β 10(5)(14.14)(15)πΆππ 2 112.5 π΄ = β(12.07)(17.07)(7.93)(7.07) β 10(5)(14.14)(15)πΆππ 2 112.5 π¨ = ππ. πππππ Quadrilateral Inscribed in a Circle π¨ = β(π β π)(π β π)(π β π)(π β π
) Ptolemyβs Theorem π¨π©(π«πͺ) + π¨π«(π©πͺ) = π¨πͺ(π©π«) Example: Find the area of quadrilateral ABCD inscribed in a circle if AB=90m, DA=50m, CD=70m, BD=101.76m and AC=97.29m.
A
B
π΄π΅(π·πΆ) + π΄π·(π΅πΆ) = π΄πΆ(π΅π·) 90(70) + 50(π΅πΆ) = 97.29(101.76) π΅πΆ = 72π π =
π+π+π+π 90 + 70 + 72 + 50 = = 141 2 2
C D
π΄ = β(π β π)(π β π)(π β π)(π β π) π΄ = β(141 β 90)(141 β 70)(141 β 72)(141 β 50) π¨ = ππππ. ππππ
Page 12 of 25
Cyclic Quadrilateral Circumscribing a Circle π¨ = βππππ
Example: Find the area of a cyclic quadrilateral circumscribing a circle having sides of 48m, 32m, 54m and 38m respectively. π΄ = βππππ = β48(32)(54)(38) = ππππ. ππππ
Volume = It is the amount of space occupied by a three dimensional object as measured in cubic units. 1. Cube = It is a regular solid of six equal squares sides. π½ = ππ π¨π» = πππ
s s
π
π¨π³ = ππ
s
Example: Solve for the volume and lateral area of the cube whose side is 10cm. π = π 3 = 103 = πππππππ π΄ π = 6π 2 = 6(10)2 = 6(100) = ππππππ π΄πΏ = 4π 2 = 4(10)2 = 4(100) = ππππππ
Page 13 of 25
2. Prism = It is a polyhedron with two polygonal faces lying in parallel planes and with the other faces parallelograms. π½ = πππ π¨π» = π(ππ + ππ + ππ) π½ = π¨π© π
Example: a. Solve for the volume and lateral area of the prism whose base is 10cm, width 15cm and a height of 5cm. π = ππ€β = 10(15)(5) = πππππ
π
π΄πΏ = 2(πβ + βπ€) = 2(10(5) + 5(15)) = ππππππ π΄ π = 2(πβ + ππ€ + βπ€) = 2(10(5) + 10(15) + 5(15)) = ππππππ
b. Solve for the volume and lateral area of the prism whose base area is an equilateral triangle with side of 10cm and a height of 5cm π=
ππππππβ 102 πππ60(5) = = πππ. πππππ 2 2
π΄πΏ = 3(10)(5) = πππ πππ π΄ π = 150 + 2 (
102 πππ60 ) = πππ. π πππ 2
3. Cylinder = It is a prism whose base area is a circle π½ = π
ππ π π¨π³ = ππ
ππ π¨π» = ππ
ππ + ππ
ππ = ππ
π(π + π)
Page 14 of 25
Example: Solve for the volume and total area of the cylinder whose radius is 10cm and altitude of 5cm. π = ππ 2 β = π(10)2 (5) = ππππ. ππππ π΄πΏ = 2ππβ = 2π(10)(5) = πππ. πππππ π΄ π = 2ππ(π + β) = 2π(10)(10 + 5) = πππ. πππππ
4. Pyramid = It is a polyhedron whose base is a polygon and triangular faces with a common vertex. π½=
πππ π¨π© π = π π
For square base π¨π³ = πππ π¨π» = πππ + ππ For rectangular base π¨π³ = πππ + πππ π¨π» = πππ + πππ + ππ For triangular base πππ π¨π³ = π πππ ππ πΊππππ π¨π» = + π π Example: Solve for the volume and total area of the pyramid whose base area is square with side of 10cm and altitude of 5cm. ππ€β π΄π΅ β 10(10)(5) π= = = = πππ. πππππ 3 3 3 π΄πΏ = 2π π = 2(10)(5β2 ) = πππ. πππππ π΄ π = 2π π + π 2 = 141.42 + 102 = πππ. πππππ
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5. Cone = It is a solid bounded by a circular plane base and the surface formed by line segments joining every point of the boundary of the base to a common vertex.
π΄πΏ = πππΏ
Example: a. The lateral area of a right circular cone is 47.12sq.cm. Find the volume of the cone if it has a slant height of 5cm. π΄πΏ = πππΏ 47.12 = ππ(5) π = 3ππ β = βπΏ2 β π 2 = β52 β 32 = 4ππ π=
ππ 2 β π(3)2 (4) = = ππ. ππππ 3 3
Relationship of volume and height π½π ππ = π½π» ππ» Example: a. Given the volume of water in a cone is only Β½ of that of the cone. If the height of water is 6cm, what is the height of the cone? ππ π1 = 2 π1 63 = 3 ππ βπ ππ 3 2 = 6 ππ βπ 3 1 63 = 3 2 βπ ππ = π. ππππ
Page 16 of 25
b. A circular cone having an altitude of 9m is divided into 2 segments having the same vertex. If the smaller altitude is 6m, find the ratio of the volume of small cone to the big cone. π1 β1 3 = ππ βπ 3 π1 63 = 3 = π. ππ ππ 9
Cone constructed from a sector π½πΉ πΊ = πΉπ½, π= πππ Where: S = arc length of the sector or the circumference of the circle of the cone formed from the sector R = radius of the sector r = radius of the circle base of the cone ΞΈ = central angle of the sector Example: a. Find the volume of a cone to be constructed from a sector having a diameter of 72cm and a central angle of 210o. πππ 210π(36) π= = = 131.95 180 180 π=
ππ
210(36) = = 21ππ 360 360
β = β362 β 212 β = 29.24ππ π=
π΄π΅ β 131.95(29.24) = = ππππ. πππππ 3 3
b. The axis of the cone makes an angle of 60o with the horizontal. If the length of the axis is 30cm and its base radius 20cm, compute for the volume of the curve.
30
h
60
Page 17 of 25
β 30 β = 25.98ππ πππ60 =
π=
ππ 2 β π(20)2 (25.98) = = ππ, πππ. πππππ 3 3
6. Sphere = It is a solid bounded by a surface consisting of all points at a given distance from a point constituting its center. ππ
ππ π π¨π³ = ππ
ππ π½=
Example: What is the lateral area and volume of a sphere whose radius is 10cm? π΄πΏ = 4ππ 2 = 4π(10)2 = ππππ. πππππ π=
4ππ 3 4π(10)3 = = ππππ. πππππ 3 3
Spherical Wedge and Spherical Lune π
π½ππ π½πΊπΎ = πππ π
π½ππ π¨πΊπ³ = ππ Example: a. The area of a lune is 30sq.m. If the area of a sphere is 120sq.m., what is the angle of the lune? 120 = 4ππ 2 π = 3.09π πππ 2 π΄ππΏ = 90 ππ(3.09)2 30 = 90 π½ = ππ. ππ b. Find the volume and area of a spherical wedge whose central angle is Ο/5 radians and a radius of 6cm.
Page 18 of 25
πππ =
πππ 3 π(36)(6)3 = = ππ. πππππ 270 270
π΄ππΏ =
πππ 2 π(36)(6)2 = = ππ. πππππ 90 90
Spherical Segment π
ππ (ππ β π) π½= π π½=
π
π[π(πππ + πππ ) + ππ ] π
π¨π = ππ
ππ Example: a. The area of a zone of a spherical segment is 180Ο sq.m. If the radius of the sphere is 15m, compute the volume of the spherical segment. π΄π = 2ππβ 180π = 2π(15)β β = 6π π=
πβ2 (3π β β) π(6)2 (3(15) β 6) = = ππππ. πππππ 3 3
b. Find the volume of a spherical segment the radii of whose bases are 4m and 5m respectively with an altitude of 6m. π=
πβ[3(π12 + π22 ) + β2 ] π(6)[3(42 + 52 ) + 62 ] = = πππ. πππππ 6 6
c. A mixture compound from equal parts of two liquids, one white and the other black, was placed in a hemispherical bowl. The total depth of the two liquids is 6 inches. After standing for a short time the mixture separated, the white liquid settling below the black liquid if the thickness of the segment of the black liquid is 2 inches, find the radius of the bowl in inches. πβ2 (3π β β) π= 3 π(6)2 (3π β 6) 2π(4)2 (3π β 4) = 3 3 3(3π β 6) = 2(16)(3π β 4) 27π β 54 = 24π β 32 π = π. ππππ
Page 19 of 25
d. A sphere having diameter of 30cm is cut into 2 segments. The altitude of the first segment is 6cm. What is the ratio of the area of the second segment to that of the first. π΄2 2πππ» 24 = = =4 π΄1 2ππβ 6
e. A bowl in the form of a spherical segment with 2 bases has a height of 0.10m. The upper base is a great circle with a diameter of 0.60m. Compute the capacity of the bowl in cu.cm. 4ππ 3 πβ2 (3π β β) π= β 6 3 3 4π(0.3) π(0.2)2 (3(0.3) β 0.2) β = π. ππππ 6 3
Frustum of a Cone π
π(πΉπ + ππ + πΉπ) π½= π π¨π³ = π
(πΉ + π)π³ Example: A frustum of a cone has an upper base radius of 3m and a lower base radius of 6m. If the altitude of the frustum is 9m, compute the volume of the frustum and its lateral area. π=
πβ(π
2 + π 2 + π
π) π(9)(62 + 32 + 6(3)) = = 3π(36 + 9 + 18) = πππ. ππππ 3 3
πΏ = β92 + 1.52 = 9.49 π΄πΏ = π(π
+ π)πΏ = π(6 + 3)(9.49) = πππ. ππππ Frustum of a Regular Pyramid π(π + π© + βπ©π) π½= π Example: a. The volume of the frustum of a regular triangular pyramid is 135 cu.m. The lower and upper bases are both equilateral triangle with sides equal to 9m and 3m respectively. Find its altitude. 32 πππ60 = π. πππ 2 92 πππ60 π΄π΅ = = ππ. ππππ 2 π΄π =
Page 20 of 25
β(π + π΅ + βπ΅π) 3 β (3.9 + 35.07 + β35.07(3.9)) 135 = 3 405 β= = ππ 50.66 π=
b. A frustum of a regular pyramid has a upper base of 8m x 10m and a lower base of 10m x 100m and an altitude of 5m. Find the volume of the pyramid. π΄π = 10π₯8 = ππππ π΄πΏ = 10π₯100 = ππππππ
π=
5 (80 + 1000 + β80(1000)) β(π + π΅ + βπ΅π) = = ππππ. ππππ 3 3
Centroids Straight line ππ + ππ π ππ + ππ ππ = π ππ =
Example: A(3,-7) & B(-4,6) 3 + β4 βπ π₯π = = 2 π π¦π =
β7 + 6 βπ = 2 π
Plane Figures Square π ππ = π π ππ = π Rectangle π ππ = π π ππ = π Page 21 of 25
Triangle From the right angle π ππ = π π ππ = π
Spandrel
n yc xc ππ π+π π ππ = π ππ =
Quarter Circle
yc xc
π¨=
π
ππ ππ ππ , ππ = , ππ = π ππ
ππ
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Quarter Circle
yc r π¨=
π
ππ , π
ππ =
ππ ππ
Parabola
b yc a ππ =
ππ π
π¨=
π(ππ)(π) π
a
Semi-Ellipse b xc yc b b
a
2a π¨=
π
ππ π
ππ =
ππ ππ
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ππ =
ππ ππ
π¨π» Μ
π = β π¨π ππ Μ
= β π¨π π π¨π» π
Example:
6
5
A3 A1 A2 7
3
π΄1 = 5π₯7 = 35π π’ π₯ = 3.5 π¦ = 2.5 1 π΄2 = π₯3π₯5 = 7.5π π’ 2 1 π₯ = 7 + (3) = 8 3 1 5 π¦ = (5) = 3 3 1 π΄3 = π₯7π₯6 = 21π π’ 2 2 14 π₯ = (7) = 3 3 1 π¦ = (6) + 5 = 7 3 Page 24 of 25
π΄ π = 35 + 7.5 + 21 = 63.5π π’ π΄ π π₯Μ
= β π΄π π₯π 14 63.5π₯Μ
= 35(3.5) + 7.5(8) + 21 ( ) 3 Μ
= π. πππ π 5 63.5π¦Μ
= 35(2.5) + 7.5 ( ) + 21(7) 3 Μ
= π. πππ π
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