Solid Mensuration 1 2.docx

  • Uploaded by: Peter Lay
  • 0
  • 0
  • June 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Solid Mensuration 1 2.docx as PDF for free.

More details

  • Words: 4,275
  • Pages: 25
Solid Mensuration Circle = It is a closed figure with set of points that is equidistant to a fixed point called center. π…π’…πŸ π‘¨πœŸ = π…π’“πŸ = πŸ’ π‘ͺ𝑢 = πŸπ…π’“ Example: Solve for the area and circumference of a circle whose radius is 10cm. π‘¨πœŸ = πœ‹π‘Ÿ 2 = πœ‹(10)2 = πŸ‘πŸπŸ’. πŸπŸ”π’„π’ŽπŸ 𝐢𝑂 = 2πœ‹π‘Ÿ = 2πœ‹(10) = πŸ”πŸ. πŸ–πŸ‘π’„π’Ž Arc Length = It is a set of definite points that are equidistant to a fixed point called center. 𝑺𝒂𝒓𝒄 =

π…πœ½π’“ πŸπŸ–πŸŽ

Sector = A closed figure bounded by an arc and two radii. π‘¨πœŸ = 𝑷=

π…πœ½π’“πŸ πŸ‘πŸ”πŸŽ

π…πœ½π’“ π…πœ½ + πŸπ’“ = 𝒓 (𝟐 + ) πŸπŸ–πŸŽ πŸπŸ–πŸŽ

Example: Solve for the area and perimeter of a sector whose radius is 10cm and its central angle is 36o. π‘¨πœŸ =

πœ‹πœƒπ‘Ÿ 2 πœ‹(36)(10)2 = = πŸ‘πŸ. πŸ’πŸπ’„π’ŽπŸ 360 360

𝑃 = π‘Ÿ (2 +

πœ‹πœƒ 36πœ‹ ) = 10 (2 + ) = πŸπŸ”. πŸπŸ–π’„π’Ž 180 180

Segment = A closed figure bounded by an arc and a chord. π…πœ½π’“πŸ π’“πŸ π‘Ίπ’Šπ’πœ½ π…πœ½ π‘Ίπ’Šπ’πœ½ π‘¨πœŸ = βˆ’ = π’“πŸ ( βˆ’ ) πŸ‘πŸ”πŸŽ 𝟐 πŸ‘πŸ”πŸŽ 𝟐 Example: Solve for the area of a segment whose radius is 10cm and its central angle is 36o. π‘¨πœŸ = π’“πŸ (

π…πœ½ π‘Ίπ’Šπ’πœ½ πŸ‘πŸ”π… π‘Ίπ’Šπ’πŸ‘πŸ” βˆ’ ) = 𝟏𝟎𝟐 ( βˆ’ ) = 𝟐. πŸŽπŸ‘π’„π’ŽπŸ πŸ‘πŸ”πŸŽ 𝟐 πŸ‘πŸ”πŸŽ 𝟐

Page 1 of 25

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 𝒃𝒉 π‘¨πœŸ = 𝟐 𝑷= 𝒂+𝒃+𝒄 Example: Solve the area and perimeter of a right triangle whose base is 15cm and its altitude is 5cm. 𝒃𝒉 πŸ“(πŸπŸ“) 𝑨= = = πŸ‘πŸ•. πŸ“π’„π’ŽπŸ 𝟐 𝟐 𝑐 = √52 + 152 = √25 + 225 = √250 = πŸπŸ“. πŸ–πŸπ’„π’Ž 𝑃 = π‘Ž + 𝑏 + 𝑐 = 5 + 15 + 15.81 = πŸ‘πŸ“. πŸ–πŸπ’„π’Ž C

b

a A B c 1. Heron’s Formula: 𝟐

π‘¨πœŸ = βˆšπ’”(𝒔 βˆ’ 𝒂)(𝒔 βˆ’ 𝒃)(𝒔 βˆ’ 𝒄) Where: AΞ” = Area of the triangle a, b, c = sides of the oblique triangle 𝒂+𝒃+𝒄 𝟐 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) 𝑃 = π‘Ž + 𝑏 + 𝑐 = 7.21 + 9.43 + 9.22 = πŸπŸ“. πŸ–πŸ”π’„π’Ž 2 2 2 2 𝐴𝐡 = √(4 βˆ’ βˆ’2) + (5 βˆ’ 1) = √6 + 4 = √52 = πŸ•. 𝟐𝟏 𝑃 25.86 𝑠= = = 𝟏𝟐. πŸ—πŸ‘ 𝐡𝐢 = √(βˆ’2 βˆ’ 6)2 + (1 βˆ’ βˆ’4)2 = √82 + 52 2 2 = √89 = πŸ—. πŸ’πŸ‘ 2 𝐢𝐴 = √(6 βˆ’ 4)2 + (βˆ’4 βˆ’ 5)2 = √22 + 92 π‘¨πœŸ = βˆšπ‘ (𝑠 βˆ’ π‘Ž)(𝑠 βˆ’ 𝑏)(𝑠 βˆ’ 𝑐) = √85 = πŸ—. 𝟐𝟐 = √12.93(5.72)(3.5)(3.71) = πŸ‘πŸŽ. πŸ—πŸ—π’”π’– 𝒔=

Page 2 of 25

2. Coordinate Method: π‘¨πœŸ =

𝟏 π’™πŸ [ 𝟐 π’šπŸ

π’™πŸ π’™πŸ‘ π’šπŸ π’šπŸ‘

π’™πŸ π’šπŸ ]

Example: Solve the previous example by coordinate method. π‘¨πœŸ =

𝟏 πŸ’ βˆ’πŸ πŸ” πŸ’ 𝟏 [ ] = ((πŸ’ + πŸ– + πŸ‘πŸŽ)β€” (βˆ’πŸπŸŽ + πŸ” βˆ’ πŸπŸ”)) πŸ“ 𝟏 βˆ’πŸ’ πŸ“ 𝟐 𝟐 =

𝟏 ((πŸ’πŸ)β€” (βˆ’πŸπŸŽ)) = πŸ‘πŸπ’”π’– 𝟐

3. Two sides and the included angle known π‘¨πœŸ =

π’‚π’„π‘Ίπ’Šπ’ 𝑩 𝟐

Example: In triangle ABC, a = 40m, c = 50m and angle A = 53o. Find the area of the triangle? 𝑆𝑖𝑛𝐢 𝑆𝑖𝑛53 = 50 40 50𝑆𝑖𝑛53 𝐢1 = π‘†π‘–π‘›βˆ’1 ( ) = 86.65 40 𝐢2 = 180 βˆ’ 86.65 = 93.35 𝐡1 = 180 βˆ’ 86.65 βˆ’ 53 = 40.35 𝐡2 = 180 βˆ’ 93.35 βˆ’ 53 = 33.65 π‘¨πœŸ =

π‘Žπ‘π‘†π‘–π‘› 𝐡 40(50)𝑆𝑖𝑛40.35 = = πŸ”πŸ’πŸ•. πŸ’πŸ”π’ŽπŸ 2 2

π‘¨πœŸ =

π‘Žπ‘π‘†π‘–π‘› 𝐡 40(50)𝑆𝑖𝑛33.65 = = πŸ“πŸ“πŸ’. πŸπŸπ’ŽπŸ 2 2

Page 3 of 25

4. One side and two adjacent angles known π’ƒπŸ π‘Ίπ’Šπ’π‘¨ π‘Ίπ’Šπ’ π‘ͺ π‘¨πœŸ = πŸπ‘Ίπ’Šπ’ 𝑩 Example: The area of a triangle is 8346m2 and two of its interior angles are 37o25’ and 56o17’. What is the length of the longest side? 𝐡 = 180 βˆ’ 37Β°25β€² βˆ’ 56Β°17β€² = πŸ–πŸ”. πŸ‘ π‘¨πœŸ =

π’ƒπŸ π‘Ίπ’Šπ’π‘¨ π‘Ίπ’Šπ’ π‘ͺ πŸπ‘Ίπ’Šπ’ 𝑩

8346 = 𝑏2 =

𝑏 2 𝑆𝑖𝑛37Β°25β€² 𝑆𝑖𝑛 56Β°17β€² 2𝑆𝑖𝑛 86.3

8346(2)𝑆𝑖𝑛 86.3 𝑆𝑖𝑛37Β°25β€² 𝑆𝑖𝑛 56Β°17β€²

8346(2)𝑆𝑖𝑛 86.3 𝑏=√ = 181.54π‘š 𝑆𝑖𝑛37Β°25β€² 𝑆𝑖𝑛 56Β°17β€² 5. Triangle inscribe in a circle π‘¨πœŸ =

𝒂𝒃𝒄 πŸ’π’“

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. 𝑠=

8+10+14 2

=

32 2

= πŸπŸ”

2

𝐴π›₯ = βˆšπ‘ (𝑠 βˆ’ π‘Ž)(𝑠 βˆ’ 𝑏)(𝑠 βˆ’ 𝑐) 2

√16(16 βˆ’ 8)(16 βˆ’ 10)(16 βˆ’ 14)

= √16(8)(6)(2) = πŸ‘πŸ—. πŸπŸ—π’„π’ŽπŸ π‘¨πœŸ =

𝒂𝒃𝒄 πŸ’π’“

39.19 =

π‘Ÿ=

8(10)(14) 4π‘Ÿ

8(10)(14) = 7.14 4(39.19)

Page 4 of 25

6. The circle is inscribed in a triangle 𝑨 = 𝒓𝒔 Example: The sides of the triangle are 8, 10 and 14 cm. Find the radius of the circle inscribed in the triangle. 𝑠=

8 + 10 + 14 32 = = πŸπŸ” 2 2

𝐴π›₯ = 39.19π‘π‘š2 𝐴π›₯ = π‘Ÿπ‘  39.19 = π‘Ÿ(16) π‘Ÿ=

39.19 = 16

2.45cm

7. Triangle with an escribed circle 𝑨 = 𝒓(𝑺 βˆ’ 𝒂) 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. 𝑨 = 𝒓(𝑺 βˆ’ 𝒂) πŸπŸπŸ” = 𝟏𝟐(𝑺 βˆ’ πŸπŸ–) 𝑠=

216 + 18 = πŸ‘πŸ” 12

𝑠=

π‘Ž+𝑏+𝑐 2

36 =

18 + 24 + 𝑐 2

𝑐 = 36(2) βˆ’ 18 βˆ’ 24 = πŸ‘πŸŽ

Page 5 of 25

8. Length of the median of a triangle 𝒉𝒄 =

𝟏𝟐 βˆšπŸπ’‚πŸ + πŸπ’ƒπŸ βˆ’ π’„πŸ 𝟐

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 β„Žπ‘ =

12 √2π‘Ž2 + 2𝑏 2 βˆ’ 𝑐 2 2

β„Žπ‘ =

12 √2(18)2 + 2(24)2 βˆ’ (15)2 = 19.84π‘π‘š 2

9. Length of bisector of an angle of a triangle to opposite side πŸβˆšπ’ƒπ’„π’”(𝒔 βˆ’ 𝒂) 𝒉𝒄 = 𝒃+𝒄 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. 𝑠=

π‘Ž+𝑏+𝑐 2

𝑠=

14 + 24 + 18 = 28π‘π‘šπ‘  2

β„Žπ‘ =

2βˆšπ‘π‘π‘ (𝑠 βˆ’ π‘Ž) 𝑏+𝑐

β„Žπ‘ =

2√24(14)(28)(28 βˆ’ 18) = 16.14π‘π‘š 24 + 14

Page 6 of 25

10. Polygon Inscribed in a circle πŸ‘πŸ”πŸŽ 𝒏 Where: Ο† = central angle of an isosceles triangle formed from a polygon n = number of sides of a polygon βˆ…=

πŸπŸ–πŸŽ βˆ’ βˆ… 𝟐 Where: 𝛽 = interior angle of an isosceles triangle formed from a polygon 𝜷=

𝑨 = π’π’“πŸ π‘Ίπ’Šπ’πœ·π‘ͺπ’π’”πœ· 𝑷 = πŸπ’π’“π‘ͺπ’π’”πœ· Example: Compute the area and perimeter of a pentagon inscribed in a circle, r =10cm. βˆ…=

360 360 = = πŸ•πŸ 𝑛 5

𝛽=

180 βˆ’ βˆ… 180 βˆ’ 72 = = πŸ“πŸ’ 2 2

𝐴 = π‘›π‘Ÿ 2 π‘†π‘–π‘›π›½πΆπ‘œπ‘ π›½ = 5(10)2 𝑆𝑖𝑛54πΆπ‘œπ‘ 54 = πŸπŸ‘πŸ•. πŸ•πŸ”π’„π’ŽπŸ 𝑃 = 2π‘›π‘ŸπΆπ‘œπ‘ π›½ = 2(5)(10)πΆπ‘œπ‘ 54 = πŸ“πŸ–. πŸ•πŸ–π’„π’Ž 11. Polygon Circumscribing a Circle π’π’“πŸ π‘»π’‚π’πœ· πŸπ’π’“ 𝑷= π‘»π’‚π’πœ· 𝑨=

Example: Compute the area and perimeter of a pentagon circumscribing a circle, r = 10cm. βˆ…=

360 360 = = πŸ•πŸ 𝑛 5

𝛽=

180 βˆ’ βˆ… 180 βˆ’ 72 = = πŸ“πŸ’ 2 2

𝑃=

2π‘›π‘Ÿ 2(5)(10) = = πŸ•πŸ. πŸ”πŸ“π’„π’Ž π‘‡π‘Žπ‘›π›½ π‘‡π‘Žπ‘›54

𝐴=

π‘›π‘Ÿ 2 5(10)2 = = πŸ‘πŸ”πŸ‘. πŸπŸ•π’„π’ŽπŸ π‘‡π‘Žπ‘›π›½ π‘‡π‘Žπ‘›54

Page 7 of 25

Name: Course/Year & Section: Compute the area and perimeter of the following if r = 15cm. 1.a. Inscribed Octagon βˆ…=

360 360 = = πŸ’πŸ“ 𝑛 8

𝛽=

180 βˆ’ βˆ… 180 βˆ’ 72 = = πŸ”πŸ•. πŸ“ 2 2

𝐴 = π‘›π‘Ÿ 2 π‘†π‘–π‘›π›½πΆπ‘œπ‘ π›½ = 8(15)2 𝑆𝑖𝑛67.5πΆπ‘œπ‘ 67.5 = πŸ”πŸ‘πŸ”. πŸ’πŸŽπ’„π’ŽπŸ 𝑃 = 2π‘›π‘ŸπΆπ‘œπ‘ π›½ = 2(8)(15)πΆπ‘œπ‘ 67.5 = πŸ—πŸ. πŸ–πŸ“π’„π’Ž 1.b. Circumscribing Octagon 𝑃=

2π‘›π‘Ÿ 2(8)(15) = = πŸ—πŸ—. πŸ’πŸπ’„π’Ž π‘‡π‘Žπ‘›π›½ π‘‡π‘Žπ‘›67.5

𝐴=

π‘›π‘Ÿ 2 8(15)2 = = πŸ•πŸ’πŸ“. πŸ“πŸ–π’„π’ŽπŸ π‘‡π‘Žπ‘›π›½ π‘‡π‘Žπ‘›67.5

2.a. Inscribed Decagon βˆ…=

360 360 = = πŸ‘πŸ” 𝑛 10

𝛽=

180 βˆ’ βˆ… 180 βˆ’ 36 = = πŸ•πŸ 2 2

𝐴 = π‘›π‘Ÿ 2 π‘†π‘–π‘›π›½πΆπ‘œπ‘ π›½ = 10(15)2 𝑆𝑖𝑛72πΆπ‘œπ‘ 72 = πŸ”πŸ”πŸ. πŸπŸ”π’„π’ŽπŸ 𝑃 = 2π‘›π‘ŸπΆπ‘œπ‘ π›½ = 2(10)(15)πΆπ‘œπ‘ 72 = πŸ—πŸ. πŸ•πŸπ’„π’Ž 2.b. Circumscribing Decagon 𝑃=

2π‘›π‘Ÿ 2(10)(15) = = πŸ—πŸ•. πŸ’πŸ–π’„π’Ž π‘‡π‘Žπ‘›π›½ π‘‡π‘Žπ‘›72

𝐴=

π‘›π‘Ÿ 2 10(15)2 = = πŸ•πŸ‘πŸ. πŸŽπŸ•π’„π’ŽπŸ π‘‡π‘Žπ‘›π›½ π‘‡π‘Žπ‘›72

Page 8 of 25

Ellipse 𝑨 = 𝝅𝒂𝒃 (π’‚πŸ + π’ƒπŸ ) 𝑷 = πŸπ…βˆš 𝟐 Example: a = 5cm, b = 4cm 𝐴 = πœ‹π‘Žπ‘ = πœ‹(5)(4) = 20πœ‹ = πŸ”πŸ. πŸ–πŸ‘π’„π’ŽπŸ (π‘Ž2 + 𝑏 2 ) (52 + 42 ) 25 + 16 41 𝑃 = 2πœ‹βˆš = 2πœ‹βˆš = 2πœ‹βˆš = 2πœ‹βˆš = πŸπŸ–. πŸ’πŸ“π’„π’Ž 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 𝑨=

n 0 1 2 or >

h

𝒃𝒉 𝒏+𝟏

n = degree of the curve

b

Example: Determine the area of a second degree curve whose base is 10cm and height 5cm. π‘β„Ž 10(5) 50 𝐴= = = = πŸπŸ”. πŸ”πŸ•π’„π’ŽπŸ 𝑛+1 2+1 3

Area of Quadrilaterals Square = A quadrilateral whose sides are equal and consecutive sides are perpendicular to each other 𝑨 = π’”πŸ s P = 4s s Example: Solve the area and perimeter of a square whose side is 5cm each. 𝑨 = π’”πŸ = πŸ“πŸ = πŸπŸ“π’„π’ŽπŸ 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. 𝐴 = π‘ β„Ž = 𝑠 2 π‘†π‘–π‘›βˆ… 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. 𝐴 = π‘ β„Ž = 𝑠 2 π‘†π‘–π‘›βˆ… = 102 𝑆𝑖𝑛30 = 100(0.5) = πŸ“πŸŽπ’„π’ŽπŸ 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Ÿ π’…πŸ 𝟐 12𝑑2 168 = 2 𝑑2 = 28π‘π‘š 𝑨=

π‘₯ = √62 + 142 = πŸπŸ“. πŸπŸ‘π’„π’Ž Rhomboid= A quadrilateral whose opposite sides are equal in length but the consecutive sides are not perpendicular to each other.

s

h

𝑨 = 𝒃𝒉 = π’ƒπ’”π‘Ίπ’Šπ’βˆ… 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. 𝐴 = π‘π‘ π‘†π‘–π‘›βˆ… = 10(5)𝑆𝑖𝑛30 = πŸπŸ“π’„π’ŽπŸ 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œƒ = πΆπ‘œπ‘  βˆ’1 (

122 βˆ’ 152 βˆ’ 92 ) = 53.13 βˆ’2(15)(9)

π’…πŸ π’…πŸ π‘Ίπ’Šπ’πœ½ 𝟐 18(30)𝑆𝑖𝑛53.13 𝐴= = πŸπŸπŸ”π’„π’ŽπŸ 2 Trapezoid = A quadrilateral whose opposite bases are parallel. a 𝑨=

𝑨=𝒉 c

h

d

(𝒂+𝒃) 𝟐

= π’Žπ’‰ =

π’ƒπŸ βˆ’ π’‚πŸ 𝟐(π‘ͺπ’π’•πœ½ + π‘ͺπ’π’•πœ·)

=

(π’ƒπŸ βˆ’ π’‚πŸ )π‘»π’‚π’πœ½π‘»π’‚π’πœ· 𝟐(π‘»π’‚π’πœ½+π‘»π’‚π’πœ·)

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. 𝐴 = π‘šβ„Ž = 32(6) = πŸπŸ—πŸπ’ŽπŸ 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

(𝑏 2 βˆ’ π‘Ž2 )π‘‡π‘Žπ‘›πœƒπ‘‡π‘Žπ‘›π›½ (2502 βˆ’ 1002 )π‘‡π‘Žπ‘›75π‘‡π‘Žπ‘›60 = = πŸ‘πŸ, πŸŽπŸ“πŸ’. πŸŽπŸ–π’ŽπŸ 2(π‘‡π‘Žπ‘›πœƒ + π‘‡π‘Žπ‘›π›½) 2(π‘‡π‘Žπ‘›75 + π‘‡π‘Žπ‘›60) π‘₯ 150 𝑦 150 = = 𝑆𝑖𝑛60 𝑆𝑖𝑛45 𝑆𝑖𝑛75 𝑆𝑖𝑛45 𝐴=

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. 𝑨 = √(𝒔 βˆ’ 𝒂)(𝒔 βˆ’ 𝒃)(𝒔 βˆ’ 𝒄)(𝒔 βˆ’ 𝒅) βˆ’ 𝒂𝒃𝒄𝒅π‘ͺπ’π’”πŸ 𝜽

Page 11 of 25

𝒔=

𝒂+𝒃+𝒄+𝒅 𝟐

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‘Ž+𝑏+𝑐+𝑑 10 + 5 + 14.14 + 15 𝑠= = = 22.07 2 2 πœƒ=

225 = 112.5 2

𝐴 = √(𝑠 βˆ’ π‘Ž)(𝑠 βˆ’ 𝑏)(𝑠 βˆ’ 𝑐)(𝑠 βˆ’ 𝑑) βˆ’ π‘Žπ‘π‘π‘‘πΆπ‘œπ‘  2 πœƒ 𝐴 = √(22.07 βˆ’ 10)(22.07 βˆ’ 5)(22.07 βˆ’ 14.14)(22.07 βˆ’ 15) βˆ’ 10(5)(14.14)(15)πΆπ‘œπ‘  2 112.5 𝐴 = √(12.07)(17.07)(7.93)(7.07) βˆ’ 10(5)(14.14)(15)πΆπ‘œπ‘  2 112.5 𝑨 = πŸ—πŸ—. πŸ—πŸ—π’„π’ŽπŸ Quadrilateral Inscribed in a Circle 𝑨 = √(𝒔 βˆ’ 𝒂)(𝒔 βˆ’ 𝒃)(𝒔 βˆ’ 𝒄)(𝒔 βˆ’ 𝒅) Ptolemy’s Theorem 𝑨𝑩(𝑫π‘ͺ) + 𝑨𝑫(𝑩π‘ͺ) = 𝑨π‘ͺ(𝑩𝑫) 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

𝐴𝐡(𝐷𝐢) + 𝐴𝐷(𝐡𝐢) = 𝐴𝐢(𝐡𝐷) 90(70) + 50(𝐡𝐢) = 97.29(101.76) 𝐡𝐢 = 72π‘š 𝑠=

π‘Ž+𝑏+𝑐+𝑑 90 + 70 + 72 + 50 = = 141 2 2

C D

𝐴 = √(𝑠 βˆ’ π‘Ž)(𝑠 βˆ’ 𝑏)(𝑠 βˆ’ 𝑐)(𝑠 βˆ’ 𝑑) 𝐴 = √(141 βˆ’ 90)(141 βˆ’ 70)(141 βˆ’ 72)(141 βˆ’ 50) 𝑨 = πŸ’πŸ•πŸ”πŸ–. πŸπŸ”π’ŽπŸ

Page 12 of 25

Cyclic Quadrilateral Circumscribing a Circle 𝑨 = βˆšπ’‚π’ƒπ’„π’… Example: Find the area of a cyclic quadrilateral circumscribing a circle having sides of 48m, 32m, 54m and 38m respectively. 𝐴 = βˆšπ‘Žπ‘π‘π‘‘ = √48(32)(54)(38) = πŸπŸ•πŸ•πŸ“. πŸ‘πŸ“π’ŽπŸ

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. 𝑽 = π’”πŸ‘ 𝑨𝑻 = πŸ”π’”πŸ

s s

𝟐

𝑨𝑳 = πŸ’π’”

s

Example: Solve for the volume and lateral area of the cube whose side is 10cm. 𝑉 = 𝑠 3 = 103 = πŸπŸŽπŸŽπŸŽπ’„π’ŽπŸ‘ 𝐴 𝑇 = 6𝑠 2 = 6(10)2 = 6(100) = πŸ”πŸŽπŸŽπ’„π’ŽπŸ 𝐴𝐿 = 4𝑠 2 = 4(10)2 = 4(100) = πŸ’πŸŽπŸŽπ’„π’ŽπŸ

Page 13 of 25

2. Prism = It is a polyhedron with two polygonal faces lying in parallel planes and with the other faces parallelograms. 𝑽 = π’ƒπ’˜π’‰ 𝑨𝑻 = 𝟐(𝒃𝒉 + π’ƒπ’˜ + π’‰π’˜) 𝑽 = 𝑨𝑩 𝒉

Example: a. Solve for the volume and lateral area of the prism whose base is 10cm, width 15cm and a height of 5cm. 𝑉 = π‘π‘€β„Ž = 10(15)(5) = πŸ•πŸ“πŸŽπ’„π’Ž

πŸ‘

𝐴𝐿 = 2(π‘β„Ž + β„Žπ‘€) = 2(10(5) + 5(15)) = πŸπŸ“πŸŽπ’„π’ŽπŸ 𝐴 𝑇 = 2(π‘β„Ž + 𝑏𝑀 + β„Žπ‘€) = 2(10(5) + 10(15) + 5(15)) = πŸ“πŸ“πŸŽπ’„π’ŽπŸ

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 𝑉=

π‘Žπ‘π‘†π‘–π‘›πœƒβ„Ž 102 𝑆𝑖𝑛60(5) = = πŸπŸπŸ”. πŸ“πŸπ’„π’ŽπŸ‘ 2 2

𝐴𝐿 = 3(10)(5) = πŸπŸ“πŸŽ π’„π’ŽπŸ 𝐴 𝑇 = 150 + 2 (

102 𝑆𝑖𝑛60 ) = πŸπŸ‘πŸ”. πŸ” π’„π’ŽπŸ 2

3. Cylinder = It is a prism whose base area is a circle 𝑽 = π…π’“πŸ 𝒉 𝑨𝑳 = πŸπ…π’“π’‰ 𝑨𝑻 = πŸπ…π’“π’‰ + πŸπ…π’“πŸ = πŸπ…π’“(𝒓 + 𝒉)

Page 14 of 25

Example: Solve for the volume and total area of the cylinder whose radius is 10cm and altitude of 5cm. 𝑉 = πœ‹π‘Ÿ 2 β„Ž = πœ‹(10)2 (5) = πŸπŸ“πŸ•πŸŽ. πŸ–π’„π’ŽπŸ‘ 𝐴𝐿 = 2πœ‹π‘Ÿβ„Ž = 2πœ‹(10)(5) = πŸ—πŸ’πŸ. πŸ’πŸ–π’„π’ŽπŸ 𝐴 𝑇 = 2πœ‹π‘Ÿ(π‘Ÿ + β„Ž) = 2πœ‹(10)(10 + 5) = πŸ—πŸ’πŸ. πŸ’πŸ–π’„π’ŽπŸ

4. Pyramid = It is a polyhedron whose base is a polygon and triangular faces with a common vertex. 𝑽=

π’ƒπ’˜π’‰ 𝑨𝑩 𝒉 = πŸ‘ πŸ‘

For square base 𝑨𝑳 = πŸπ’”π’ 𝑨𝑻 = πŸπ’”π’ + π’”πŸ For rectangular base 𝑨𝑳 = π’ƒπ’πŸ + π’˜π’πŸ 𝑨𝑻 = π’ƒπ’πŸ + π’˜π’πŸ + π’ƒπ’˜ For triangular base πŸ‘π’”π’ 𝑨𝑳 = 𝟐 πŸ‘π’”π’ π’”πŸ π‘Ίπ’Šπ’πŸ”πŸŽ 𝑨𝑻 = + 𝟐 𝟐 Example: Solve for the volume and total area of the pyramid whose base area is square with side of 10cm and altitude of 5cm. π‘π‘€β„Ž 𝐴𝐡 β„Ž 10(10)(5) 𝑉= = = = πŸπŸ”πŸ”. πŸ”πŸ•π’„π’ŽπŸ‘ 3 3 3 𝐴𝐿 = 2𝑠𝑙 = 2(10)(5√2 ) = πŸπŸ’πŸ. πŸ’πŸπ’„π’ŽπŸ 𝐴 𝑇 = 2𝑠𝑙 + 𝑠 2 = 141.42 + 102 = πŸπŸ’πŸ. πŸ’πŸπ’„π’ŽπŸ

Page 15 of 25

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.

𝐴𝐿 = πœ‹π‘ŸπΏ

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. 𝐴𝐿 = πœ‹π‘ŸπΏ 47.12 = πœ‹π‘Ÿ(5) π‘Ÿ = 3π‘π‘š β„Ž = √𝐿2 βˆ’ π‘Ÿ 2 = √52 βˆ’ 32 = 4π‘π‘š 𝑉=

πœ‹π‘Ÿ 2 β„Ž πœ‹(3)2 (4) = = πŸ‘πŸ•. πŸ•π’„π’ŽπŸ‘ 3 3

Relationship of volume and height π‘½πŸ π’‰πŸ = 𝑽𝑻 𝒉𝑻 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? 𝑉𝑇 𝑉1 = 2 𝑉1 63 = 3 𝑉𝑇 β„Žπ‘ 𝑉𝑇 3 2 = 6 𝑉𝑇 β„Žπ‘ 3 1 63 = 3 2 β„Žπ‘ 𝒉𝒄 = πŸ•. πŸ“πŸ”π’„π’Ž

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. 𝑉1 β„Ž1 3 = 𝑉𝑇 β„Žπ‘‡ 3 𝑉1 63 = 3 = 𝟎. πŸ‘πŸŽ 𝑉𝑇 9

Cone constructed from a sector πœ½π‘Ή 𝑺 = π‘Ήπœ½, 𝒓= πŸ‘πŸ”πŸŽ 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. πœƒπœ‹π‘Ÿ 210πœ‹(36) 𝑆= = = 131.95 180 180 π‘Ÿ=

πœƒπ‘… 210(36) = = 21π‘π‘š 360 360

β„Ž = √362 βˆ’ 212 β„Ž = 29.24π‘π‘š 𝑉=

𝐴𝐡 β„Ž 131.95(29.24) = = πŸπŸπŸ–πŸ”. πŸŽπŸ•π’„π’ŽπŸ 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„Ž 30 β„Ž = 25.98π‘π‘š 𝑆𝑖𝑛60 =

𝑉=

πœ‹π‘Ÿ 2 β„Ž πœ‹(20)2 (25.98) = = 𝟏𝟎, πŸ–πŸ–πŸ. πŸ’πŸ–π’„π’ŽπŸ‘ 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. πŸ’π…π’“πŸ‘ πŸ‘ 𝑨𝑳 = πŸ’π…π’“πŸ 𝑽=

Example: What is the lateral area and volume of a sphere whose radius is 10cm? 𝐴𝐿 = 4πœ‹π‘Ÿ 2 = 4πœ‹(10)2 = πŸπŸπŸ“πŸ”. πŸ”πŸ’π’„π’ŽπŸ 𝑉=

4πœ‹π‘Ÿ 3 4πœ‹(10)3 = = πŸ’πŸπŸ–πŸ–. πŸ•πŸ—π’„π’ŽπŸ‘ 3 3

Spherical Wedge and Spherical Lune π…πœ½π’“πŸ‘ 𝑽𝑺𝑾 = πŸπŸ•πŸŽ π…πœ½π’“πŸ 𝑨𝑺𝑳 = πŸ—πŸŽ 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 = 4πœ‹π‘Ÿ 2 π‘Ÿ = 3.09π‘š πœ‹πœƒπ‘Ÿ 2 𝐴𝑆𝐿 = 90 πœ‹πœƒ(3.09)2 30 = 90 𝜽 = πŸ—πŸŽ. 𝟎𝟏 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

π‘‰π‘†π‘Š =

πœ‹πœƒπ‘Ÿ 3 πœ‹(36)(6)3 = = πŸ—πŸŽ. πŸ’πŸ–π’„π’ŽπŸ‘ 270 270

𝐴𝑆𝐿 =

πœ‹πœƒπ‘Ÿ 2 πœ‹(36)(6)2 = = πŸ’πŸ“. πŸπŸ’π’„π’ŽπŸ‘ 90 90

Spherical Segment π…π’‰πŸ (πŸ‘π’“ βˆ’ 𝒉) 𝑽= πŸ‘ 𝑽=

𝝅𝒉[πŸ‘(π’“πŸπŸ + π’“πŸπŸ ) + π’‰πŸ ] πŸ”

𝑨𝒁 = πŸπ…π’“π’‰ 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. 𝐴𝑍 = 2πœ‹π‘Ÿβ„Ž 180πœ‹ = 2πœ‹(15)β„Ž β„Ž = 6π‘š 𝑉=

πœ‹β„Ž2 (3π‘Ÿ βˆ’ β„Ž) πœ‹(6)2 (3(15) βˆ’ 6) = = πŸπŸ’πŸ•πŸŽ. πŸπŸ•π’„π’ŽπŸ‘ 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. 𝑉=

πœ‹β„Ž[3(π‘Ÿ12 + π‘Ÿ22 ) + β„Ž2 ] πœ‹(6)[3(42 + 52 ) + 62 ] = = πŸ’πŸ—πŸ—. πŸ“πŸπ’„π’ŽπŸ‘ 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. πœ‹β„Ž2 (3π‘Ÿ βˆ’ β„Ž) 𝑉= 3 πœ‹(6)2 (3π‘Ÿ βˆ’ 6) 2πœ‹(4)2 (3π‘Ÿ βˆ’ 4) = 3 3 3(3π‘Ÿ βˆ’ 6) = 2(16)(3π‘Ÿ βˆ’ 4) 27π‘Ÿ βˆ’ 54 = 24π‘Ÿ βˆ’ 32 𝒓 = πŸ•. πŸ‘πŸ‘π’Šπ’

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. 𝐴2 2πœ‹π‘Ÿπ» 24 = = =4 𝐴1 2πœ‹π‘Ÿβ„Ž 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. 4πœ‹π‘Ÿ 3 πœ‹β„Ž2 (3π‘Ÿ βˆ’ β„Ž) 𝑉= βˆ’ 6 3 3 4πœ‹(0.3) πœ‹(0.2)2 (3(0.3) βˆ’ 0.2) βˆ’ = 𝟎. πŸπŸ”π’ŽπŸ‘ 6 3

Frustum of a Cone 𝝅𝒉(π‘ΉπŸ + π’“πŸ + 𝑹𝒓) 𝑽= πŸ‘ 𝑨𝑳 = 𝝅(𝑹 + 𝒓)𝑳 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. 𝑉=

πœ‹β„Ž(𝑅 2 + π‘Ÿ 2 + π‘…π‘Ÿ) πœ‹(9)(62 + 32 + 6(3)) = = 3πœ‹(36 + 9 + 18) = πŸ“πŸ—πŸ‘. πŸ•πŸ”π’ŽπŸ‘ 3 3

𝐿 = √92 + 1.52 = 9.49 𝐴𝐿 = πœ‹(𝑅 + π‘Ÿ)𝐿 = πœ‹(6 + 3)(9.49) = πŸπŸ”πŸ–. πŸ‘πŸπ’ŽπŸ Frustum of a Regular Pyramid 𝒉(𝒃 + 𝑩 + βˆšπ‘©π’ƒ) 𝑽= πŸ‘ 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 𝑆𝑖𝑛60 = πŸ‘. πŸ—π’ŽπŸ 2 92 𝑆𝑖𝑛60 𝐴𝐡 = = πŸ‘πŸ“. πŸŽπŸ•π’ŽπŸ 2 𝐴𝑆 =

Page 20 of 25

β„Ž(𝑏 + 𝐡 + βˆšπ΅π‘) 3 β„Ž (3.9 + 35.07 + √35.07(3.9)) 135 = 3 405 β„Ž= = πŸ–π’Ž 50.66 𝑉=

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‘ˆ = 10π‘₯8 = πŸ–πŸŽπ’ŽπŸ 𝐴𝐿 = 10π‘₯100 = πŸπŸŽπŸŽπŸŽπ’ŽπŸ

𝑉=

5 (80 + 1000 + √80(1000)) β„Ž(𝑏 + 𝐡 + βˆšπ΅π‘) = = πŸπŸπŸ•πŸ. πŸ’πŸŽπ’ŽπŸ‘ 3 3

Centroids Straight line π’™πŸ + π’™πŸ 𝟐 π’šπŸ + π’šπŸ π’šπ’„ = 𝟐 𝒙𝒄 =

Example: A(3,-7) & B(-4,6) 3 + βˆ’4 βˆ’πŸ π‘₯𝑐 = = 2 𝟐 𝑦𝑐 =

βˆ’7 + 6 βˆ’πŸ = 2 𝟐

Plane Figures Square 𝒃 𝒙𝒄 = 𝟐 𝒉 π’šπ’„ = 𝟐 Rectangle 𝒃 𝒙𝒄 = 𝟐 𝒉 π’šπ’„ = 𝟐 Page 21 of 25

Triangle From the right angle 𝒃 𝒙𝒄 = πŸ‘ 𝒉 π’šπ’„ = πŸ‘

Spandrel

n yc xc πŸπ’ƒ 𝒏+𝟐 𝒉 π’šπ’„ = πŸ‘ 𝒙𝒄 =

Quarter Circle

yc xc

𝑨=

π…π’“πŸ πŸ’π’“ πŸ’π’“ , 𝒙𝒄 = , π’šπ’„ = πŸ’ πŸ‘π… πŸ‘π…

Page 22 of 25

Quarter Circle

yc r 𝑨=

π…π’“πŸ , πŸ’

π’šπ’„ =

πŸ’π’“ πŸ‘π…

Parabola

b yc a π’šπ’„ =

πŸπ’ƒ πŸ“

𝑨=

𝟐(πŸπ’‚)(𝒃) πŸ‘

a

Semi-Ellipse b xc yc b b

a

2a 𝑨=

𝝅𝒂𝒃 𝟐

π’šπ’„ =

πŸ’π’ƒ πŸ‘π…

Page 23 of 25

𝒙𝒄 =

πŸ’π’‚ πŸ‘π…

𝑨𝑻 Μ… 𝒙 = βˆ‘ π‘¨π’Š π’™π’Š Μ… = βˆ‘ π‘¨π’Š π’š 𝑨𝑻 π’š

Example:

6

5

A3 A1 A2 7

3

𝐴1 = 5π‘₯7 = 35𝑠𝑒 π‘₯ = 3.5 𝑦 = 2.5 1 𝐴2 = π‘₯3π‘₯5 = 7.5𝑠𝑒 2 1 π‘₯ = 7 + (3) = 8 3 1 5 𝑦 = (5) = 3 3 1 𝐴3 = π‘₯7π‘₯6 = 21𝑠𝑒 2 2 14 π‘₯ = (7) = 3 3 1 𝑦 = (6) + 5 = 7 3 Page 24 of 25

𝐴 𝑇 = 35 + 7.5 + 21 = 63.5𝑠𝑒 𝐴 𝑇 π‘₯Μ… = βˆ‘ 𝐴𝑖 π‘₯𝑖 14 63.5π‘₯Μ… = 35(3.5) + 7.5(8) + 21 ( ) 3 Μ… = πŸ’. πŸ’πŸπ’– 𝒙 5 63.5𝑦̅ = 35(2.5) + 7.5 ( ) + 21(7) 3 Μ… = πŸ‘. πŸ–πŸ—π’– π’š

Page 25 of 25

Related Documents


More Documents from ""

Untitled.pdf
May 2020 3
Questionaire '10.docx
June 2020 4
Filipino Rb.docx
May 2020 10
Filipino.docx
May 2020 13