Special case: Square. The same formula works, but you can also use A= Side x Side or A= (Side) 2


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1 Geometry Chapter 11/12 Review Shape: Rectangle Formula A= Base x Height Special case: Square. The same formula works, but you can also use A= Side x Side or A= (Side) 2 Height = 6 Base = 8 Area = 8 x 6 = 48 Square with side of 7 Area = 7x7=49 Shape: Parallelogram Formula: A= Base x Height side is not the height. Major difficulty: If one side is the base, then the other side=7 height=6 base= Area = x6=60. Note: The side=7 is not used ! Draw an altitude from the upper left vertex This forms a triangle with the short leg 4 and the long leg = height = 4!3. This makes the area =20 x 4!3 = 80!3 4 20
2 Shape : Triangle First formula : A=Base x Height / 2 Major difficulty: Must have a right angle or perpendicular relationship between the height and base. A= (B x H) / 2 Height Base h 8 Then the area = 6! / 2 = 3! Find the area by first drawing an altitude. Then solve for h: h 2 +9=64 h=! Shape : Trapezoid First formula : Ê A = b + b 1 2 ˆ h Ë 2 Second formula : A = Mh where M = median b 2 =8 b 1 = h=6 Area = (8+)(6)=4 2 M=8 Area = 8 x = 40 h= 14 14! Draw a height from the upper left vertex, making a triangle This makes the small leg, the big leg!3. A=(1/2)(14+20)(!3)= 8!3
3 Shape : Kite Formula: A= (d 1 x d 2 ) / 2 considered as a kite for this formula. d 1 Note: Both a square and a rhombus can be d 2 For the problem below, realize that the diagonals are perpendicular. That creates and triangles. That gives the parts of the diagonals as shown. With that one diagonal = 4+4=8 and the other =4 +4!3 which does not reduce. The area = 8(4 + 4!3)/2 = 4(4 + 4!3) = ! !3 Shape: Equilateral triangle. Formula A= s 2!3/4 Example: The perimeter of an equilateral triangle = 30. Find the area. Since the perimeter = 30, each side = 30/3 =. The area = 2!3/4=0!3/4=2!3.
4 Shape : Regular polygon. Formula A= (a x p)/2 where a is the apothem and p the perimeter. Note: A regular polygon has all sides congruent and all angles congruent. a Example : Find the area of a regular hexagon which has each side being. a a=!3 Draw the segment as shown in the 2nd hexagon. This makes a triangle. The for the side of the hexagon is split to. This makes the apothem!3. The perimeter = 6 x = 60. The area = (1/2)(!3)(60) = (30)(!3) = 10!3. Shape: Circle. Formula: A= " r 2. Example diameter =. The radius = /2 =. The area = " x 2 =2".
5 Shape: Sector of a circle. r= 80 The formula is a modification of the circle area formula. Find the area of the whole circle and then find the fraction that the sector makes up. The area of the whole circle = " x 2 = 0". The fraction is 80 /360 = 2/9. 2/9 x 0" = 200"/9. Shape : Segment of a circle. r= 60 To find the area first find the area of the circle, then the sector, and then the triangle. The segment's area = the sector area  the triangle area. Area of the circle = " x 2 = 0". Area of the sector = ((1/6) x 0" = 0"/3. Area of the triangle = 2!3/4=2!3. Area of the segment = 0"/32!3.
6 Shape: Triangle Formula: Hero's formula. This will find the area of a triangle even if you know none of the angles. A = s(s  a)(s  b)(s  c) The sides of a triangle can be represented as a,b, and c. The s stands for the semiperimeter of the triangle. Add up all the sides and divide by 2 and you will have the semiperimeter. Example problem: What is the area of a triangle with sides 11,13 and 8? Find the semiperimeter s = ( )/2 = 16. A = 16(1611)(1613)(168) = 16()(3)(8) = 4 x3x4x2 = 4x2 x3x2 = 8 30 Sometimes, you will need to break the numbers int o smaller pieces. 8x1x21 = 4x2 x x3 x 3x7 = 4 x 3x3 x 2xx7 = 2x3 70 = 6 70 Shape: Inscribed quadrilateral. Formula: Brahmaguptas formula. A = (s  a)(s  b)(s  c)(s  d) The formula only works if the quadrilateral can be inscribed in a quadrilateral. Other than that, it works very much like Hero's formula. Example: What is the area of an inscribed quadrilateral with sides 4,,7, and? s= (4++7+)/2= 13. A = (134)(13  )(137)(13 ) = (9)(8)(6)(3) = 3 (4x2)(2x3)(3) = 3x2 (2x2)(3x3) = 3x2x2x3 = 36.
7 Shape: Prism Surface Area Formula: Surface Area = Base Area + Lateral Area This is abbreviated as S.A. = B.A. + L.A. The base area consists of the two triangles. The lateral area is the 3 rectangles. The rectangles are drawn as parallelograms to give a three dimensional effect. To find the base area draw an altitude from the 16 side of either triangle. h h= Find the h with the Pythagorean theorem. This gives h as 6. The triangle area = (1/2)(16 x 6) = 48. B.A. = 2 x 48 = 96. The lateral area consists of the three rectangles 16 by 20, by 20, and by 20. L.A. = 16 x 20 + x 20 + x 20 = = 720. The Surface Area = S.A. = = 816.
8 Another prism example. The surface area has 6 faces: 2 of 6 by, 2 of 6 by, and 2 of by. S.A.= 2( 6 x + 6 x + x ) = Shape: Pyramid Surface Area = L.A. + B.A. The L.A. consists of triangle sides. The base area varies, but it is typically a square. Example: Find the surface area of a regular square pyramid with a lateral edge of 13 and base edge of SL. 12. First draw a slant height. This splits the base edge to and. Find the slant height with the triangle with the Pythagorean Theorem: 2 +SL 2 =13 2. This gives the slant height as 12. Each triangle has an area of (1/2)( x 12) = 60. L.A. = 4 x 60 =240. B.A. = x = 0 S.A. = = 340.
9 Shape: Cylinder. Surface Area Formula: B.A. = 2"r 2 (The two circles.) L.A. = 2"rh. Example: Find the surface area of a cylinder with radius of 3 and height of 8. r=3 h = 8 B.A. = 2 " x 3 2 = 18" L.A. = 2 " x 3 x 8 = 48" S.A. = 18" + 48" = 66" Shape: Cone. Surface Area Formula: B.A. = "r 2 where L is the slant height. (The circle at the base) L.A. = "rl Example; Find the surface area of a cone with radius and a height of 12. By the Pythagorean theorem, L = 13. B.A. = " x 2 = 2" L.A. = " x x 13 = 6" S.A. = 2" + 6" = 90". h L Shape: Sphere. Surface Area Formula: S.A. = 4"r 2. The sphere has no base at all. r Example: Find the surface area of a sphere with radius of 6. S.A. = 4" x 6 2 = 4" x 36 = 144". radius
10 Shape: Prism, Box, or Cylinder Volume: V = Bh where B is the base area. To find the base area draw an altitude from the 16 side of either triangle. h h= Find the h with the Pythagorean theorem. This gives h as 6. The triangle area = (1/2)(16 x 6) = 48. V=Bh = 48 x 20 = 960 Box example. Use any face as the base. B= x 6 = 60 V= 60 x = 300 6
11 Cylinder example r= 3 h= B= " x 3 2 =9" V= Bh = 9" x = 90" Shape: Pyramid or Cone. Formula V= (1/3) Bh where B is the base area. Cone example: Find the volume of a cone with radius 6 and slant height of. Using the Pythagorean theorem, we have 6 2 +h 2 = 2 which solves to h = 8. L B= " x 6 2 =36" V= (1/3) B h = (1/3) x 36" x 8= 12" x 8 = 96" h r
12 Pyramid example: Find the volume of a regular square pyramid with base edges of and a lateral height of First use the square base to find the segments marked. Then use the Pythagorean theorem to find the altitude of 12. V= (1/3)( x ) (12) = 1200/3 = 400. Shape: Sphere. Formula: V = (4/3)" r 3 Example: Find the volume of a sphere with radius 6. V = (4/3)" x 6 3 = (4/3) " x 216 = 4 " x 72 =288 "
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