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Perimeter (P) and Area (A)

Perimeter (P) and Area (A). TRIANGLE. RECTANGLE. SQUARE. RIGHT TRIANGLE. Pythagorean Theorem c 2 = a 2 + b 2. c. b. c. b. w. h. l. s. a. a. P = 2l+2w A = l ·w. P = 4 s A = s 2. P = a+b+c A = ½ a ·h.

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Perimeter (P) and Area (A)

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  1. Perimeter (P) and Area (A) TRIANGLE RECTANGLE SQUARE RIGHT TRIANGLE • Pythagorean Theorem • c2 = a2 + b2 c b c b w h l s a a P = 2l+2w A = l·w P = 4sA = s2 P = a+b+cA = ½a·h P = a+b+c A = ½ a·b CIRCLE PARALLELOGRAM TRAPEZOID b2 a c r h b h a b1 P = 2a+2b A = ½b·h P = a+ b1+ c+ b2A = ½ (b1+ b2 )·h Circumference = 2π r A = π r2 Return to Table

  2. Perimeter and Area Example 1: Find the area of a right triangle having hypotenuse 26m and one side of 24m. 26 Area of triangle = ½ b·h = ½ (24)x, where x is the other side of the triangle (see figure). x Pythagorean Theorem asserts that: 262 = 242 + x2 So, 676 = 576 + x2  100 = x2 x = ± 10 and discard the negative value for x. Thus we have x =10 andArea of triangle = ½ (24)(10) =240 m2. Example 2: The perimeter of a rectangle is 196 in. and the length is 8 in. more than the wide. Find the area. If we call w the wide, then l = w+8 (see figure) 24 w l = w+8 But perimeter = 2l+ 2w =196 So, 2(w+8) + 2w = 196  2w + 16 + 2w = 196  4w = 180 So w = 45 in. and l = 53 in So area = l ·w = (53)(45) = 2385 in2. Return to Table

  3. Volume (V) & Total Surface Area (TSA) PYRAMID RECTANGULAR SOLID CUBE h h w e B l V = e3TSA = 6e2 V = l·w·h TSA = 2(l w+w h+h l) V = (1/3)·B·h Area of Base CONE CYLINDER SPHERE r s r h h r r V = π r2 hTSA = 2π r(r+h) V = (4/3) πr3 TSA = 4 π r2 V = (1/3) π r 2 h TSA = π r (r + s) Return to Table

  4. TSA and Volume Example 1: Find the TSA and the Volume for the figure on the right (see fig.) a) SA for cone = π r s = π 5 s , but s=(62+52) = 61 = π 5 61 SA for cylinder = 2 π r h = 2 π (5)10 =100 π Base area = π r 2 = π 5 2 = 25 π So, TSA = π561+100π +25 π = 5π(25+61 ) ft2 b) Volume = Vol of cylinder + Vol of Cone = π r 2 h + (1/3) π r 2 h = π 5 2 (10) +(1/3)π 5 2(6) = 300 π ft3 Example 2:A cylinder has radius 4 in, and height18 in. Find the radius of a sphere having the same volume. Cylinder volume = π r 2 h = π (4)2 18 in2 = 288 π in2 Sphere volume = (4/3) π r 3 So, (4/3) π r 3 = 288 π  r 3 =216  r = 6 in. For another example click here Return to Table

  5. a) Pick VM, the height of equilateral triangle BCV, since BM=MC=10, Example 3: The pyramid as shown in the figure below has a square base and is equilateral (all edges are equals). Find its TSA and Volume if the edge measure is 20ft. Using Pythagorean Theorem on right triangle BMV : 202 = 102 +(VM)2  (BM)2 =300  VM =10 3 C So, area BCV =(1/2)20 (10 3 ) =100 3 TSA = Base+ 4 (Triangle) = 202 + 4(100 3) = 400(1+ 3) ft2 M • Let VH be the high of the pyramid. Using Pythagorean • Theorem on AHB and the fact AH=BH, we get: 202 = (AH)2 + (BH)2 200 = (AH)2  AH = 102 Now Pythagorean Theorem on AHV give us: 202 = (10 2)2 + (VH)2  VH = 102 Volume = (1/3)Base h = (1/3)400(10 2) = (4000/3) 2 ft 3 Return to Table

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