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Perimeter & Area Perimeter & Area Graphs & Formulas Volume (V) & Total Surface Area (TSA) Surface Area (TSA) Graphs & Formulas Example 1 Example 2 Example.

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Presentation on theme: "Perimeter & Area Perimeter & Area Graphs & Formulas Volume (V) & Total Surface Area (TSA) Surface Area (TSA) Graphs & Formulas Example 1 Example 2 Example."— Presentation transcript:

1 Perimeter & Area Perimeter & Area Graphs & Formulas Volume (V) & Total Surface Area (TSA) Surface Area (TSA) Graphs & Formulas Example 1 Example 2 Example 3

2 Perimeter (P) and Area (A) P = 2l+2w A = l·w w l RECTANGLESQUARE P = 4s A = s 2 s TRIANGLE h a P = a+b+c A = ½ a·h c b a b c RIGHT TRIANGLE P = a+b+c A = ½ a·b CIRCLE Circumference = 2π r A = π r 2 PARALLELOGRAM b h a P = 2a+2b A = ½ b·h TRAPEZOID b1b1 b2b2 h r P = a+ b 1 + c+ b 2 A = ½ (b 1 + b 2 )·h ac Pythagorean Theorem c 2 = a 2 + b 2 Return to Table

3 Perimeter and Area Pythagorean Theorem asserts that: 26 2 = x 2 So, 676 = x 2  100 = x 2  x = ± 10 and discard the negative value for x. Thus we have x =10 and Area of triangle = ½ (24)(10) =240 m 2. 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) Example 1 : Find the area of a right triangle having hypotenuse 26m and one side of 24m. Area of triangle = ½ b·h = ½ (24)x, where x is the other side of the triangle (see figure). 26 x 24 w l = w+8 But perimeter = 2l+ 2w =196 So, 2(w+8) + 2w = 196  2w w = 196  4w = 180 So w = 45 in. and l = 53 in So area = l ·w = (53)(45) = 2385 in 2. Return to Table

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

5 TSA and Volume Example 1: Find the TSA and the Volume for the figure on the right (see fig.) Cylinder volume = π r 2 h = π (4) 2 18 in 2 = 288 π in 2 a) SA for cone = π r s = π 5 s, but s=  ( ) =  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  π +25 π = 5π(25+  61 ) ft 2 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 ft 3 Sphere volume = (4/3) π r 3 So, (4/3) π r 3 = 288 π  r 3 =216  r = 6 in. Example 2: A cylinder has radius 4 in, and height18 in. Find the radius of a sphere having the same volume. For another example click hereclick here Return to Table

6 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. a) Pick VM, the height of equilateral triangle  BCV, since BM=MC=10, Using Pythagorean Theorem on right triangle  BMV : 20 2 = (VM) 2  (BM) 2 =300  VM =10  3 So, area  BCV =(1/2)20 (10  3 ) =100  3 b)Let VH be the high of the pyramid. Using Pythagorean Theorem on  AHB and the fact AH=BH, we get: Volume = (1/3)Base h = (1/3)400(10  2) = (4000/3)  2 ft 3 M TSA = Base+ 4 (Triangle) = (100  3) = 400(1+  3) ft 2 C Now Pythagorean Theorem on  AHV give us: 20 2 = (10  2)2 + (VH) 2  VH = 10  = (AH) 2 + (BH) 2  200 = (AH) 2  AH = 10  2 Return to Table


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