Download presentation

Presentation is loading. Please wait.

Published byNestor Greg Modified over 3 years ago

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 = 24 2 + x 2 So, 676 = 576 + 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 + 16 + 2w = 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= (6 2 +5 2 ) = 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 ) 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 = 10 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) = 20 2 + 4(100 3) = 400(1+ 3) ft 2 C Now Pythagorean Theorem on AHV give us: 20 2 = (10 2)2 + (VH) 2 VH = 10 2 20 2 = (AH) 2 + (BH) 2 200 = (AH) 2 AH = 10 2 Return to Table

Similar presentations

Presentation is loading. Please wait....

OK

Geometry Jeopardy.

Geometry Jeopardy.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Spleen anatomy and physiology ppt on cells Ppt on social contract theory hobbes Ppt on conservation of momentum ppt Ppt on excessive use of chemicals in our daily life Ppt on anti corruption movements Ppt on email etiquettes presentation tips Ppt on power transmission elements Ppt on leverages definition Ppt on exploring fibonacci numbers Ppt on depth first search algorithms