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Geometry Formulas and Examples for Perimeter, Area, Volume, and TSA

Explore formulas and practical examples for triangles, rectangles, circles, pyramids, cylinders, cones, spheres, and more in geometry. Learn how to find perimeters, areas, volumes, and total surface areas for various shapes.

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Geometry Formulas and Examples for Perimeter, Area, Volume, and TSA

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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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