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3.3 Problem Solving in Geometry

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Presentation on theme: "3.3 Problem Solving in Geometry"— Presentation transcript:

1 3.3 Problem Solving in Geometry

2 Perimeter (Circumference), Area, and Volume
Ex. Find the area: 26m 18m m m 37m A = ½h(a + b) A = ½(18)( ) A = ½(18)(63) A = 9(63) A = 567 The area is 567 m2

3 Ex. A rectangle has a width of 46cm and a perimeter of 208cm
Ex. A rectangle has a width of 46cm and a perimeter of 208cm. What is its length? x = length width = 46 cm length = x P = 2L + 2W 208 = 2(x) + 2(46) 208 = 2x – 92 = 2x + 92 – = 2x = x The length is 58 cm

4 Circles diameter – distance across circle through the center radius – distance from center to edge of circle (half the diameter) Note: π is approximately 3.14 d r

5 Ex. Find the area and circumference. (a) in terms of π
Ex. Find the area and circumference (a) in terms of π (b) rounded to the nearest whole number 9m A = πr2 A = π(9)2 A = π81 A = 81π (a) area is 81π m2 A ≈ 81(3.14) A ≈ (b) area is approximately 254 m2 C = 2πr C = 2π(9) C = 18π (a) circumference is 18π m C ≈ 18(3.14) C ≈ m (b) circumference is approximately 57 m

6 Ex. A cylinder with radius 3 inches and height 4 inches has its radius tripled. How many times greater is the volume of the larger cylinder than the smaller cylinder?

7 V1 = πr2h V1 = π(3)2(4) V1 = π(9) (4) V1 = 36π in3
Ex. A cylinder with radius 3 inches and height 4 inches has its radius tripled. How many times greater is the volume of the larger cylinder than the smaller cylinder? V1 = πr2h V1 = π(3)2(4) V1 = π(9) (4) V1 = 36π in3 V2 = πr2h V2 = π(9)2(4) V2 = π(81) (4) V2 = 324π in3 9 times 4in 4in 3in 9in 9 1

8 Ex. A water reservoir is shaped like a rectangular solid with a base that is 50 yards by 30 yards, and a vertical height of 20 yards. At the start of a three-month period of no rain, the reservoir was completely full. At the end of this period, the height of the water was down to 6 yards. How much water was used in the three-month period?

9 Vstart = lwh Vstart = (50)(30)(20) Vstart = 30,000 yd3
Ex. A water reservoir is shaped like a rectangular solid with a base that is 50 yards by 30 yards, and a vertical height of 20 yards. At the start of a three-month period of no rain, the reservoir was completely full. At the end of this period, the height of the water was down to 6 yards. How much water was used in the three-month period? Vstart = lwh Vstart = (50)(30)(20) Vstart = 30,000 yd3 Vend = lwh Vend = (50)(30)(6) Vend = 9,000 yd3 Vstart – Vend = 30,000 – 9,000 = 21,000 yd3 20yd 6yd 30yd 50yd

10 Angles of a Triangle C A B The sum of the interior angles of a triangle is 180°. A° + B° + C° = 180°

11 Ex. One angle of a triangle is three times as large as another
Ex. One angle of a triangle is three times as large as another. The measure of the third angle is 40°more than that of the smallest angle. Find the measure of each angle.

12 Ex. One angle of a triangle is three times as large as another
Ex. One angle of a triangle is three times as large as another. The measure of the third angle is 40°more than that of the smallest angle. Find the measure of each angle. 3x x x A° + B° + C° = 180° x + 3x + x + 40 = 180 5x + 40 = 180 5x + 40 – 40 = 180 – 40 5x = x = 28 x = 28° 3x = 3(28) = 84° x + 40 = = 68° 28°, 84°and 68°

13 straight angle 180° right angle 90°

14 complementary angles – 2 angles whose sum is 90° If one angle is x B complementary angle is 90 – x A A + B = 90° supplementary angles – 2 angles whose sum is 180° B A supplementary angle is 180 – x A + B = 180° angle comp supp 50° 40° 130° 17° 73° 163° x° (90 – x)° (180 – x)°

15 Ex. Find the measure of an angle whose supplement measures 39° more than twice its complement.

16 Ex. Find the measure of an angle whose supplement measures 39° more than twice its complement.
angle = x comp = 90 – x supp = 180 – x 180 – x = 2(90 – x) – x = 180 – 2x – x = 219 – 2x 180 – x + 2x = 219 – 2x + 2x x = x – 180 = 219 – 180 x = 39 The angle is 39°

17 Groups Page 212 – 213: 31, 47, 53, 61


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