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1 Standards 8, 10, 11 Classifying Solids PROBLEM 1 PROBLEM 2 Surface Area of Cones Volume of a Right Cone PROBLEM 3 PROBLEM 4 PROBLEM 5 END SHOW PRESENTATION.

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Presentation on theme: "1 Standards 8, 10, 11 Classifying Solids PROBLEM 1 PROBLEM 2 Surface Area of Cones Volume of a Right Cone PROBLEM 3 PROBLEM 4 PROBLEM 5 END SHOW PRESENTATION."— Presentation transcript:

1 1 Standards 8, 10, 11 Classifying Solids PROBLEM 1 PROBLEM 2 Surface Area of Cones Volume of a Right Cone PROBLEM 3 PROBLEM 4 PROBLEM 5 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 2 Standard 8: Students know, derive, and solve problems involving perimeter, circumference, area, volume, lateral area, and surface area of common geometric figures. Estándar 8: Los estudiantes saben, derivan, y resuelven problemas involucrando perímetros, circunferencia, área, volumen, área lateral, y superficie de área de figuras geométricas comunes. Standard 10: Students compute areas of polygons including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. Estándar 10: Los estudiantes calculan áreas de polígonos incluyendo rectángulos, triángulos escalenos, triángulos equiláteros, rombos, paralelogramos, y trapezoides. Standard 11: Students determine how changes in dimensions affect the perimeter, area, and volume of common geomegtric figures and solids. Estándar 11: Los estudiantes determinan cambios en dimensiones que afectan perímetro, área, y volumen de figuras geométricas comunes y sólidos. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 3 PRISM PYRAMID CYLINDER SPHERE CONE Standards 8, 10, 11 SOLIDS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 4 h r l l 2 r r 2 r L= area of sector 2 l area of sector area of circle perimeter of cone’s base perimeter of circle = 2 C= l 2 l = area of sector 2 r C= Area of Circle 2 l 2 r C= perimeter of cone’s base r 2 l l = area of sector 2 l 2 l TOTAL SURFACE AREA: T = area of sector + area of cone’s base 2 r B= area of sector l r =L= T = L + B T= 2 r l r + SURFACE AREA OF A RIGHT CIRCULAR CONE Standards 8, 10, 11 h= height r = radius l = slant height Lateral Area PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 5 h r VOLUME OF A RIGHT CIRCULAR CONE Standards 8, 10, 11 2 r B= V = Bh 1 3 V = 2 r 1 3 h h= height r = radius PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 6 Standards 8, 10, 11 Find the lateral area, the surface area and volume of a right cone with a height of 26 cm and a radius of 12 cm. Round your answers to the nearest tenth. h r =26 cm 12 cm = Lateral Area: l r L= we need to find the slant height, using the Pythagorean Theorem: l l = 26 + 12 2 2 2 l = 676 + 144 2 l = 820 2 l 28.6 cm Calculating the base area: 2 r B= 2 r 12 cm B= ( ) 2 B= 452.2 cm 2 Calculating surface area: T = L + B L= ( )( ) 12 cm 28.6 cm L = 1077.7 cm 2 T = 1077.7 cm + 452.2 cm 2 2 T = 1529.9 cm 2 Calculating the volume: 2 r V = 1 3 h 26 cm V = 1 3 ( ) 2 12 cm V = 3918.7 cm 3 B = 144 V = 1 3 ( ) 26 cm 144 cm 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 7 Standards 8, 10, 11 Find the lateral area and the surface area and volume of a right cone whose slant height is 9 ft and whose circumference at the base is 4 ft. Round your answers to the nearest tenth. h r l We need to find the radius: C=2 r 2 2 r= C 2 2 r= 2 ft 4 2 ft = we need to find the height, using the Pythagorean Theorem: 9 = h + 2 222 81 = h + 4 2 -4 h = 77 2 h = 8.8 ft Lateral Area: l r L= L= ( )( ) 2 ft 9.0 ft L = 56.5 ft 2 Calculating the base area: 2 r B= 2 ft B= ( ) 2 B= 12.6 ft 2 B = 4 Calculating surface area: T = L + B T = 56.5 ft + 12.6 ft 2 2 T = 69.1 ft 2 Calculating the volume: 2 r V = 1 3 h 8.8 ft V = 1 3 ( ) 2 2 ft V = 36.8 ft 3 V = 1 3 ( ) 8.8 ft 4 ft 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 8 Standards 8, 10, 11 Find the lateral area, the surface area, and the volume of a right cone whose height is 18 m and whose slant height is 22 m. Round your answers to the nearest unit. h = 18 m r l =22 m we need to find the radius, using the Pythagorean Theorem: 22 = r + 18 2 2 2 484= r + 324 2 -324 r = 160 2 r = 13 m Lateral Area: l r L= L= ( )( ) 13 m 22 m L = 898 m 2 Calculating the base area: 2 r B= 13 m B= ( ) 2 B= 531 m 2 B = 169 Calculating surface area: T = L + B T = 898 m + 531 m 2 2 T = 1429 m 2 Calculating the volume: 2 r V = 1 3 h 18 m V = 1 3 ( ) 2 13 m V = 3184 m 3 V = 1 3 ( ) 18m 169 m 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 9 Standards 8, 10, 11 Find the lateral surface of a cone whose volume is 900 mm and whose radius is 15 mm. Round your answers to the closest tenth. 3 2 r V = 1 3 h ( ) = 1 3 ( ) h 2 900 15 900 = 1 3 ( 225 ) h (3) 2700 = 225 h 225 225 h= 2700 706.5 h = 3.8 mm Now we draw the cone: h r 15= we need to find the slant height, using the Pythagorean Theorem: l = 3.8 + 15 2 2 2 l = 14.4 + 225 2 l = 239.4 2 l 15.5 mm Lateral Area: l r L= L= ( )( ) 15mm 15.5 mm L 730 mm 2 =3.8 Calculating the height: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 10 Standards 8, 10, 11 The ratio of the radii of two similar cones is 3:8. If the volume of the larger cone is 2090 units, what is the approximate volume of the smaller cone? 3 VOLUME 1 > VOLUME 2 2 r V = 1 3 h Volume: VOLUME 1 VOLUME 2 IF V = h 2 r 1 3 1 11 h 2 r 1 3 2 22 THEN V 1 h 2 r 1 3 11 V 2 h 2 r 1 3 22 = AND r 1 r 2 = h 1 h 2 = 8 3 V r h = 1 1 1 2 2 2 2 2 V 2 = 88 3 3 2090 2 V 2 = 648 93 2090 2090 512 V 2 27 = (27)(2090) = 512V 2 512 = 1 1 1 2 2 2 V r h 2 Substituting values: THEN AND IF They are similar V 110 units 2 3 What can you conclude about the ratio of the volumes and the ratio of the radii? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


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