Presentation is loading. Please wait.

Presentation is loading. Please wait.

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 PRESENTATION CREATED.

Similar presentations


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 PRESENTATION CREATED."— 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 PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved END SHOW

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.

3 3 PRISM PYRAMID CYLINDER SPHERE CONE Standards 8, 10, 11 SOLIDS

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

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

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 = l = l = l 28.6 cm Calculating the base area: 2 r B= 2 r 12 cm B= ( ) 2 B= cm 2 Calculating surface area: T = L + B L= ( )( ) 12 cm 28.6 cm L = cm 2 T = cm cm 2 2 T = cm 2 Calculating the volume: 2 r V = 1 3 h 26 cm V = 1 3 ( ) 2 12 cm V = cm 3 B = 144 V = 1 3 ( ) 26 cm 144 cm 2

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 = h 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 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

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 = r r = 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 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

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 = 1 3 ( 225 ) h (3) 2700 = 225 h h= h = 3.8 mm Now we draw the cone: h r 15= we need to find the slant height, using the Pythagorean Theorem: l = l = l = l 15.5 mm Lateral Area: l r L= L= ( )( ) 15mm 15.5 mm L 730 mm 2 =3.8 Calculating the height:

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 h 2 r THEN V 1 h 2 r V 2 h 2 r = AND r 1 r 2 = h 1 h 2 = 8 3 V r h = V 2 = V 2 = V 2 27 = (27)(2090) = 512V = 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?


Download ppt "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 PRESENTATION CREATED."

Similar presentations


Ads by Google