Can someone check where my mistakes are please? Measurement Questions (1 Viewer)

[kpad5991]

Question 1

Find the surface area (to one decimal place)


SA top = 1/2 x 9 x 12 = 54 cm^2

SA bottom = 1/2 x 9 x 12 = 54 cm^2

SA front = 1/2 x 12 x 5 = 30 cm^2

SA back = 1/2 x 15 x 5 = 37.5 cm^2 (used Pythagoras to get base of 15 cm)

Total SA= 175.5 cm^2

The actual answer says it is 180cm^2 (to one decimal place)


Question 2

Find the surface area (to one decimal place)



SA front = (1/2 x 20 x 6) + (20 x 15) = 360 cm^2
SA back = 360 cm^2
SA bottom = 30 x 20 = 600 cm^2
SA RHS = (30 x 15) + (11.66 x 30) used Pythagoras to find width of rectangle w^2 = 10^2 +6^2, w = 11.66 cm
= 799.86 cm^2
SA LHS = 799.86 cm^2

Total SA = 2919.7 cm^2 (to one decimal place)

The actual answer says it is 2819.2 cm^2 (to one decimal place)

Question 3

Find the surface area (to one decimal place)



SA RHS = 16.1 x 5.9 = 94.99 mm^2
SA LHS = 94.99 mm^2
SA front = 16.1 x 4 = 64.4 mm^2
SA back = 9.2 x 16.1 = 148.12 mm^2
SA top = 2.79/2 (4+9.2) = 18.414 mm^2 used Pythagoras to get perpendicular height of trapezium h^2 = 5.9^2 - 5.2^2, h = 2.79
SA bottom = 18.414 mm^2
Total SA = 439.3 mm^2 (to one decimal place)

The actual answer says it is 512.0 mm^2 (to one decimal place)


Question 4

Find the volume of the following truncated shape


Volume of large pyramid = 1/3 (4 x 4) x 15 = 80 cm^3

Volume of small pyramid = 1/3 (2 x 2 ) x 8 = 32/3 cm^3

Volume of truncated shape = 80 - 32/3 = 69.3 cm^3

The actual answer is 112 cm^3

Question 5

Find the area of the shaded area correct to 1 decimal place


Area of entire sector = 1/6 x pi x 14^2
= 98/3 pi cm^2

Area of small sector = 1/6 x pi x 3^2
= 3/2 pi cm^2

Area of shaded region = 98/3 pi - 3/2 pi = 97.9 cm^2 (to one decimal place)

The actual answer is 39. 3 cm^2

 

[cossine]

Question 1

Find the surface area (to one decimal place)
View attachment 31970

SA top = 1/2 x 9 x 12 = 54 cm^2

SA bottom = 1/2 x 9 x 12 = 54 cm^2

SA front = 1/2 x 12 x 5 = 30 cm^2

SA back = 1/2 x 15 x 5 = 37.5 cm^2 (used Pythagoras to get base of 15 cm)

Total SA= 175.5 cm^2

The actual answer says it is 180cm^2 (to one decimal place)


Question 2

Find the surface area (to one decimal place)

View attachment 31971

SA front = (1/2 x 20 x 6) + (20 x 15) = 360 cm^2
SA back = 360 cm^2
SA bottom = 30 x 20 = 600 cm^2
SA RHS = (30 x 15) + (11.66 x 30) used Pythagoras to find width of rectangle w^2 = 10^2 +6^2, w = 11.66 cm
= 799.86 cm^2
SA LHS = 799.86 cm^2

Total SA = 2919.7 cm^2 (to one decimal place)

The actual answer says it is 2819.2 cm^2 (to one decimal place)

Question 3

Find the surface area (to one decimal place)

View attachment 31972

SA RHS = 16.1 x 5.9 = 94.99 mm^2
SA LHS = 94.99 mm^2
SA front = 16.1 x 4 = 64.4 mm^2
SA back = 9.2 x 16.1 = 148.12 mm^2
SA top = 2.79/2 (4+9.2) = 18.414 mm^2 used Pythagoras to get perpendicular height of trapezium h^2 = 5.9^2 - 5.2^2, h = 2.79
SA bottom = 18.414 mm^2
Total SA = 439.3 mm^2 (to one decimal place)

The actual answer says it is 512.0 mm^2 (to one decimal place)


Question 4

Find the volume of the following truncated shape
View attachment 31973

Volume of large pyramid = 1/3 (4 x 4) x 15 = 80 cm^3

Volume of small pyramid = 1/3 (2 x 2 ) x 8 = 32/3 cm^3

Volume of truncated shape = 80 - 32/3 = 69.3 cm^3

The actual answer is 112 cm^3

Question 5

Find the area of the shaded area correct to 1 decimal place

View attachment 31974
Area of entire sector = 1/6 x pi x 14^2
= 98/3 pi cm^2

Area of small sector = 1/6 x pi x 3^2
= 3/2 pi cm^2

Area of shaded region = 98/3 pi - 3/2 pi = 97.9 cm^2 (to one decimal place)

The actual answer is 39. 3 cm^2

SA top

1/2 * 13 * 9 = 58.5

 

[CM_Tutor]

With question 1, the top triangle has sides of 9 cm, 13 cm, and cm, which is a Pythagorean triad, so its area is and the total surface area is 180 cm².

 

[Eagle Mum]

Q3
SA RHS = 16.1 x 5.9 = 94.99 mm^2
SA LHS does not = RHS
Use Pythagoras to determine length (L) of unknown side: L^2 = 5.9^2 + (9.2-4)^2
L = 7.864
SA LHS = 7.864 x 16.1 = 126.6
SA front = 16.1 x 4 = 64.4 mm^2
SA back = 9.2 x 16.1 = 148.12 mm^2
SA top = 5.9*(4+9.2)/2 = 38.94 Perpendicular height of trapezium = 5.9 since corner angles are right angles
SA bottom = 38.94 mm^2
Total SA = 511.99 or 512.0 (to 2 do)


Q4
Volume of large pyramid = 1/3 (4 x 4) x (15+8) = 122.67 cm^3
Volume of small pyramid = 1/3 (2 x 2 ) x 8 = 32/3 = 10.67 cm^3
Volume of truncated shape = 122.67 - 10.67 = 112 cm^3


Q5
Your answer is correct based on their diagram. You correctly interpreted that the radius of the small sector = R-r = 14-11= 3
They’ve arrived at their solution by taking the radius of the small sector as 11 which is not what their diagram shows.

 

[kpad5991]

Q3
SA RHS = 16.1 x 5.9 = 94.99 mm^2
SA LHS does not = RHS
Use Pythagoras to determine length (L) of unknown side: L^2 = 5.9^2 + (9.2-4)^2
L = 7.864
SA LHS = 7.864 x 16.1 = 126.6
SA front = 16.1 x 4 = 64.4 mm^2
SA back = 9.2 x 16.1 = 148.12 mm^2
SA top = 5.9*(4+9.2)/2 = 38.94 Perpendicular height of trapezium = 5.9 since corner angles are right angles
SA bottom = 38.94 mm^2
Total SA = 511.99 or 512.0 (to 2 do)


Q4
Volume of large pyramid = 1/3 (4 x 4) x (15+8) = 122.67 cm^3
Volume of small pyramid = 1/3 (2 x 2 ) x 8 = 32/3 = 10.67 cm^3
Volume of truncated shape = 122.67 - 10.67 = 112 cm^3


Q5
Your answer is correct based on their diagram. You correctly interpreted that the radius of the small sector = R-r = 14-11= 3
They’ve arrived at their solution by taking the radius of the small sector as 11 which is not what their diagram shows.

Thank you so so so much

If possible can you help with this question. At least the steps that I need to do. I can do the maths just not sure where to start.