Area, SMB-001
Area, SMB-008 MC
Area, SMB-006
A child's toy has the following design.
Find the area of the shaded region to the nearest square centimetre. (3 marks)
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A target is drawn as follows
What is the area of the entire target?
Round your answer to the nearest square centimetre. (2 marks)
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`284\ text(cm)^2`
`text(Area)` | `= pi r^2` |
`= pi (2 + 1.5 xx 5)^2` | |
`= pi (9.5)^2` | |
`= 283.52…` | |
`= 284\ text(cm)^2` |
Find the area of the shaded part in the diagram below, giving your answer in square metres to 1 decimal place. (3 marks)
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\(62.8\ \text{m}^2 \)
\(\text{Shaded area}\) | \(=\ \text{Area of large sector – Area of small sector}\) | |
\(= \dfrac{30}{360} \times \pi R^2-\dfrac{30}{360} \times \pi r^2 \) | ||
\(= \dfrac{1}{12} \pi (16^2-4^2) \) | ||
\(= 62.83… \) | ||
\(=62.8\ \text{m}^2\ \ \text{(1 d.p.)} \) |
Four identical circles of radius `r` are drawn inside a square, as shown in the diagram below
The region enclosed by the circles has been shaded in the diagram.
The shaded area can be found using
`B`
`text(Area of square) = 4r xx 4r = 16r^2`
`text(Area of circles) = 4 pi r^2`
`text(Shaded Area)`
`= 1/4 xx (16r^2-4 pi r^2)`
`= 4 r^2-pi r^2`
`=> B`
A child's toy has the following design.
Find the area of the shaded region to the nearest square centimetre. (3 marks)
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`27\ text(cm²)`
`text{Circle radius = 3 cm}`
`text(Consider the rectangle starting from the middle of the left circle.)`
`text{Shaded Area}` | `= text{Area rectangle}-3.5 xx text{Area circle}` |
`= (21 xx 6)-3.5 xx pi xx 3^2` | |
`= 27.03 …` | |
`=27\ text{cm² (nearest whole)}` |
A path 1.5 metres wide surrounds a circular lawn of radius 3 metres.
Find the area of the path, correct to the nearest square metre. (2 marks)
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`35\ text{m²}`
`text(Area of annulus)`
`= pi (R^2-r^2)`
`= pi (4.5^2-3^2)`
`= pi (11.25)`
`=35.3…`
`=35\ text{m² (nearest m²)}`
The diagram shows the floor of a shower. The drain in the floor is a circle with a diameter of 10 cm.
Find the area of the shower floor, excluding the drain, to the nearest square centimetre. (3 marks)
`9921\ text(cm²)`
`text(Area)` | `=\ text(Square – Circle)` |
`= (100 xx 100)-(pi xx 5^2)` | |
`= 10\ 000-78.5398…` | |
`= 9921.46…\ text(cm²)` | |
`= 9921\ text{cm² (nearest cm²)}` |
The shaded region shows a quadrant with a rectangle removed.
Find the area of the shaded region, to the nearest cm2. (2 marks)
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`52\ text(cm²)`
`text(Shaded area)` | `=\ text(Area of segment – Area of rectangle)` |
`=1/4 pi r^2-(6xx2)` | |
`=1/4 pi xx9^2-12` | |
`=51.617…` | |
`=52\ text(cm²)` |
`42\ text{cm²}`
`text(Area)` | `=\ text(Area of Sector – Area of triangle)` |
`= (theta/360 xx pi r^2)-(1/2 l xx h)` | |
`= (90/360 xx pi xx 8^2)-(1/2 xx 4 xx 4)` | |
`= 50.2654…-8` | |
`= 42.265…\ text(cm²)` | |
`= 42\ text{cm² (nearest cm²)}` |
In the circle, centre `O`, the area of the quadrant is 100 cm².
What is the arc length `l`, correct to one decimal place?
`C`
`text(Find)\ r:`
`text(Area)` | `= 1/4 pir^2` |
`100` | `= 1/4 pir^2` |
`r^2` | `= 400/pi` |
`:. r` | `= 11.283…\ text(cm)` |
`text(Arc length)` | `= theta/360 xx 2pir` |
`= 90/360 xx 2pi xx 11.283…` | |
`= 17.724…` | |
`= 17.7\ text(cm)` |
`=> C`
A logo is designed using half of an annulus.
What is the area of the logo, to the nearest cm²?
`B`
`text(Area)` | `=1/2xxpi(R^2-r^2)` |
`=1/2xxpi(5^2-2^2)` | |
`=21/2pi` | |
`=33\ text(cm²)\ \ \ text{(nearest cm²)}` |
`=>\ B`