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Secondary 1 | Maths
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pls help
Date Posted: 2 years ago
Views: 359
Area of each right angle
= "1/2 times base times height"
= 1/2 x 1.2 x 1.8
= 1.08 cm²
So, the two right angles have a total area of 2.16 cm².
The circle has a diameter of 0.9 cm, or equivalently a radius of 0.45 cm.
Area of that circle
= "πr²"
= π x 0.45 x 0.45
= 0.2025π cm²
Area of that surface of the pendant
= (1.08 - 0.2025π) cm²
We multiply this area by the thickness (similar to "height" in things like cubes and cuboids) to get our volume.
Volume of the pendant
= 0.2 x (1.08 - 0.2025π) cm³
Mass of this pendant
= 10.5 x 0.2 x (1.08 - 0.2025π) g
Price of this pendant
= $ [0.49 x 10.5 x 0.2 x (1.08 - 0.2025π)
= $0.4566984848
~ $0.46
= "1/2 times base times height"
= 1/2 x 1.2 x 1.8
= 1.08 cm²
So, the two right angles have a total area of 2.16 cm².
The circle has a diameter of 0.9 cm, or equivalently a radius of 0.45 cm.
Area of that circle
= "πr²"
= π x 0.45 x 0.45
= 0.2025π cm²
Area of that surface of the pendant
= (1.08 - 0.2025π) cm²
We multiply this area by the thickness (similar to "height" in things like cubes and cuboids) to get our volume.
Volume of the pendant
= 0.2 x (1.08 - 0.2025π) cm³
Mass of this pendant
= 10.5 x 0.2 x (1.08 - 0.2025π) g
Price of this pendant
= $ [0.49 x 10.5 x 0.2 x (1.08 - 0.2025π)
= $0.4566984848
~ $0.46
See 1 Answer
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Steps:
1) Find area of the ornament (2 triangle - 1 circle)
2) Find volume of the ornament (Area X thickness)
3) Find mass of silver in ornament (Volume X silver mass in 1 cubic cm)
4) Find value of silver (Mass of slver in ornament X cost of 1 g silver)
1) Find area of the ornament (2 triangle - 1 circle)
2) Find volume of the ornament (Area X thickness)
3) Find mass of silver in ornament (Volume X silver mass in 1 cubic cm)
4) Find value of silver (Mass of slver in ornament X cost of 1 g silver)
Date Posted:
2 years ago
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