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- Types of angles and triangles
- Interior and exterior angle theorem of a triangle.
- Congruency of triangles
- Similar triangles
- Perimeter and area of a triangle
- Pythagoras theorem
- Perimeter and area of polygons
- Circumference and area of a circle
- Cylinders, cones and spheres
- Equation of a straight line
Our proven 4-step learning approach enables students to understand mathematical concepts and apply these concepts to successfully solve practical problems. We equip you with tips for tricky questions, and teach shortcuts to improve testing speed and mental calculations.
Our examples below illustrate our comprehensive explanations with all steps. This means: Better understanding and Greater success in Geometry for YOU
| 1. |
The total surface area of a hollow cylinder, open at both ends, of external radius 8 feet and height 10 feet is 338 sq feet. Find the thickness of the metal in the cylinder.
Let R and r be respectively the external radius and the internal radius, and h be the height.
We have been given that R= 8 feet and h = 10 feet.
We know the formula to find the total surface area of a hollow cylinder
i.e.
| Total surface area = |
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Also given that the total surface area = 338 sq feet
Therefore,
So, the internal radius = 5 feet.
Therefore,
Thickness of metal in the cylinder = R - r
= 8 - 5
= 3 feet
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| 2. |
A rectangular tank is 225 m by 162 m at the base. With what speed must water flow into it through an aperture 60 cm by 45 cm so that the water level in the tank reaches 20 cm in five hours?
Since the level of water rises by 20 cm in 5 hours. Therefore, the volume of water that has to flow into the tank in 5 hours 
Therefore, the volume of water that will flow into the tank in one
| hour |
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 ...(1) |
| Area of cross-section of the aperture |
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Let the speed of the water be x meters per hour.
Then,
Volume of water that will flow into the tank in one hour = (Area of cross-section of the Aperture) x speed in meter per hour
= ....(2)
Then from (1) and (2), we get
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| 3. |
How many cubic meters of earth must be dug out to sink a well 22.5 m deep and of diameter 7 m? Also find the cost of plastering the inner curved surface at $10 per square meter.
We know that,
The volume of earth dugout = volume of the well
Area of the inner curved surface = 
Therefore,
The cost of plastering the inner curved surface = $
= $4950
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