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primary 6 | Maths
| Data Analysis
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Circles. Please help me
Date Posted: 5 years ago
Views: 495
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You need to know the method to find the area of a sector . Wish it helps .
Date Posted:
5 years ago
The area of the equilateral triangle is incorrect. It's height isn't 9cm. If you correctly used Pythagoras' theorem you will get 7.79cm (3s.f) instead.
The final answer ended up being correct because the total area calculated equalled the area of two equivalent sectors, and the area of 1 sector was correctly calculated in your first step.
The final answer ended up being correct because the total area calculated equalled the area of two equivalent sectors, and the area of 1 sector was correctly calculated in your first step.
Alternative method :
Shift the bottom curved segment and join it to the top of the left equilateral triangle.
Since all angles in an equilateral triangle are 60° each, the two combined angles of the two triangles (from the bottom left side of the left triangle to the bottom right side of the right triangle is 120°)
A full circle is considered to have an angle of 360°.
Since 120° ÷ 360° = ⅓, the shaded area is ⅓ of a full circle. So its area is also ⅓ of a full circle.
⅓ x 3.14 x 9 x 9 = 84.78
Shift the bottom curved segment and join it to the top of the left equilateral triangle.
Since all angles in an equilateral triangle are 60° each, the two combined angles of the two triangles (from the bottom left side of the left triangle to the bottom right side of the right triangle is 120°)
A full circle is considered to have an angle of 360°.
Since 120° ÷ 360° = ⅓, the shaded area is ⅓ of a full circle. So its area is also ⅓ of a full circle.
⅓ x 3.14 x 9 x 9 = 84.78
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Here
Date Posted:
5 years ago
Can you see another sector below the first sector? The 9 cm line? Imagine it moved up and it will match the triangle . So both triangle's area is 40.5 .
You might want clarify which part you are shifting up. Because the area of the equilateral triangle isn't 40.5 no matter what you shift
Another way of calculating the area of the triangle without using Pythagoras' Theorem is ½ab sin c , where
a and b are the sides of the triangle, 9cm each.
c is the angle subtended by these two sides. It's 60° since it's an angle of an equilateral triangle.
½ x 9 x 9 sin60°
= ½ x 81 x √3/2
= 81/4 x √3
= 35.07 (4s.f)
Another way to see why it isn't 40.5 :
Take the line joining the two equilateral triangles as the base of the right equilateral triangle.
Its height would then be the horizontal line drawn from the middle of that line to the rightmost vertex.
Now this line is parallel to the horizontal sides of the square. However, it is shorter than them as shown in the diagram. Which means it is shorter than 9cm.
So the height of the triangle cannot be 9cm. And thus its area cannot be ½ x 9 x 9
and does not equal 40.5
Another way of calculating the area of the triangle without using Pythagoras' Theorem is ½ab sin c , where
a and b are the sides of the triangle, 9cm each.
c is the angle subtended by these two sides. It's 60° since it's an angle of an equilateral triangle.
½ x 9 x 9 sin60°
= ½ x 81 x √3/2
= 81/4 x √3
= 35.07 (4s.f)
Another way to see why it isn't 40.5 :
Take the line joining the two equilateral triangles as the base of the right equilateral triangle.
Its height would then be the horizontal line drawn from the middle of that line to the rightmost vertex.
Now this line is parallel to the horizontal sides of the square. However, it is shorter than them as shown in the diagram. Which means it is shorter than 9cm.
So the height of the triangle cannot be 9cm. And thus its area cannot be ½ x 9 x 9
and does not equal 40.5
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The length of the square is 9cm so you can move the line up . That's why i assumed that the length of the triangle is 9cm .
Date Posted:
5 years ago
Yes, that's one way to get one side of the triangle, since the side of one square and the side of the triangle is equal to the radius of the quarter circle.
It is actually quite intuitive if you recognise that the square and the equilateral triangle on the right share a common side.
Since the square's sides are 9cm long, the triangle also has its sides as 9cm since in an equilateral triangle, all sides are the same length.
But this does not mean the triangle's height is 9cm also
It is actually quite intuitive if you recognise that the square and the equilateral triangle on the right share a common side.
Since the square's sides are 9cm long, the triangle also has its sides as 9cm since in an equilateral triangle, all sides are the same length.
But this does not mean the triangle's height is 9cm also
Bro, mathematics has a theory of achieving to a same target by different ways or methods.
Yes, but in your case getting 40.5 for the area of the triangle is conceptually incorrect because the height of the triangle is not 9cm. You have to ensure all your steps are mathematically and conceptually correct.
Even though you got the right final answer, it will be marked down and commented as a fluke if the working you provided was used in any exam.
Even though you got the right final answer, it will be marked down and commented as a fluke if the working you provided was used in any exam.
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