Area Problem - Find the Answer | Susanto

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In summary: However, they should not simply be summed.- You should have 2 triangles of 7x14 instead of 1.- There is only half of a circle, so its area should be divided by 2.- The half circle is something that has been removed from the object. So instead of adding it, it should be subtracted.Another way to approach this would be to find the area $A_T$ of the trapezoid (all areas in $\text{cm}^2$):A_T=\frac{h}{2}(B+b)Now find the area $A_S$ of the semi-circle:A_S=\
  • #1
susanto3311
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hi all...

View attachment 4099

please, see my picture, the area is...

a) 217 cm2
b) 434 cm2
c) 154 cm2
d) 294 cm2

thanks in advance...

susanto
 

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  • #2
susanto3311 said:
hi all...

please, see my picture, the area is...

a) 217 cm2
b) 434 cm2
c) 154 cm2
d) 294 cm2

thanks in advance...

susanto

Hey susanto3311! ;)

What would be the radius of the circle?
 
  • #3
I like Serena said:
Hey susanto3311! ;)

What would be the radius of the circle?

the radius is 28 - (2x7) = 14 cm..

it's true?
 
  • #4
susanto3311 said:
the radius is 28 - (2x7) = 14 cm..

it's true?

Almost. That's the diameter. The radius is half of that.

What would be the height? (Wondering)
 
  • #5
I like Serena said:
Almost. That's the diameter. The radius is half of that.

What would be the height? (Wondering)

hi Serena, i don't know?could you show me...
 
  • #6
susanto3311 said:
hi Serena, i don't know?could you show me...

The cross stripes seem to indicate a height of 2x7=14 cm.

Now which parts does the area consist of and what are their areas?
I think you should be able to find those by now...
 
  • #7
I like Serena said:
The cross stripes seem to indicate a height of 2x7=14 cm.

Now which parts does the area consist of and what are their areas?
I think you should be able to find those by now...

i still confuse...how do make more easy?
 
  • #8
susanto3311 said:
i still confuse...how do make more easy?

Please, could you give some indication of where you are stuck?
Or what you are thinking?
 
  • #9
I like Serena said:
Please, could you give some indication of where you are stuck?
Or what you are thinking?

hello Serena,

here my problem..
View attachment 4100
 

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  • #10
susanto3311 said:
hello Serena,

here my problem..

Good! (Smile)

You have already found that the diameter at the bottom is 14 cm.
So the distance at the top should also be 14 cm.

As for the diagonals of the triangles, you don't need them.
But if you want, you can calculate them using the Pythagorean Theorem $a^2+b^2=c^2$. Or more to the point: $c = \sqrt{a^2+b^2}$.

On the left you have a right triangle with base 7 and height 14.
Do you know how to calculate its area? (Wondering)
 
  • #11
I like Serena said:
Good! (Smile)

You have already found that the diameter at the bottom is 14 cm.
So the distance at the top should also be 14 cm.

As for the diagonals of the triangles, you don't need them.
But if you want, you can calculate them using the Pythagorean Theorem $a^2+b^2=c^2$. Or more to the point: $c = \sqrt{a^2+b^2}$.

On the left you have a right triangle with base 7 and height 14.
Do you know how to calculate its area? (Wondering)

hi Serena,

i think like this..

it's all true answer?
 

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  • #12
susanto3311 said:
hi Serena,

i think like this..

it's all true answer?

You have the right parts with the right areas. Good!

However, they should not simply be summed.
- You should have 2 triangles of 7x14 instead of 1.
- There is only half of a circle, so its area should be divided by 2.
- The half circle is something that has been removed from the object. So instead of adding it, it should be subtracted.
(Thinking)
 
  • #13
Another way to approach this would be to find the area $A_T$ of the trapezoid (all areas in $\text{cm}^2$):

\(\displaystyle A_T=\frac{h}{2}(B+b)\)

Now find the area $A_S$ of the semi-circle:

\(\displaystyle A_S=\frac{\pi}{8}d^2\)

And so the area $A$ of the given figure is:

\(\displaystyle A=A_T-A_S=\frac{h}{2}(B+b)-\frac{\pi}{8}d^2\)

You know all of the following:

$h$ = height of trapezoid

$B$ = "big" base of trapezoid

$b$ = "little" base of trapezoid

$d$ = diameter of semi-circle

So now just plug the numbers in, and round to the nearest integer. What do you find?
 
  • #14
maybe, i have try..
 

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  • area final_xxx.png
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  • #15
susanto3311 said:
maybe, i have try..

Yep! That's it! (Happy)

One improvement: for area 1 you should have {(7x14)/2}x2 = 49x2 = 98, but that's what you used anyway.
 
  • #16
hi..

why different result with 2 method :

like this ..

where is something wrong?
 

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  • area final yyy.png
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  • #17
susanto3311 said:
hi..

why different result with 2 method :

like this ..

where is something wrong?

The symbols are mixed up.
The diameter d should be 14 cm.
The "big" base B should be 28 cm.
And the "little" base b should be 14 cm.
 

FAQ: Area Problem - Find the Answer | Susanto

What is the "Area Problem" in mathematics?

The "Area Problem" is a mathematical concept that involves finding the amount of space inside a two-dimensional shape. This can be done by calculating the length and width of the shape and multiplying them together, or by using more complex formulas for irregular shapes.

How do you solve the "Area Problem"?

The "Area Problem" can be solved by using different methods depending on the shape. For squares and rectangles, the area can be found by multiplying the length by the width. For triangles, the area can be found by multiplying the base by the height and dividing by 2. For circles, the area can be found by using the formula A = πr^2, where r is the radius of the circle.

What is the importance of understanding the "Area Problem" in real life?

The concept of finding the area of a shape is important in many real-life situations, such as calculating the amount of material needed for a construction project, determining the size of a piece of land, or measuring the capacity of a container. It is also used in various fields of science, such as physics and engineering.

Can the "Area Problem" be solved for three-dimensional shapes as well?

Yes, the "Area Problem" can also be extended to three-dimensional shapes. In this case, it is called the "Volume Problem" and involves finding the amount of space inside a three-dimensional object. The formulas for calculating volume are similar to those for finding area, but they also take into account the height or depth of the shape.

Are there any real-life applications of the "Area Problem" outside of mathematics?

Yes, the concept of finding the area of a shape has various real-life applications outside of mathematics. For example, in art and design, understanding the concept of area is important for creating visually appealing compositions. In geography, it is used to measure the size of countries and continents. In computer science, it is used for image processing and pattern recognition.

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