# Help in my O Levels~

MHB Math Helper

#### leylamew

##### New member
You are supposed to at least show some effort and tell us where are you stuck ?
I'd upload the sheet where I've scribbled the logic, but I don;t have a scanner.
Here's my thinking process:
OB=OA=OP=8 cm and as OQ is x cm, MN should be...something. But I can't understand for the life of me, how to find MN. I don't understand what I'm supposed to do with the x there.

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I'd upload the sheet where I've scribbled the logic, but I don;t have a scanner.
Here's my thinking process:
OB=OA=OP=8 cm and as OQ is x cm, MN should be...something. But I can't understand for the life of me, how to find MN. I don't understand what I'm supposed to do with the x there.
And, I don't understand what the questions are asking of me in the others.

#### leylamew

##### New member
I'd upload the sheet where I've scribbled the logic, but I don;t have a scanner.
Here's my thinking process:
OB=OA=OP=8 cm and as OQ is x cm, MN should be...something. But I can't understand for the life of me, how to find MN. I don't understand what I'm supposed to do with the x there.

- - - Updated - - -

And, I don't understand what the questions are asking of me in the others.
All right, so I put my ruler to use and hoping that the diagram was to scale, I measured MQ and OQ. They're the same, so I safely assumed that MQ is x and so, therefore, MN is 2x, as MQ is 1/2MN. Did I do it right?

#### MarkFL

Staff member
All right, so I put my ruler to use and hoping that the diagram was to scale, I measured MQ and OQ. They're the same, so I safely assumed that MQ is x and so, therefore, MN is 2x, as MQ is 1/2MN. Did I do it right?
Yes, as triangles $MOQ$ and $NOQ$ are congruent, and are right isosceles triangles, we know that:

$$\displaystyle \overline{MQ}=x$$

$$\displaystyle \overline{QN}=x$$

$$\displaystyle \overline{MQ}+\overline{QN}=2x$$

And we know that $$\displaystyle \overline{MQ}+\overline{QN}=\overline{MN}$$, hence:

$$\displaystyle \overline{MN}=2x$$

Now, for the next part of the problem, find the total area of the cross-section using the formula for the area of a circular sector (or simply from the fact that it is a quarter circle) and then subtract away the area of triangle $MON$. What do you find?

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#### topsquark

##### Well-known member
MHB Math Helper
$$\displaystyle \bar{MQ}=x$$
@MarkFL: Just a comment about the LaTeX coding: \overline{AB} gives $$\overline{AB}$$. I think that looks like a better operation here than \bar.

-Dan

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#### MarkFL

Staff member
Thank you, that does look better!

#### MarkFL

Staff member
...
Now, for the next part of the problem, find the total area of the cross-section using the formula for the area of a circular sector (or simply from the fact that it is a quarter circle) and then subtract away the area of triangle $MON$. What do you find?
Hints:

The area $A_S$ of a circular sector is:

$$\displaystyle A_S=\frac{1}{2}r^2\theta$$

and you know $r$ and $\theta$.

The area $A_T$ of a triangle is:

$$\displaystyle A_T=\frac{1}{2}bh$$

and you know the base $b$ and the height $h$ in terms of $x$. So you now need to compute:

$$\displaystyle A=A_S-A_T=?$$