amygdaleon305

Consider this labelled diagram

It's given that \(OC=32 \)  and \(CB=38\)

⇒\(OB = OC+CB\)

           \(=32+38\)

           \(=70 \)

Radius of semi-circle (3) is \(70\). 

And, CB is diameter of semi-circle (2).

So, radius of semi-circle is \(38/2=19\). 

From figure, we can infer that

Ar. (semi-circle 1) = Ar. (semi-circle 3) - Ar. (semi-circle 2) 

                               \(= {4900\pi \over 2} - {361\pi \over 2}\)

                               \(={4539\pi \over 2}\)

                              \(=7126.23\) sq. units 

Thus, the area of circle with centre K is \(7126.23\) sq. units. 

~Amy 

THANKS A LOT!! Melody and Ginger  These last two explanations actually helped me with the problem here 

Thank you so much Ginger and Melody for your explanations :) I understood the concept even though the question's a bit different from the one you solved. 

By Pythagoras' Theorem, we can find side AC

\(AC^2=AB^2+BC^2\)

⇒\(AC^2=25+64\)

\(AC^2=89\)

\(AC=\sqrt {89}\)

So now we know the side of square is \(\sqrt {89}\)

∴ Area  \(=(\sqrt {89})^2\)

            \(=89 sq. units\)

1. 16 times, here you go :- 

\((2x+{3\over 2}(4x-6))-4(-(2-x)+{x\over 2})\)

                       \(=(2x+6x-9)-4(-2+x+{x\over 2})\)

                       \(=8x-9+8-6x\)

                        \(=2x-1\)

~Amy 

Suppose she has \($x\) at first.

Amt. of money she gives away = \(${2x\over 3}\)

⇒ Amt. of money she keeps = \(${x\over 3}\)

 ∴ She keeps \($50\)  in her left pocket and   \($({x\over 3}-50)\)  in her right pocket.                          [it is not really relevant] 

Now, if she gives away \($200\), 

Amt. of money she keeps = \($(x-200)\) 

According to question, 

\(x-200>{x\over 3}\)

⇒\({x\over 3}

⇒\(x<3x-600\)                            

⇒\(2x<600\)

⇒\(x<300\)

Also, Jeri has atleast \($200\) but less than \($300\)

∴ The possible values of x are (200,300). 

P.S. Corrections and suggestions are welcome. 

~Amy 

No, its absolutely right.

• Look, we arrived at the solution that Hugo had \(88\) with him.  
• He gave 1/4 of his money & an additional 1 to Calvin i.e. \(23\) 
• Now he has \(88-23=65\) remaining with him. 
• Then he gave 1/5 of remaining and an additional 4 to Kenneth i.e \(17\)
• He has \(65-17=48\) remaining with him
• Now its given that "amount of money Hugo had left in the end was $31 more than the amount of money he gave to Kenneth"
• So, Hugo gave \(17\) to Kenneth and he has \(48\) remaining with him. 

I did reverse answer check here and nothing came out wrong

See? everything fits into the question now what's wrong?

Okay, what's wrong?

There's a slight correction in the given GP, shouldn't it be \(2+r+{r^2\over 2}+{r^3\over 4}+...?\)

As per the question, 

First term of the GP   \(a=2\)

and common ratio  \(={r\over 2}\)

We know that sum of infinite  GP 

             S_{∞ \(={a\over r-1}\)}

⇒ S_∞  \(={2\over {r\over 2}-1}\)

            \(={4 \over r-2}\)

Now, S_∞ is defined for \(-1 < r <1 \)

∴ All possible values of r are in the interval  \((-1,1)\). 

~Amy