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FIGURE IN profile

In the fig, OABC is a square. If the area of circle O is 2pi, what is the area of the shaded region?

A- pi/2-1
B- 2-pi/2
c- pi-2
d- pi-1/2
e- 2pi-1/4

My question is what not get area of circle subtract from area of sq, i did that and it didnt work but the book gives another solution, bu ti still dunt get why i cant do it the way i want it. Sum1 explain plz?

By Fairyofwind (Fairyofwind) on Wednesday, August 27, 2003 - 01:26 pm: Edit

If the area of the circle is 2*Pi. Pi*r^2=2*Pi, r^2=2. Area of OABC=r^2=2. Area of quarter-circle=(1/4)*(2*Pi)=Pi/2. Thus, the area of the shaded region is 2-(Pi/2). B.

By Xiggi (Xiggi) on Wednesday, August 27, 2003 - 01:38 pm: Edit

Aka~

Did you remember to subtract ONLY 1/4 of the circle? Incidently, B is the only answer that includes a 2 minus something. You could "almost" stop as soon as you identify the area of the square OABC. However, it does not take too long to figure the area of 1/4 of the circle.

By Akaflex (Akaflex) on Wednesday, August 27, 2003 - 01:42 pm: Edit

heres another question + pic is in profile gotta rd tis explanation thanks guys, also heres the answer choices for the nxt question that confused me

A- 24-4pi
B- 24-4radical 2+4pi
C- 16-4radical 2+4pi
D- 16+4pi
E-16radical 2 + 8 pi

By Akaflex (Akaflex) on Wednesday, August 27, 2003 - 01:45 pm: Edit

oh, thats my mistake, i thought by taking out the whole circle ur left with a shaded area, but wouldnt that make sense. If i grab that whole circle out of it wouldnt it give me the area?

By Serene (Serene) on Wednesday, August 27, 2003 - 02:30 pm: Edit

radius = OB = AC = 8. try it from there. and note the rectangle is a square since the triangle is isoceles.

By Xiggi (Xiggi) on Wednesday, August 27, 2003 - 02:53 pm: Edit

Aka~

Regarding the first question, look at the area of the square. Only 1/4 of the circle is inscribed in it. The shaded area is the square minus the area of the circle that is IN the same square.

For the second question, start by calculating the length of the arc PBQ, you will need that. Use the radius size given by Serene. Then try to evaluate the PA, BC and AB, CQ in terms of the radius. Hint: What is the size OP and OQ?

By Akaflex (Akaflex) on Wednesday, August 27, 2003 - 03:24 pm: Edit

yah i realized that, but first time id id it i didnt get why uc ant juss subtract, the perimeter of POQ - the perimeter of rectangle?

By Serene (Serene) on Wednesday, August 27, 2003 - 03:33 pm: Edit

because perimeter of POQ - perimeter of OABC is


(PA+AO+OC+CQ+PQ)-(AO+AB+BC+OC)
= (PA+AO+OC+CQ+PQ)-(AO+OC+AO+OC)
= PA-AO-OC+CQ+PQ... which is really nothing.

instead you're trying to get PA+AO+OC+CQ+PQ

By Xiggi (Xiggi) on Wednesday, August 27, 2003 - 04:45 pm: Edit

Aka~

What you did would apply to areas, for instance the area of PAB+BCQ would the area of POQ - AOBC. But here, we are dealing with the size of the segments.

If you can calculate the perimeter of POQ, you are done

By Serene (Serene) on Wednesday, August 27, 2003 - 05:40 pm: Edit

yup, since the answer IS equal to perimeter of POQ.