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 By: apulita (offline)  Friday, August 03 2012 @ 10:54 PM CDT (Read 6128 times)
apulita

Newbie

Registered: 12/31/06
Posts: 2

 By: echeewh (offline)  Friday, August 24 2012 @ 04:55 PM CDT
echeewh

hey apulita,

Following pls find the worked solutions to Q7, 8 and 6. ( worked solution for Q4 to be provided later )
Notes: Q4 solution was posted on 26 Aug (see attached)

Q7

= (2 x Quadrant) - SemiCircle - Square

= [2 x (pi/4) x 14 x 14] - [(pi/2) x 7 x 7] - (14 x 14)

= 98 pi - (49 pi/2) - 196

= 35 cm2

==================

Q8

= (2 x Quadrant) - Square

= [2 x (pi/4) x 20 x 20] - (20 x 20)

= 200 pi - 400

= 228 cm2

====================

Q6

Observe that from centre O, there are 4 identical big quadrants of radius 15 cm each.

For each big quadrant, work out the perimeter and area of the unshaded parts.

Note that there are 2 identical quarter arcs (that form 2 identical small quadrants) in each big quadrant

P (of 1 unshaded part) = 15 + 2 quarter arcs of small quadrants

= 15 + [2 x (pi/4) x 15]

= 15 + (15 pi/2)

Total P (of 4 unshaded parts) = 4 x [15 + (15 pi/2)]

= 60 + 30 pi

= 154 cm

(note: enclosed area of 1 quarter arc of 1 small quadrant is Unshaded , while that of the other quarter arc is shaded. As such, we can cut & paste the enclosed shaded area of 1 of the quarter arcs to the identical enclosed unshaded area of the other quarter arc. So it will be left with an Unshaded square of side / radius 15/2 cm)

Area (of 1 Unshaded part) = (15/2) x (15/2)

= (225/4)

Area (of 4 unshaded parts) = 4 x (225/4)

= 225 cm2

==========================

Trust the above help.

Do let me know again should there be any clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: apulita (offline)  Friday, August 24 2012 @ 10:40 PM CDT
apulita

Thanks alot!!!

Newbie

Registered: 12/31/06
Posts: 2

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