Maths P4 doubt

[yathubhai]

Plzz cud sum1 tell me how 2 solve Q9, pg 8 in jun.03 ppr4?!?!(both parts)
the direct link to the paper is http://www.freeexampapers.us/IGCSE/Maths/CIE/2003%20Jun/0580_s03_qp_4.pdf

plzz help!!

[eightAs]

(a) For the triangle
Area shaded = 1*h*.5 Total Area=4*h*.5
%age = 25% (after manipulations)
second figure Area shaded = 0.5*1*h
Total area = 4*h
%age = 12.5%
Third figure: one of the diagonals bisects the other in a kite. Look at the figure and you'll know which one. Let the equally bisected sides be called h
Shaded area=0.5*3*h
Total area = 2(0.5*4*h)
%age u do it

[yathubhai]

wts d area of a kite?

[yathubhai]

wts d area of a kite?

is it d1*d2, where d1 and d2 are diagonals??

[holychalice]

i think its 0.5 x diagonal 1 x diagonal 2

[eightAs]

ABE ~ GFE ~ CDE
Let the area of ABE be k
GFE will also be k since the question says both are congruent.
Look at the area of CDE
Ar.(CDE):Ar.(GFE) = (2/1)^2 = 4
hence ar(CDE)=4k
Shaded area = 4k-k = 3k
Total area = CDE+ABE = 5k
%age = 60%/

[yathubhai]

thankz loaddzzz eightAs!!
i understud it!!

[@d!_†oX!©]

i think its 0.5 x diagonal 1 x diagonal 2

ya ur right!!

[@d!_†oX!©]

hey yathubhai!!!
check this out...these are the model answers to the same question...
shud be helpful!!!
http://www.freeexampapers.us/IGCSE/Maths/CIE/2003%20Jun/IGCSE%20Maths%20Paper%204%20model%20answers.pdf

[@d!_†oX!©]

hey did u say both parts???
i just read it on ur first post...
so here u go!!
Diagram 5 :-
shaded area = (pi r²20)/360
whole area = (pi r²180)/360
thus shaded area/whole area gives 20/180 => 1/9

Diagram 6 :-
shaded area = pi 1²
whole area = pi 5²
thus, shaded area/whole area gives 1²/5² => 1/25

Diagram 7 :-
shaded area = [(pi 3²theta)/360] - [(pi 2²theta)/360]
shaded area => (5 pi theta)/360
whole area = (pi 3²theta)/360
thus, shaded area/whole area gives 5/9


 

[yathubhai]

thanks adi...ab poora samajh aa gaya...

[@d!_†oX!©]

ur welcome!!