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Mathematics: Post your doubts here!

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(a)
(i)
the sum og interior angles of any polygon is "180.(n-2)" where n is the total number of sides of a polygon.
so here the sum of interior angles will be, 180(5-2) = 540
there are total 5 interior angles of a pentagon, and one of them id BCD.
so BCD = 540/5 = 108
(ii)
as the pentagon is a regular one (i.e. all sides are equal) the triangle BCD is an isosceles triangle because BC = CD.
so, angle CBD = angle CDB
2.CBD + BCD = 180
2.CBD + 108 = 180
2.CBD = 180 - 108
CBD = 72/2 = 36
(iii)
radius of the circle = OA = OB
so triangle OAB is also an isosceles triangle
108/2 = OAB = OBA = 54
got it?
(iv)
the angle b/w radius (OA in this case) and tangent (WA in this case) is 90.
OAB + WAB = 90
54 + WAB = 90
WAB = 36.
(v)
two intersecting tangents are always equal, means WA = WB
so triangle WAB is isisceles,
2.WAB + AWB = 180
AWB = 108

(b)
sum of int. angles of hexagon = 180 - (6-2) = 720
----> the ratio and proportion method.
smallest sized angle = [3/(3+4+4+4+4+5)] * 720 = 90.
 
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(a)
(i)
the sum og interior angles of any polygon is "180.(n-2)" where n is the total number of sides of a polygon.
so here the sum of interior angles will be, 180(5-2) = 540
there are total 5 interior angles of a pentagon, and one of them id BCD.
so BCD = 540/5 = 108
(ii)
as the pentagon is a regular one (i.e. all sides are equal) the triangle BCD is an isosceles triangle because BC = CD.
so, angle CBD = angle CDB
2.CBD + BCD = 180
2.CBD + 108 = 180
2.CBD = 180 - 108
CBD = 72/2 = 36
(iii)
radius of the circle = OA = OB
so triangle OAB is also an isosceles triangle
108/2 = OAB = OBA = 54
got it?
(iv)
the angle b/w radius (OA in this case) and tangent (WA in this case) is 90.
OAB + WAB = 90
54 + WAB = 90
WAB = 36.
(v)
two intersecting tangents are always equal, means WA = WB
so triangle WAB is isisceles,
2.WAB + AWB = 180
AWB = 108

(b)
sum of int. angles of hexagon = 180 - (6-2) = 720
----> the ratio and proportion method.
smallest sized angle = [3/(3+4+4+4+4+5)] * 720 = 90.
Thanks alot! <3 will try to figure it now! :D
 
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AsSalamoAlaikum Wr Wb...

Stuck somewhere in Maths?? Post your queries here! Members around will help you InshaAllah.

NOTE: If you have any doubt in the pastpper questions, then kindly post the link to the paper!

May Allah give us all success in this world as well as the HereAfter...Aameen!!

SEQUENCES-Points to remember:
Many people find it hard, but to be honest it's just more of logic..that's all!


Sequences!

Using log for indices.
Hi... I would like to get more details about the newly added topics in IGCSE Mathematics (0580) for the year 2015. If anyone knows, please let me know more about it and is there any resources available?
 
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Hi... I would like to get more details about the newly added topics in IGCSE Mathematics (0580) for the year 2015. If anyone knows, please let me know more about it and is there any resources available?
 
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I'm an IGCSE student and I was solving a past year paper and came across a cylinder volume similarity question. I tried a lot but couldn't solve it... Here Is the question -->
There is a large mug in the shape of a cylinder, open at the top.
The internal radius of the mug is 8 cm and the internal height is 12 cm.

The mug shown in the diagram is mathematically similar to a smaller mug.
The volume of the smaller mug is 1/8 of the volume of the larger one.
Find the radius of the smaller mug.

Now this is the only part of the question where i have a difficulty...
however this is the original paper... I have a problem in Q.9 of this paper... Thank you so much!!

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/International Mathematics (0607)/0607_s12_qp_42.pdf
 
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plz have an exam on sunday need ans fast
z plz tell me how to do Oct nov variant 21 2011 0580 Question 4 and specially question 15 i have tried area under graph it doesnt work my answer come 157 while the answer is 156 plz write the method with done how to do it

link plz?
 
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I'm an IGCSE student and I was solving a past year paper and came across a cylinder volume similarity question. I tried a lot but couldn't solve it... Here Is the question -->
There is a large mug in the shape of a cylinder, open at the top.
The internal radius of the mug is 8 cm and the internal height is 12 cm.

The mug shown in the diagram is mathematically similar to a smaller mug.
The volume of the smaller mug is 1/8 of the volume of the larger one.
Find the radius of the smaller mug.

Now this is the only part of the question where i have a difficulty...
however this is the original paper... I have a problem in Q.9 of this paper... Thank you so much!!

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/International Mathematics (0607)/0607_s12_qp_42.pdf
linear ratio of the two mugs = x -------> (we dont know)
Area ratio or mugs = x^2
Volumetric ratio of mugs = x^3 = 1/8
x = cube root of 1/8 = 1/2

radius (a linear measurement) of the small mug = 1/2 * radius of larger one = 1/2 * 8 = 4 cm

hope you got it.
 
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