Problem Solving

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yup!!

i too got the same answer for both the problem..
1st problem is a good one..

Ans

problem 1 : 2/3
Problem 2 : 10 pi

Thanks for posting the problems

I want to know how to solve them

Thanks

1) for the square problem

let the length of square M be m

let the length of square M be n

we know that m + n = 2 (Can be inferred from the diagram)

m^2 = 4n^2 (given in the Question )

taking squareroot on both sides

squareroot (m^2) = squareroot (4n^2)

==> m = 2n

substitude the value of m in m + n = 2
2n + n =2
3n = 2

Therefore, n = 2/3



2) Need to find the lenght of the arc

===> 30 /360 *(2*pi* r)
30/360 * 2*pi*60
1/12 * 2*pi*60
2*pi*5 = 10 pi

The Arc problem

The arc is a segment of the circle with radius r=60 cm

In order to find the length of the arc, you first need to figure out the perimeter of the circle L; L=2*r*pi=2 * 60 pi= 120 pi

Because the angle of the arc is 30 degrees, and a circle has 360 degrees in total --> the arc is 30/360= 1/12 of the entire circle

therefore, the length of the arc is 1/12*120 pi= 10 pi

N:Dure wrote:??

Area M^2=area N^2 or M^2=4N^2 or M=2N
so 2=M+N=3N so N=2/3
the answer is D

for the second question....
we will solve it by the fomular : 30*2pr/360= 30*2*60*p/360=10p
read the OG11, review maths section... it said about this