I'm starting a new thread as the discussions will happen on this new thread for next 50 questions and the sectional test :-) Happy solving :smiley:

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15 + 11 + 21 + 16 - 3*9 = 36

We can get this using venn-diagram representation, where we can use four overlapping circles to represent "no hockey", "no cricket", "no volleyball" and "no football".

I think 36 is the right answer. So option (a).

Cheers Apple

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@varun_nakra1
720

Let's solve them on similar lines:(I)A figure is formed by joining the midpoints of a rectangle (whose length and width are a and b),along with 4 other triangles.Then again another figure is formed in the similar fashion along with 4 other triangles.This continued uptil infinity1)Find the sum of areas of all the figures(quadrilaterals) so formed2)Find the sum of areas of all the triangles so formed3)Find the sum of areas of all the figures and triangles so formed(II)Now this one is from Problem Book Of Mathematics,which I had in 9th standardEvery1 knows that the figure formed by joining the midpoints of any quadrilateral is a parallelogram.We need to prove that the area of that parallelogram is half the area of the original quadrilateral.

No answers yet???????

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@varun_nakra1
720

rani145541 SaysI am not sure about the sum of the area of the triangles. I am getting first right angled triangles then equilateral triangles and then again right angel triangles. is that rt?

Nopes!!the triangles are right angled in nature..dats fine but in the second iteration dey r nt equilateral..hw did u find dem to be equilateral??Its on the similar lines of today's question..try again..its fairly simple..

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me also getting (b). 14

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Question # 63The mid-points of triangle ABC are joined to form another triangle PQR along with three other triangles. Again the mid-points of the triangle PQR are joined to form another triangle along with three other triangles. This process is repeated infinite times. If the area of triangle ABC is 6 sq. cm then what is the sum of the area of all the triangles formed in the figure?(a) 12 sq. cm (b) 14 sq. cm (c) 18 sq. cm (d) none of the foregoing

14 sq. cm..option(b)

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@rani145541
316

I am not sure about the sum of the area of the triangles. I am getting first right angled triangles then equilateral triangles and then again right angel triangles. is that rt?

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i am getting 14 too

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@varun_nakra1
720

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Same method same answer i.e 14

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@amar_kashyap
444

Area of all the triangles is asked.

1st iteration = A(tri ABC)=6

2nd " =A(tri pqr)*4=6

3rd iteration =A(tri PQR/4)*4=6/4 then 6/4^2 , then 6/4^3 n so on till infinity.

So summation ll happen as 6 + 6+6/4+6/4^2 .... that comes to 6+6(1-{1-1/4}) =6 + 6*4/3 = 6+8=14 ,option b

My answer is option (b) 14 sq cm

Same here..I am also getting 14..

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The answer is option b i.e. 14

CHEERS

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Me getting 14. option (b)

6 + = 14.

Cheers Apple

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My answer is (b) as well.

Area = 6 + = 14

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Question # 63

The mid-points of triangle ABC are joined to form another triangle PQR along with three other triangles. Again the mid-points of the triangle PQR are joined to form another triangle along with three other triangles. This process is repeated infinite times. If the area of triangle ABC is 6 sq. cm then what is the sum of the area of all the triangles formed in the figure?

(a) 12 sq. cm (b) 14 sq. cm (c) 18 sq. cm (d) none of the foregoing

Area of all the triangles is asked.

1st iteration = A(tri ABC)=6

2nd " =A(tri pqr)*4=6

3rd iteration =A(tri PQR/4)*4=6/4 then 6/4^2 , then 6/4^3 n so on till infinity.

So summation ll happen as 6 + 6+6/4+6/4^2 .... that comes to 6+6(1-{1-1/4}) =6 + 6*4/3 = 6+8=14 ,option b

My answer is option (b) 14 sq cm

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