Subject :IMO Class : Class 7
Class : Class 8
it should be 20 because the formula for number of diagonals from a vertices is n-3, so for one vertice is 10 diagonals so two is gonna be 20
Class : Class 9
ANSWER MUST BE 20 SINCE FORMULA IS N-3 SO FOR ONE VERTICE IT IS 10 SO TWO WILL BE 20
Subject :IMO Class : Class 4
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Subject :IMO Class : Class 4
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Subject :IMO Class : Class 5
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Subject :IMO Class : Class 8
Fill in the blanks.
(i) The perimeter of circle and its diameter vary __P__ with each other.
(ii) If two quantities p and q vary inversely with each other then __Q__ of their corresponding values remain constant.
(iii) When x and y are in indirect proportion and if y doubles then x becomes __R__
P | Q | R | |
A | Inversely | Ratio | Double |
B | Directly | Product | Half |
C | Inversely | Ratio | Double |
D | Directly | Product | Half |
A and c , b and d are same.
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Subject :IMO Class : Class 7
Ans 2:
Class : Class 7
it is very simple and can be solved with Pythagoras.Length=2y Breadth=yroot of [(2y)square ysquare]=9 x root of 5root of [4ysquare ysquare]=9 x root of 5root of 5ysquare=9 x root of 5 = y x root of 5thus y = 9 and 2y = 18p=54
Ans 5:
Class : Class 7
First of all,diagonal of a rectanhle=square root of l^2 b^2,so 9×square root of 5 is square root of 81 × 5 that is 405 so here 2x is l and x is b.if i m taking l is 18 and b is 9 and squaring them we get 405 so perimeter of the rectangle is 2(18 9) that is 54.
Ans 6:
Class : Class 7
w=xl=2xd=9root52(l b)=?2(x 2x)=?using pythogras theorem,xsquare 4 x square=(9root5)squared5xsquare=405x square=81x=9hence substitute values with x,perimeter=54
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Subject :IMO Class : Class 5
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Subject :IMO Class : Class 6
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Subject :IMO Class : Class 4