please answer my question..... no spam answers for getting points please..... please answer soon....23 24 26(last)(below or)
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23).
Given,
Parallel sides are 20 m and 10 m of trapezium ABCD respectively.
In order to find Area.
Construction :- Draw DM || CB.
Now, In figure there are two figures i.e. Isosceles triangle and a ||gm.
Now, In case of ||gm.
=> Ar(||gm MBCD) = Base x Height
=> Ar (||gm) = (10 x 13) m^2
=> Ar(||gm) = 130 m^2.
Again, In case of Triangle.
=> S = a+b+c/2 = 13+13+10/2 = 36/2 m = 18 m
Now,
Area of Trapezium = Ar(||gm)+ Area of Triangle
=> Area = (130 + 60) m^2 = 190 m^2
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24.)
Given,
=> Volume of Sphere = 1437-1/3 cm^3
=> 4/3πr^3 = 4312/3 cm^3
=> 4/3 x 22/7 x r^3 = 4312/3 cm^3
=> 88/21 x r^3 = 4312/3 cm^3
=> r^3 = 4312/3 x 21/88
=> r^3 = 343 cm^3
=> r = 7 cm
Again,
=> Surface area = 4πr^2
=> S.A = 4 x 22/7 x 7 x 7
=> S.A = 616 cm^2
For the Solution of question number 26 just refer to the attachment.
Given,
Parallel sides are 20 m and 10 m of trapezium ABCD respectively.
In order to find Area.
Construction :- Draw DM || CB.
Now, In figure there are two figures i.e. Isosceles triangle and a ||gm.
Now, In case of ||gm.
=> Ar(||gm MBCD) = Base x Height
=> Ar (||gm) = (10 x 13) m^2
=> Ar(||gm) = 130 m^2.
Again, In case of Triangle.
=> S = a+b+c/2 = 13+13+10/2 = 36/2 m = 18 m
Now,
Area of Trapezium = Ar(||gm)+ Area of Triangle
=> Area = (130 + 60) m^2 = 190 m^2
-----------------------------------------------------------------
24.)
Given,
=> Volume of Sphere = 1437-1/3 cm^3
=> 4/3πr^3 = 4312/3 cm^3
=> 4/3 x 22/7 x r^3 = 4312/3 cm^3
=> 88/21 x r^3 = 4312/3 cm^3
=> r^3 = 4312/3 x 21/88
=> r^3 = 343 cm^3
=> r = 7 cm
Again,
=> Surface area = 4πr^2
=> S.A = 4 x 22/7 x 7 x 7
=> S.A = 616 cm^2
For the Solution of question number 26 just refer to the attachment.
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aaravshrivastwa:
sukriya Bhai
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