Question

please answer my question..... no spam answers for getting points please..... please answer soon....23 24 26(last)(below or)

Answer 1

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 

$= > area \: = \sqrt{s(s - a)(s - b)(s - c)}$
$= > area \: = \sqrt{18(18 - 13)(18 - 13)(18 - 10)}$
$= > area \: = \sqrt{18 \times 5 \times 5 \times 8}$
$= > area \: = \sqrt{2 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2}$
$= > area \: = (2 \times 2 \times 3 \times 5) \: {cm}^{2}$
$= > area \: = 60 \: {cm}^{2}$

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

$= > \: r \: = \sqrt[3]{343}$
=> 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.