Question

solve 4 question number and explain it .​

Answer 1

$\setlength{\unitlength}{4mm}\begin{picture}(10,6)\thicklines\put(0,1){\line(0,1){10}}\put(0,1){\line(1,0){10}}\put(10,1){\line(0,1){10}}\put(0,11){\line(1,0){10}}\put(0,11){\line(1,1){5}}\put(10,11){\line(1,1){5}}\put(10,1){\line(1,1){5}}\put(0,1){\line(1,1){5}}\put(5,6){\line(1,0){10}}\put(5,6){\line(0,1){10}}\put(5,16){\line(1,0){10}}\put(15,6){\line(0,1){10}}\put(4.6,-0.5){\bf\large a cm}\put(13.5,3){\bf\large a cm}\put(-4,5.8){\bf\large a cm}\end{picture}$

Let us assume that the side of the cube is a cm .

TSA :

> 6 a² .

The cost @10 paisa / cm² = Rs. 264.60 = 26460 paisa

> TSA = 2646 cm² .

> 6 a² = 2646

> a² = 441

> a = 21 cm .

Volume of cube = a³ = 9261 cm³ .

Hence , option B is correct .

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Answer 2

Answer:

Required Answer :-

Given :-

• Cost of painting a SA of cube = ₹264.60
• Cost of 1 cm = 10 paise

To Find :-

Let the edge of cube be a cm

As we know that

TSA of cube = 6a² cubic units

Now,

₹ 264.60 = 264.60 × 100 = 26460 paise

TSA = 26460/10

TSA = 2646 cm³

Now,

=> 6a² = 2646

=> a² = 2646/6

=> a² = 441

=> a = √441

=> a = 21 cm

Now,

Volume = a³ = 21³ = 9261 cm³

Hence :-

Option B is correct

$\rule {333}{5}$

Know More :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = (4/3)πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²