Question

solve this question no spam .​

Answer 1

Answer :- (i)

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf Let -  \begin{cases} &\sf{length \: of \: cuboid \: = 4x} \\ &\sf{breadth \: of \: cuboid \: = 3x}\\ &\sf{height \: of \: cuboid \: = 2x} \end{cases}\end{gathered}\end{gathered}$

Surface Area = 2548

We know,

${{ \boxed{{\bold\green{Total \: Surface \: Area_{(Cuboid)}\: = 2(lb + bh + hl)}}}}}$

$\sf \: 2548 = 2((4x)(3x) + (3x)(2x) + (2x)(4x))$

$:  \implies \sf \: 2548 = 2(12 {x}^{2} + {6x}^{2} + {8x}^{2} )$

$:  \implies \sf \: 2548 = 2(26 {x}^{2} )$

$:  \implies \sf \: 2548 = {56x}^{2}$

$:  \implies \sf \: {x}^{2} = \dfrac{2548}{56}$

$:  \implies \bf \: {x}^{2} = 49$

$:  \implies \bf \: x = 7$

$\begin{gathered}\begin{gathered}\bf \: Hence -  \begin{cases} &\sf{length \: of \: cuboid \: = 4 \times 7 = 28 \: cm} \\ &\sf{breadth \: of \: cuboid \: = 3 \times 7 = 21 \: cm}\\ &\sf{height \: of \: cuboid \: = 2 \times 7 = 14 \: cm} \end{cases}\end{gathered}\end{gathered}$

$\boxed{ \red{ \bf \: Volume_{(cuboid)} =l \times b \times h}}$

$:  \implies \tt \: Volume_{(cuboid)} = \: 28 \times 21 \times 14$

$:  \implies \boxed{ \pink{ \bf \: Volume_{(cuboid)} = \: 8232 \: {cm}^{3} }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

Answer :- (ii)

$\begin{gathered}\begin{gathered}\bf Let -  \begin{cases} &\sf{length \: of \: cuboid \: = 4x} \\ &\sf{breadth \: of \: cuboid \: = 3x}\\ &\sf{height \: of \: cuboid \: = 2x} \end{cases}\end{gathered}\end{gathered}$

Volume of Cuboid = 3000

We know that

$\boxed{ \red{ \bf \: Volume_{(cuboid)} =l \times b \times h}}$

$:  \implies \bf \: 3000 = (4x)(3x)(2x)$

$:  \implies \bf \: 3000 = {24x}^{3}$

$:  \implies \bf \: {x}^{3} = 125$

$:  \implies \bf \: x = 5$

$\begin{gathered}\begin{gathered}\bf Hence -  \begin{cases} &\sf{length \: of \: cuboid \: = 4 \times 5 = 20 \: cm} \\ &\sf{breadth \: of \: cuboid \: = 3 \times 5 = 15 \: cm}\\ &\sf{height \: of \: cuboid \: = 2 \times 5 = 10 \: cm} \end{cases}\end{gathered}\end{gathered}$

${{ \boxed{{\bold\green{Total \: Surface \: Area_{(Cuboid)}\: = 2(lb + bh + hl)}}}}}$

$\bf \: TSA_{(cuboid)} = 2(20 \times 15 + 15 \times 10 + 20 \times 10)$

$:  \implies \bf \: TSA_{(cuboid)} = 2(300 + 150 + 200)$

$:  \implies \bf \: TSA_{(cuboid)} = 1300 \: {cm}^{2}$

─━─━─━─━─━─━─━─━─━─━─━─━─

Additional Information

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth²)

Area of square = side²

Perimeter of square = 4× side

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 = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²