Question

The lateral surface area of a solid rectangular box is 520 cm2

and its length , breadth

and height

are in the ratio 9:4:5. Find the total surface area of the box.​

Answer 1

Answer:

TSA is 808

Step-by-step explanation:

pls mark as brainlist and rate the answer

Answer 2

Given:

• The lateral surface area of a solid rectangular box is 520 cm²

• It's length,breadth and it's hieght are in the ratio 9:4:5

To Find:

• The total surface area of the cubiod

Solution:

❍ Let's Assume that,

• Length of the cubiod is ${\pink{\bf{9a}}}$

• Breath of the cubiod is ${\blue{\bf{4a}}}$

• hieght of the cubiod is ${\gray{\bf{5a}}}$

↦ As we Know that,

$\star \: { \underline{ \boxed{ \sf{lateral \: surface \: area \: of \: a \: cuboid = 2h(l + b)}}}}$

Where,

• H stands for ${\pink{\bf{Hieght}}}$

• L stands for ${\gray{\bf{Length}}}$

• B stands for ${\blue{\bf{Breadth}}}$

Now, let's substitute the assumesd values:

$\longrightarrow \sf \: 520 {cm}^{2} = 2 \times 5a (9a + 4a) \\ \\ \\ \longrightarrow \sf \: 520 {cm}^{2} = 10a \times 13a \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf520 {cm}^{2} = 130 { a}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: {a}^{2} = \cancel\frac{520}{130} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \\ \\ \\ \longrightarrow \sf {a}^{2} = 4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: a = \sqrt{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow { \orange{\sf{a = 2\bigstar}} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

❍ Therefore,

• $\leadsto {\sf{\pink{Length = 9 \times a = 18cm}}}$

• $\leadsto {\sf{\pink{Breadth = 4 \times a = 8cm}}}$

• $\leadsto {\sf{\pink{Length = 5 \times a = 10cm}}}$

As we've found the dimensions let's find the total surface area now,

↦ As we know that,

$\star \: { \underline{ \boxed{ \sf{total \: surface \: area \: of \: a \: cuboid = 2(lb + bh + hl)}}}}$

Let's substitute,

$\longrightarrow \tt \: T.S.A = 2(18 \times 8 + 8 \times 10 + 10 \times 18) \\ \\ \\ \longrightarrow \tt \: T.S.A = 2(144 + 80 + 180) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt \: \: T.S.A = 2(404) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow{ \orange{ \tt \: T.S.A = 808 {cm}^{2} \bigstar}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence,

${\longrightarrow }\sf the\: total \:surface \:area\:of \:the\:cuboid \:is \:{\pink {\sf{808{cm}^{2} }}\bigstar}$