Question

3. The sum of length, breadth and height of a cuboid is 12 cm. The length of the diagonal is
5√2 cm. Find the surface area of the cuboid.​

Answer 1

Answer:

94 cm²

Step-by-step explanation:

Let the length, breadth and height of the cuboid be l, b and h respectively.

A.T.Q.

Sum of length, breadth and height = 12 cm

l + b + h = 12 ...(i)

Also,

Diagonal of cuboid = 5√2 cm

${\sf{ {\sqrt{l^2 + b^2 + h^2}} }}$ = 5√2 ...(ii)

On further solving (ii), we get

→ ${\sf{ {\sqrt{l^2 + b^2 + h^2}} }}$ = 5√2

→ l² + b² + h² = (5√2)²

→ l² + b² + h² = 50 ...(iii)

Now,

Squaring both sides of (i), we get

→ (l + b + h)² = (12)²

• { Identity : (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca }

Here, a = l, b = b, c = h

→ l² + b² + h² + 2lb + 2bh + 2hl = 144

→ l² + b² + h² + 2(lb + bh + hl) = 144

→ 50 + 2(lb + bh + hl) = 144

→ 2(lb + bh + hl) = 144 - 50

→ 2(lb + bh + hl) = 94

As we know that,

Surface Area of cuboid = 2(lb + bh + hl), where l, b and h are length, breath and height respectively.

Therefore,

Surface Area of cuboid = 94 cm²

Answer 2

Answer:

$\bold\green{Surface\:area}$ = ${\boxed{\mathfrak\red{94\:{cm}^{2} }}}$

Explanation:

Given:-

• Sum of length, breadth and height of cuboid = 12 cm
• Length of diagonal = 5√2 cm

• Surface area of cuboid = ?

${\underline{\underline{\bold{According\:to\:question:-}}}}$

$\tt \: \: 1st\: case:- \\\\\Rightarrow\tt l + b + h = 12 \\\\\tt \: \: *Squaring\:both\:sides:- \\\\\Rightarrow\tt {(l + b + h)}^{2} = {12}^{2} \\\\\Rightarrow\tt {l}^{2} + {b}^{2} + {h}^{2} + 2(lb + bh + hl) = 144 ...........(i) \\\\\tt \: \: \: [\because {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) \\\\\\\tt \: \: 2nd\: case:- \\\\\Rightarrow\tt \sqrt{ {l}^{2} + {b}^{2} + {h}^{2} } = 5\sqrt{2} \\\\\tt \: \: *Squaring\:both\:sides:- \\\\\Rightarrow\tt {\sqrt{ {l}^{2} + {b}^{2} + {h}^{2} } }^{2} = {(5\sqrt{2})}^{2} \\\\\Rightarrow\tt {l}^{2} + {b}^{2} + {h}^{2} = 25 \times 2 \\\\\Rightarrow\tt {l}^{2} + {b}^{2} + {h}^{2} = 50..........(ii)$

We know that,

${\boxed{\bold{Surface\:area = 2(lb + bh + hl) }}}$

$\tt *Now,putting\:value\:of\:(ii)\:in\:(i)\: :- \\\\\Rightarrow\tt 50 + 2(lb + bh + hl) = 144 \\\\\Rightarrow\tt 2(lb + bh + hl) = 144 - 50 \\\\\therefore\tt 2(lb + bh + hl) = 94\:{cm}^{2}$

$\therefore{\boxed{\bold{Surface\:area=94\:{cm}^{2} }}}$