Question

2. A narrow box is 6V2cm long and 4V2cm wide. Find
its height if the length of the diagonal is 4v7cm.​

Answer 1

Answer:

66 y 4th Y6ea3 3afxju t was iritis

Answer 2

Appropriate Question:

• A narrow box is 6√2cm long and 4√2 cm wide. Find its height if the length of the diagonal is 4√7cm.

Given:

• Length of box, l =62cm
• Breadth of box, b=42cm
• Length of diagonal of box, d=47cm

To find:

• Height of box?

Required Formula:

$\begin{gathered} {\color{purple}\bigstar}\;{\boxed{\frak{{Diagonal_{\;(cuboid)} = \sqrt{(length)^2 + (breadth)^2 + (height)^2}}}}}\\\\\end{gathered} \\ \begin{gathered}:\implies\sf d = \sqrt{l^2 + b^2 + h^2}\\\\\end{gathered}$

Here,

• l = 6√2 cm
• b = 4√2 cm
• d = 4√7 cm

⠀⠀⠀

Solution:

$\begin{gathered}\star\;{\underline{\frak{ \color{navy}{Now,\:Putting\:values\;in\;formula,}}}}\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 4\sqrt{7} = \sqrt{(6\sqrt{2})^2 + (4\sqrt{2})^2 + h^2}\\ \\ \\ :\implies\sf 4\sqrt{7} = \sqrt{72 + 32 + h^2}\\ \\ \\\qquad:\implies\sf (4\sqrt{7})^2 = \bigg(\sqrt{72 + 32 + h^2}\bigg)^2 \qquad \qquad {{\sf\ \Big\{Squaring\:both\:sides\Big\}}}\\\ \\ \\:\implies\sf (4\sqrt{7})^2 = 72 + 32 + h^2\\ \\ \\ :\implies\sf 112 = 104 + h^2\\ \\ \\ :\implies\sf h^2 = 112 - 104\\ \\ \\ :\implies\sf h^2 = 8\\ \\ \\ :\implies\sf \sqrt{h^2} = \sqrt{8}\\ \\ \\ :\implies\sf h = \sqrt{8}\\ \\ \\:\implies{\underline{\boxed{\frak{{h = 2 \sqrt{2}\:cm}}}}}\pink\bigstar\\ \\ \\ \end{gathered}$

$\therefore{ \underline{ \sf{Hence, the \: height \: of \: box \: is \: \textsf{ \textbf{h =2√2 cm}.}}}}$

тнαηк үσυ

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