Question

The length of the longest pole that can be placed on the floor of a room is 10m and the length of the longest pole that can be placed in the room is 10root2m. The height of the room is...?

Answer 1

Given

Hypotenuse ⇒ 10√2 m

Base ⇒ 10 m

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To Find

The Height.

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Solution

For this question, we must apply the Pythagorean Theorem.

Pythagorean Theorem ⇒ Base² + Height² = Hypotenuse²

Base ⇒ 10 m

Hypotenuse ⇒ 10√2 m

Height ⇒ x

Let's solve your equation step-by-step

10² + x² = (10√2)²

Step 1: Subtract 10² from both sides of equation.

⇒ 10² + x² = (10√2)²

⇒ 10² + x² - 10² = (10√2)² - 10²

⇒ x² = (10√2)² - 10²

⇒ x² = {10² × (√2)²} - 10²

⇒ x² = (100 × 2) - 100

⇒ x² = 200 - 100

⇒ x² = 100

⇒ x = √100

⇒ x = 10

∴ The height of the room is 10 m

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Answer 2

Answer:

Given:

• Longest pole which can be placed on the floor = 10 m

• Longest pole which can be placed in the room = 10√2 m

To find:

• Height of the room

Solution:

Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(10.6,2.9){\large\sf{C}}\put(7.7,1){\large\sf{A}}\put(10.6,1){\large\sf{B}}\put(8,1){\line(1,0){2.5}}\put(10.5,1){\line(0,2){1.9}}\qbezier(8,1)(8.5,1.4)(10.5,2.9)\put(9,0.7){\sf{\large{10 m}}}\put(10.7,1.9){\sf{\large{ ? }}}\put(10.3,1){\line(0,1){0.2}}\put(10.3,1.2){\line(3,0){0.2}}\put(8,1){\circle*{.15}}\put(7.3,2){\sf{\large{West}}}\put(11.7,2){\sf{\large{East}}}\put(10,0){\sf{\large{South}}}\put(10,3.8){\sf{\large{North}}}\end{picture}$

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$\underline{\bigstar\:\textsf{By Pythagoras Theorem :}}$

$</p><p>:\implies\sf (Hypotenuse)^2 = (Perpendicular)^2+(Base)^2\\\\\\:\implies\sf (AB)^2=(BC)^2+(AC)^2\\\\ :\implies\sf \: (AB)^2 \: = {10}^{2} +( { 10\sqrt{2} })^{2} \\ \\ :\implies\sf \: (AB)^2 \: = 100 + 200 \\ \\ :\implies\sf \: (AB)^2 \: = 300 \\ \\ :\implies\sf \: AB\: = 10 \sqrt{3}$