Question

The length of longest pole that can be placed on the floor of a room is 10 m and the length of longest pole that can be placed in the room to 10 root 2 m. The height of the room is​

Answer 1

Answer:

height=10

Step-by-step explanation:

H^2=p^2+b^2

(10√2)^2=p^2+(10)^2

200=(p}^2+100

100=p^2

√100=p

10=p.

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 :

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

⠀

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

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