Question

The heights of two poles are 6m and 11m are standing on ground and the distance b,w the two ples is 12m.Then find the distance b;w the end of the poles

Answer 1

Given:

• Two poles of height 6m and 11m

• Distance between two poles is 12m

To Find:

Distance between the end of the poles

Diagram:

$\setlength{\unitlength}{0.8 cm}\begin{picture}(12,4)\thicklines\put(5.6,8.9){ C }\put(5.5,5.8){ D }\put(11.1,5.8){ B }\put(11.05,8.9){ E }\put(11.1,12.4){ A }\put(4.5,7.5){ 6\:m }\put(8.1,5.3){ 12 \:m }\put(11.5,7.5){ 11 \:m }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(11,9){\line(0,1){3.34}}\put(6,9){\line(3,2){5}}\put(6.3,9.11){\circle*{0.2}}\put(6.2,9.11){\circle*{0.1}}\put(6.16,9.09){\circle*{0.1}}\put(6.16,9.07){\circle*{0.1}}\put(6.166,9.084){\circle*{0.1}}\put(6.145,9.079){\circle*{0.1}}\put(6.145,9.07){\circle*{0.1}}\put(6.150,9.065){\circle*{0.1}}\put(6.2,9.04){\circle*{0.1}}\put(6.08,9.03){\circle*{0.08}}\end{picture}$

Solution:

Let AB be the pole of height 11m and CD be the pole of height 6m that are 12m apart and AC is the distance between the end of the poles

Now, according to diagram

⇒ CD= EB= 6M

Also,

AB= AE+EB

AB-EB= AE

AE= AB-EB

AE= 11-6

AE= 5m

Also, from diagram it is clear that

DB=CE= 12m

Now, In ΔAEC

Hypotenuse= AC

Perpendicular= AE

Base= CE

Now, on applying Pythagoras theorem in ΔAEC, we get

$\sf{(Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}}$

$\longrightarrow\sf{(AC)^{2}=(AE)^{2}+(CE)^{2}}$

$\longrightarrow\sf{(AC)^{2}=(5)^{2}+(12)^{2}}$

$\longrightarrow\sf{(AC)^{2}=25+144}$

$\longrightarrow\sf{(AC)^{2}=169}$

$\longrightarrow\sf{AC=\sqrt{169}}$

$\longrightarrow\sf{AC=13m}$

Hence, distance between the end of the poles is 13 m.