Question

solve plzzzzz........ ​

Answer 1

$\large\underline{ \underline{ \sf \maltese \: {Given:-}}}$

• Height of first pole (ED) = 58 m
• Height of second pole (AC) = 10 m
• Distance Between two poles (AD) = 14 m

$\large\underline{ \underline{ \sf \maltese \: {To \: find:-}}}$

Length of the wire (AE)

$\large\underline{ \underline{ \sf \maltese \: {Solution:-}}}$

Let:–

AE = length of the wire

$\qquad \blacksquare \: \sf{AB = CD = 14 \: m}$

$\qquad \blacksquare \: \sf{EB = BD - AC = \bigg(58 - 10 \bigg) \: m = 48 \: m}$

~ By using Pythagoras' theorem:–

$\qquad \bull \sf{(AE)^2 = (AB)^2+(EB)^2}$

${:} \implies \sf{(AE)^2=14^2 + 48^2}$

${:} \implies \sf{(AE)^2=2304+196}$

${:} \implies \sf{(AE)^2=2500}$

${:} \implies \sf{AE= \sqrt{2500}}$

${:} \implies \boxed {\frak\red{AE= 50}}$

$\qquad \therefore \: {\underline{\sf{ the \: length \: of \: wire = { 50 \: m}}}}$

Answer 2

Answer:

Height of first pole = 58m

Height of second pole = 10m

Distance Between two poles = 14m

58-10 =48

By using Pythagroes theorm

(AC)^2 = (AB)^2+(BC)^2

= (48)^2+(14)^2

= 2304+196

(AC)^2 = 2500

AC = √2500

AC = 50

Therefore, the length of wire from one pole to another pole is 50m.

hope this helps you