Question

if a 5metre and 50 metre high pole are 100metre apart then find the height of point of intersection of line joining top of each pole to foot of opposite pole​

Answer 1

Answer:

3.75 m................

Answer 2

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Here we have to find the height of PQ

Now, consider ΔABC and ΔPQC:

• ∠PQC=∠ABC=90°
• ∠PCQ=∠ACB(common)

So,ΔABC≈ΔPQC

Similarly,ΔPBQ≈ΔDBC

Now, In similar triangles,ΔABC and ΔPQC:

$\huge\frac{AB} {BC} = \frac{PQ}{QC}$

$\huge\frac{50} {100} =\frac{PQ} {QC}$

$\huge \frac{1}{2}\:\:\:=\frac{PQ}{QC}$

QC=2PQ_________(i)

Now,In similar triangles,ΔPBQ and ΔDBC

$\huge\frac{DC} {BC} = \frac{PQ}{BQ}$

$\huge\frac{5} {100} =\frac{PQ} {BQ}$

$\huge \frac{1}{20}\:\:=\frac{PQ}{BQ}$

BQ=20PQ _________(ii)

Now adding i and ii

(QC+BQ)=22PQ

BC=22PQ

100=22PQ

PQ=100/22

$\huge{=4.5metre}$