Step 1
Given,
\(\displaystyle\triangle{M}{P}{T}\) with \(\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}\)
PN = 4 cm
PR = 2.5 cm
NM = 2.9 cm
We have to find the length of \(\displaystyle\overline{{{R}{T}}}\) in centimeters.
Step 2
We have
\(\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}\)
Then, using the properties for two parallel lines cut by transversal lines
\(\displaystyle\angle{P}{N}{R}=\angle{P}{M}{T}\) ..........(1)
\(\displaystyle\angle{P}{R}{N}=\angle{P}{T}{M}\) ............(2)
Now,
In \(\displaystyle\triangle{N}{P}{R}{\quad\text{and}\quad}\triangle{M}{P}{T}\)
\(\displaystyle\angle{P}{N}{R}=\angle{P}{M}{T}\) (from (1))
\(\displaystyle\angle{P}{R}{N}=\angle{P}{T}{M}\) (from (2))
\(\displaystyle\angle{N}{P}{R}=\angle{M}{P}{T}\) (common angle )
So,
From AAA (angle-angle-angle) rule of similarity
\(\displaystyle\triangle{N}{P}{R}\sim\triangle{M}{P}{T}\)
Step 3
Now,
In similar triangle, ratio of corresponding sides will be equal.
\(\displaystyle\frac{{{N}{P}}}{{{M}{P}}}=\frac{{{P}{R}}}{{{P}{T}}}=\frac{{{N}{R}}}{{{M}{T}}}\)
Taking
\(\displaystyle\frac{{{N}{P}}}{{{M}{P}}}=\frac{{{P}{R}}}{{{P}{T}}}\)
\(\displaystyle\Rightarrow\frac{{4}}{{{M}{N}+{N}{P}}}=\frac{{2.5}}{{{P}{R}+{R}{T}}}\) (substituted the values)
\(\displaystyle\Rightarrow\frac{{4}}{{{2.9}+{4}}}=\frac{{2.5}}{{{2.5}+{R}{T}}}\)
\(\displaystyle\Rightarrow\frac{{4}}{{6.9}}=\frac{{2.5}}{{{2.5}+{R}{T}}}\)
\(\displaystyle\Rightarrow{4}{\left({2.5}+{R}{T}\right)}={6.9}\times{2.5}\)
\(\displaystyle\Rightarrow{4}{R}{T}+{10}={17.25}\)
\(\displaystyle\Rightarrow{4}{R}{T}={17.25}-{10}\)
\(\displaystyle\Rightarrow{4}{R}{T}={7.25}\)
\(\displaystyle\Rightarrow{R}{T}=\frac{{7.25}}{{4}}\)
\(\displaystyle\Rightarrow{R}{T}={1.8125}\)
So,
The lenght of RT = 1.8cm
Given,
\(\displaystyle\triangle{M}{P}{T}\) with \(\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}\)
PN = 4 cm
PR = 2.5 cm
NM = 2.9 cm
We have to find the length of \(\displaystyle\overline{{{R}{T}}}\) in centimeters.
Step 2
We have
\(\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}\)
Then, using the properties for two parallel lines cut by transversal lines
\(\displaystyle\angle{P}{N}{R}=\angle{P}{M}{T}\) ..........(1)
\(\displaystyle\angle{P}{R}{N}=\angle{P}{T}{M}\) ............(2)
Now,
In \(\displaystyle\triangle{N}{P}{R}{\quad\text{and}\quad}\triangle{M}{P}{T}\)
\(\displaystyle\angle{P}{N}{R}=\angle{P}{M}{T}\) (from (1))
\(\displaystyle\angle{P}{R}{N}=\angle{P}{T}{M}\) (from (2))
\(\displaystyle\angle{N}{P}{R}=\angle{M}{P}{T}\) (common angle )
So,
From AAA (angle-angle-angle) rule of similarity
\(\displaystyle\triangle{N}{P}{R}\sim\triangle{M}{P}{T}\)
Step 3
Now,
In similar triangle, ratio of corresponding sides will be equal.
\(\displaystyle\frac{{{N}{P}}}{{{M}{P}}}=\frac{{{P}{R}}}{{{P}{T}}}=\frac{{{N}{R}}}{{{M}{T}}}\)
Taking
\(\displaystyle\frac{{{N}{P}}}{{{M}{P}}}=\frac{{{P}{R}}}{{{P}{T}}}\)
\(\displaystyle\Rightarrow\frac{{4}}{{{M}{N}+{N}{P}}}=\frac{{2.5}}{{{P}{R}+{R}{T}}}\) (substituted the values)
\(\displaystyle\Rightarrow\frac{{4}}{{{2.9}+{4}}}=\frac{{2.5}}{{{2.5}+{R}{T}}}\)
\(\displaystyle\Rightarrow\frac{{4}}{{6.9}}=\frac{{2.5}}{{{2.5}+{R}{T}}}\)
\(\displaystyle\Rightarrow{4}{\left({2.5}+{R}{T}\right)}={6.9}\times{2.5}\)
\(\displaystyle\Rightarrow{4}{R}{T}+{10}={17.25}\)
\(\displaystyle\Rightarrow{4}{R}{T}={17.25}-{10}\)
\(\displaystyle\Rightarrow{4}{R}{T}={7.25}\)
\(\displaystyle\Rightarrow{R}{T}=\frac{{7.25}}{{4}}\)
\(\displaystyle\Rightarrow{R}{T}={1.8125}\)
So,
The lenght of RT = 1.8cm
