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