To find: The values of KP, KM, MR, ML, MN and PR. Given: 12210203131.jpg bar(PR)||bar(KL) = 9, ln = 16, PM = 2(KP) KM=KP+PM=3KP

Question
Similarity
asked 2021-01-10
To find: The values of KP, KM, MR, ML, MN and PR.
Given:
image
\(\displaystyle\overline{{{P}{R}}}{\mid}{\mid}\overline{{{K}{L}}}={9},{\ln{=}}{16},{P}{M}={2}{\left({K}{P}\right)}\)
KM=KP+PM=3KP

Answers (1)

2021-01-11
Conclusion:
By A-A similarity, it can be proved that:
\(\displaystyle\triangle{L}{M}{K}\sim\triangle{M}{N}{K}\)
Similar triangles have corresponding sides as proportional.
\(\displaystyle\frac{{{M}{K}}}{{{L}{K}}}=\frac{{{N}{K}}}{{{M}{K}}}\)
\(\displaystyle\Rightarrow\frac{{{3}{K}{P}}}{{25}}=\frac{{9}}{{{3}{K}{P}}}\)
\(\displaystyle\Rightarrow{K}{P}={5}\)
\(\displaystyle\Rightarrow{K}{M}={3}{K}{P}={15}\)
By A-A similarity, it can be proved that:
\(\displaystyle\triangle{L}{M}{K}\sim\triangle{R}{M}{P}\)
Similar triangles have corresponding sides as proportional.
\(\displaystyle\frac{{{M}{K}}}{{{L}{K}}}=\frac{{{P}{M}}}{{{R}{P}}}\)
\(\displaystyle\Rightarrow\frac{{15}}{{25}}=\frac{{10}}{{{R}{P}}}\)
\(\displaystyle\Rightarrow{K}{P}={16}\frac{{2}}{{3}}\)
Using Pythagorean Theorem in right angled triangle \(\displaystyle\triangle{L}{M}{K}\):
\(\displaystyle{H}{y}{p}{o}{t}{e}\nu{s}{e}^{{2}}={B}{a}{s}{e}^{{2}}+{P}{e}{r}{p}{e}{n}{d}{i}{c}\underline{{a}}{r}^{{2}}\)
\(\displaystyle={25}^{{2}}={15}^{{2}}+{L}{M}^{{2}}\)
\(\displaystyle\Rightarrow{L}{M}^{{2}}={400}\)
\(\displaystyle\Rightarrow{L}{M}={20}\)
In similar triangles: \(\displaystyle\triangle{L}{M}{K}\sim\triangle{M}{N}{K}\)
Ratio can be written as:
\(\displaystyle\frac{{{L}{M}}}{{{M}{K}}}=\frac{{{M}{N}}}{{{N}{K}}}\)
\(\displaystyle\Rightarrow\frac{{20}}{{15}}=\frac{{{M}{N}}}{{9}}\)
\(\displaystyle\Rightarrow{M}{N}={12}\)
\(\displaystyle{K}{P}={5},{K}{M}={15},{M}{R}={13}\frac{{1}}{{3}},{M}{L}={20},{M}{N}={12},{P}{R}={16}\frac{{2}}{{3}}\)
0

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