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  • In Fig., ABC is a triangle right angled at B and BD \perp AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.

In Fig., ABC is a triangle right angled at B and BD \perp AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.

 

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Answer: \left [ 2\sqrt{5}\; cm\; and\; 6\; cm \right ]

Given :- \angle B=90^{o} and BD\perp AC

              AD = 4 cm and CD = 5 cm

      In \Delta ABD and \Delta BDC

      \angle ADB = \angle BDC               (each equal to 90o)

     \angle BAD = \angle DBC                 (each equal to 90o-C)

    \therefore \Delta ABD\sim \Delta BDC            (by AA similarity criterion)

\Rightarrow \frac{DB}{DA}=\frac{DC}{DB}

By cross multiply we get

DB^{2}=DA.DC

DB^{2}=4 \times 5

DB=\sqrt{20}=2\sqrt{5}cm

In \Delta BDC use Pythagoras theorem

BC^{2}=BD^{2}+CD^{2}

BC^{2}=\left ( 2\sqrt{5} \right )^{2}+\left ( 5 \right )^{2}

BC^{2}=20+25=45

BC=\sqrt{45}=3\sqrt{5}

We know that  \Delta DBA\sim \Delta DBC

\therefore \frac{DB}{DC}=\frac{BA}{BC}\Rightarrow \frac{2\sqrt{5}}{5}=\frac{BA}{3\sqrt{5}}

BA=\frac{2\sqrt{5}\times 3\sqrt{5}}{5}=\frac{6 \times 5}{5}=6\; cm

Posted by

infoexpert23

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