Question

in triangle abc right angled at B and BD perpendicular to AC and AD is equal to 8 cm and CD is equal to 10 cm then AB equal to​

Answer 1

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$In\ \triangle ABC\ right-angled\ at\ B\ and\\ BD \perp AC\ and\ AD = 8\ cm\ and\\ CD = 10\ cm ,\ then\ find\ AB$

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$\boxed{\red{\bold{In \triangle ADB and \triangle ABC}}}$

$\angle ADB = \angle ABC (\because\ each\ angle\ 90^o)$

$\angle A = \angle A(common)$

$\purple{\implies \triangle ADB \sim \triangle ABC} \blue{(By\ AA\ similarity)}$

$\boxed{\red{\bold{Taking\ sides\ proportional:}}}$

$\blue{\implies \displaystyle \frac{AD}{BC} = \frac{BD}{BC} = \frac{AB}{AC}}$

$\blue {\implies}\displaystyle \frac{AD}{AB} = \frac{AB}{AC}$

${\blue{\implies}}AB^2 = AD . AC$

${\blue{\implies}}AB^2 = 8(8+10)$

${\blue{\implies}}AB^2 = 8 × 18$

${\blue{\implies}}AB^2 = 144$

$\boxed{\red{\bold{Taking\ square\ roots:}}}$

$\green{\implies AB= \sqrt{144}}$

$\green{\boxed{\implies{\boxed{AB=12}}}}$

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