Question

in the given figure D is the midpoint of the hypotenuse AC of right triangle ABC then BD is equal to​

Answer 1

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by pythagoras property

${h}^{2} = {b}^{2} + {p}^{2}$

${h}^{2} = {4}^{2} + {3}^{2} = 16 + 9 = 25 \\ h = 5$

hypotenuse = 5cm

so half of hypotenuse =2.5

$\tan(45 ) = \frac{p}{3} \\ 1 = \frac{p}{3} \\ p = 3$

so,BD(p)=3cm.