Question

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Answer 1

19.In ∆ given that , perpendicular (p)=4

hypotenuse (h)=8

so here using law of pithagoras i.e

${h}^{2} = {p}^{2} + {b}^{2} \\ = > h = \sqrt{ {p}^{2} } + {b}^{2}$

likely

$b = \sqrt{ {h}^{2} } - {p}^{2}$

$= \sqrt{ {8}^{2} } - \sqrt{ {4}^{2} } \\ = \sqrt{64 - 16} \\ = \sqrt{48 } \\ = 2 \sqrt{3}$

9.In ∆ given that,base(b)=2

perpendicular(p)=1.5

so here give out (h)

$h = \sqrt{ {2}^{2} } + \sqrt{ {1.5}^{2} } \\ = \sqrt{4 + 2.25} \\ = \sqrt{6.25} \\ = 0.25$