Question

plzz solve this plzzzzzz​

Answer 1

Answer:

By pythagoras theorem,

hypotenuse² = base²+altitude²

=BC²+AB²

=12²+9²

=144+81

=225=15²

therefore,length of hypotenuse =15 cm

Answer 2

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$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

Q1) Find the coordinates of the midpoint of the line segment joining p(0,6) and Q(12,20)

$\: \: \large \green{ \bold{ \underline{given : }}}$

$\bullet \bf{Coordinates -\: p(0,6) \:and \: Q(12,20)}$

$\star \large\blue { \bold{ \underline{ \underline{ \: step \: by \: step \: explaination}}}}$

$\\ \implies\displaystyle\sf{let, \: \: P=(0,6)=(x_1 ,y_1) \: and \: \underline{Q=(12,20) =(x_2,y_2)}}$

$\\ \implies \displaystyle \sf \: let \: r(x.y) = is \: the \: midpoint \: of \: pq$

$\: \: \orange{ \bold{ \underline{by \: using \: midpoint \: formula}}}$

$\\ \implies\displaystyle\sf{x=\dfrac{x_1+x_2}{2} \: and \: y=\dfrac{y_1+y_2}{2}}$

$\: \\ \implies\displaystyle \bf{x = \frac{0 + 12}{2} } \: \: and \: y = \frac{6 + 20}{2}$

$\: \implies \displaystyle \bf{x = \frac{12}{2} \: and \: y = \frac{26}{2} }$

$\: \: \\ \implies \displaystyle \bf{x = \cancel \frac{12}{2} = 6 \: and \: y = \cancel\frac{26}{2} = 13}$

$\star \boxed{ \underline{ \underline{ \bf{the \: midpoint \: are \: x = 6 \: and \: y = 13}}}}$

$\implies\displaystyle\bf{R(x,y)=R(6,13)}$

$\bullet \underline{\bf{the \: coordinates \: of\: midpoint \: of \: seg \: PQ \: are\: (x, y) =(6,13)}}$

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Q2) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9cm and 12cm.

$\large \green{ \bold{ \underline{ \underline{solution}}}} :$

$\implies\displaystyle\sf{\delta ABC \: \angle ABC=90\degree}$

$\implies\displaystyle\sf{AB=9cm,BC=12cm}$

$\implies\underline{\displaystyle\sf{By \: Pythagoras \: Theorem}}$

$\implies\displaystyle\sf{(AB)^2+(BC)^2=(AC)^2}$

$\implies\displaystyle\sf{(9)^2+(12)^2=(AC)^2}$

$\implies\displaystyle\sf{81+144=(AC)^2}$

$\implies\displaystyle\sf{225=(AC)^2}$

$\implies\displaystyle\sf{AC=\sqrt{225}}$

$\implies\displaystyle\sf{AC=15cm}$

$\green \bigstar \boxed{ \underline{ \underline{ \bf{the \: hypotenuse \: is \: 15cm}}}}$