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Answer 1

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Answer 2

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

$\sf \underline \green{ \therefore \: the \: \: two \: sides \: are \: 12cm \: and \: 5cm}$

$\huge \sf \underline \orange{To \: find}$

• Other two sides

$\sf \huge \underline \blue{solution : }$

$\sf \underline{Let \: the \: base \: of \: \: right \: angle \: be \: xcm}$

$\sf{Then \: altitudes \: be \: x - 7cm}$

$\: \: \: \: \sf \underline{ we \: know \: that}$

$\sf{ \boxed{ \underline{ \underline{ \red{ \tt{pythagoras \: theorem = ( {b}^{2} ) + ( {h}^{2}) = {(hp}^{2) \: }}}}}}}$

• b is base (x)

• h is height (x-7)

• hp is hypothenuse (13)

$\sf \implies{ {x}^{2} + (x - 7) {}^{2} = {13}^{2}}$

$\sf \implies{ {x}^{2} + {x}^{2} - 14x + 49 = 169 }$

$\sf \implies{ {2x}^{2} - 14x + 49 - 169 = 0}$

$\sf \implies{ {2x}^{2} - 14x - 120 = 0 }$

$\sf \implies{ {x}^{2} - 7x - 60 = 0 }$

$\sf \implies{ {x}^{2} - 12x + 5x - 60 = 0 }$

$\sf \implies{ x(x - 12) + 5(x - 12) = 0}$

$\sf \implies{ (x - 12)(x + 5) = 0}$

$\sf \implies{x - 12 = 0 \: or \: x + 5 = 0}$

$\sf { x \: cannot \: be \: negative}$

$\sf \underline{ \therefore \: x = 12}$

$\sf \underline{x - 7 = 12 - 7 = 5}$

$\sf \underline \green{ \therefore \: the \: \: two \: sides \: are \: 12cm \: and \: 5cm}$