Question

in a right angled triangle hypotenuse is 3 cm less than two times the base the third side is 3 cm less than hypotenuse if base is x find the length of other two sides​

Answer 1

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In a right angled triangle hypotenuse is 3 cm less than two times the base the third side is 3 cm less than hypotenuse if base is x find the length of other two sides?

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$\large{ \sf{hypotenuse = 15units}} \\ \\ \large{ \sf{perpendicular = 12 \: units}}$

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$\sf{ \star{ \orange{ \underline{ \underline{given \longrightarrow}}}}}$

• Base is x
• Hypotenuse is 3 less than two times of base
• third side is 3 less than hypotenuse

$\sf{ \star{ \orange{ \underline{ \underline{to \: find \: \longrightarrow}}}}}$

• Hypotenuse and third side

$\sf{ \purple{ \bold{ \underline{ \underline{according \: to \: question...}}}}} \\ \\ \large{ \sf{ \implies{ \: base \: = x}}} \\ \\ \large{ \sf{ \implies{then \: hypotenuse = 2x - 3}}} \\ \\ \large{ \sf{ \implies{third \: side =( 2x - 3) - 3}}} \\ \large{ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2x - 6}} \\ \\ \large{ \sf{ \leadsto{\: u \: must \: know}}} \\ \large{ \sf{ \boxed{ \green{ {p}^{2} + {b}^{2} = {h}^{2} }}}} \\ \large{ \sf{ \red{where \: \: p = perpendicular}}} \\ \large{ \sf{ \red{b = base}}} \\ \large{ \sf{ \red{h = hypotenuse}}} \\ \\ \large{ \sf{ \implies{by \: using \: formula...}}} \\ \\ \large{ \sf{ \implies{(2x - 6) {}^{2} + {x}^{2} = (2x - 3) {}^{2} }}} \\ \\ \large{ \sf{ \implies{4 {x}^{2} - 24x + 36 + {x}^{2} = 4 {x}^{2} - 12x + 9}}} \\ \\ \large{ \sf{ \implies{5 {x}^{2} - 24x + 36 = 4 {x}^{2} - 12x + 9}}} \\ \\ \large{ \sf{ \implies{ {x}^{2} - 12x + 27 = 0}}} \\ \\ \large{ \sf{ \implies{ {x}^{2} - 9x - 3x + 27 = 0}}} \\ \\ \large{ \sf{ \implies{x(x - 9) - 3(x - 9)}}} \\ \\ \large{ \sf{ \implies{(x - 9)(x - 3) = 0}}} \\ \\ \large{ \sf{ \implies{x \: = 9 \: \: or \: \: 3}}}$

$\large{ \sf{ \purple{ \boxed{case \: 1)}}}} \: if \: x = 3 \\ \\ then \: 2x - 3 = > 2(3) - 3 = 3 \\ \\ 2x - 6 = > 2(3) - 6 = 0 \\ \\ { \sf{ \large{the \: third \: side \: is \: 0 \: }}}\\ \\ \large{ \sf{ \ which \: is \: not \: possible}}.... \\ \\ \large{ \purple{ \sf{ \boxed{case - 2)}}}} \: if \: x = 9 \\ \\ then \: 2x - 3 = > 2(9) - 3 = 15 \\ \\ 2x - 6 = > 2(9) - 6 = 12 \\ \\ \large{ \star{ \sf{ \red{hence \: base = 9 \: units}}}} \\ \\ \large{ \sf{ \red{hypotenuse = 15 \: units}}} \\ \\ \large{ \sf{ \red{perpendicular = 2(9) - 6 = 12 \: units}}}$

$\large{ \blue{ \bold{required \: sides \: are \: 9 \: \: \: 12 \: \: \: and \: 15 \: units}}}$