The length of the hypotenuse of a right triangle is 25 cm .one of its sides is 24 cm. Find the other side ? About the author Charlotte
[tex]let \: abc \: is \: a \: right \: angle \: triangle \: \\ where \: ab \: = base = 24 \: cm \: \\ ac = hypoteneus \: = 25cm \: \\ bc = perpendicular=? \\ by \: using \: pythagoras \: theorum \: \\ = > {ab}^{2} + {bc}^{2} = {ac}^{2} \\ = > {24}^{2} + {bc}^{2} = {25}^{2} \\ = > 576 + {bc}^{2} = 625 \\ = > {bc}^{2} = 625 – 576 \\ = > {bc}^{2} = 49 \\ = > bc = \sqrt{49} \\ = > bc = 7 \: cm \\ therefore, \: the \: lenght \: of \: other \: side \: is \: 7 \: cm[/tex] Reply
⇒ Let one of the side be x and the other be x+5 By Pythagoras theorem, (25) 2 =x ^2 +(x+5) ^2 ⇒ 625=x ^2 +x ^ 2 +10x+25 ⇒ 625=2x^2 +10x+25 ⇒ 2x^2 +10x−600=0 ⇒ x^2 +5x−300=0 ⇒ x^2 +20x−15x−300=0 ⇒ x(x+20)−15(x+20)=0 ⇒ (x+20)(x−15)=0 ⇒ x+20=0 and x−15=0 ∴ x=−20 and x=15 But x cannot be negative. ⇒ Hence, one side =x=15cm ⇒ Other side x+5=15+5=20cm ⇒ The product of the length of sides =15cm×20cm=300cm 2 Reply
[tex]let \: abc \: is \: a \: right \: angle \: triangle \: \\ where \: ab \: = base = 24 \: cm \: \\ ac = hypoteneus \: = 25cm \: \\ bc = perpendicular=? \\ by \: using \: pythagoras \: theorum \: \\ = > {ab}^{2} + {bc}^{2} = {ac}^{2} \\ = > {24}^{2} + {bc}^{2} = {25}^{2} \\ = > 576 + {bc}^{2} = 625 \\ = > {bc}^{2} = 625 – 576 \\ = > {bc}^{2} = 49 \\ = > bc = \sqrt{49} \\ = > bc = 7 \: cm \\ therefore, \: the \: lenght \: of \: other \: side \: is \: 7 \: cm[/tex]
⇒ Let one of the side be x and the other be x+5
By Pythagoras theorem,
(25)
2 =x ^2 +(x+5) ^2
⇒ 625=x ^2
+x ^
2 +10x+25
⇒ 625=2x^2 +10x+25
⇒ 2x^2
+10x−600=0
⇒ x^2
+5x−300=0
⇒ x^2
+20x−15x−300=0
⇒ x(x+20)−15(x+20)=0
⇒ (x+20)(x−15)=0
⇒ x+20=0 and x−15=0
∴ x=−20 and x=15
But x cannot be negative.
⇒ Hence, one side =x=15cm
⇒ Other side x+5=15+5=20cm
⇒ The product of the length of sides =15cm×20cm=300cm
2