Math, asked by vandanabhargava1977, 6 months ago

the hypotenuse of a right angled triangle is 29 cm the difference between the other two sides is 1 cm find the other two sides​

Answers

Answered by ITZBFF
37

 \mathsf \red{let \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   } \\   \mathsf{one \: side \: be \: x \: cm \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  } \\  \\  \mathsf \red{given \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: } \\   \mathsf{hypotnuse \: of \: triangle \:  =  \: 29 \: cm} \\  \mathsf{difference \: of \: two \: sides \:  =  \: 1 \: cm} \\  \mathsf{ \therefore \: the \: other \: side \: \:  =  \: x + 1 \: cm  \:  \: } \\  \\  \mathsf \red{By \: using \:pythagoras \: thereom \:  :  } \\  \\  \mathsf{⇒ {x}^{2} +  {(x + 1)}^{2}   = {29}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  } \\  \mathsf{⇒ {x}^{2} +  {x}^{2} + 1 + 2x = 841 \:  \:  \:  \:  \:  \:  \:   } \\  \mathsf{⇒2 {x}^{2} + 2x = 841 - 1 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   } \\  \mathsf{⇒2( {x}^{2} + x) = 840 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  } \\  \mathsf{⇒ {x}^{2}  + x =  \frac{840}{2} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   }  \\ \mathsf{⇒ {x}^{2}  + x = 420 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: } \\  \mathsf{⇒  {x}^{2}   + x - 420 = 0 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   }  \\ \\   \mathsf \red{by \: using \: quadratic \:formula :   \:  \:  \:  \:  \: } \\  \boxed{  \large\mathsf \blue{x =  \frac{ - b  \: \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a} }} \\ \\   \mathsf{a = 1 \: , \: b = 1 \:,  \: c \:  =  \:  - 420} \\  \\  \mathsf \red{by \: substituting : } \\  \\  \mathsf{⇒x =  \frac{ - 1 \:  \pm \:  \sqrt{ {1}^{2}  - 4(1)( - 420)} }{2(1)} } \\  \\  \mathsf{⇒x \:  =  \:  \frac{ - 1 \:  \: \pm \:  \sqrt{1 + 1680  } }{2} } \\  \\  \mathsf{⇒x =  \frac{ - 1 \:  \pm \:  \sqrt{1681} }{2}} \\  \\  \mathsf{⇒x =  \frac{ - 1 \:  \pm \: 41}{2}  } \\  \\  \mathsf{⇒x =   \frac{ - 1 + 41}{2} , x =  \frac{ - 1 - 41}{2}  }

 \mathsf{x =  \frac{40}{2},  \frac{ - 42}{2}  } \\  \\  \mathsf{x = 20, - 21} \\ \\  \mathsf\red{distance \: cannot \: be \: negative \: so \:   :  } \\  \\  \mathsf{x = 20 \: cm \: and \: \: (x  + 1) = 21 \: cm} \:

\mathsf{}

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Answered by sagarkumar8384030664
7

Let one side be xcm.Since the difference between the two sides is 1cm

∴ the other side=x+1cm

As the given triangle is a right angled triangle with hypotenuse =29cm

By using pythagoras theorem, we get

x²+(x+1)²=29²

x²+x²+1+2x=841

2x²+2x+1=841

2x²+2x=841-1

2x²+2x=840

2(x²+x)=840

x²+x=840/2

x²+x=420

x²+ x - 420=0

A = 1

B = 1

C = -420

putting in Quadratic formula

 - b +  -  \sqrt{b { }^{2} }  - 4ac \div 2a

after solution the value of x= 20, (x+1)= 21

so we neglect negative value therefore x= 20

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