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one leg of a right triangle is 5 more than the other leg the hypotenuse is one more than twice the size of the short leg find the dimension of the triangle

Sagot :

Short leg be x

  • Perpendicular=P=x
  • Base=B=x+5
  • Hypotenuse=H=2x+1

Apply Pythagorean theorem

[tex]\\ \rm\Rrightarrow P^2+B^2=H^2[/tex]

[tex]\\ \rm\Rrightarrow x^2+(x+5)¥2=(2x+1)^2[/tex]

[tex]\\ \rm\Rrightarrow x^2+x^2+10x+25=4x^2+4x+1[/tex]

[tex]\\ \rm\Rrightarrow 2x^2+10x+25=4x^2+4x+1[/tex]

[tex]\\ \rm\Rrightarrow 4x^2-2x^2+4x-10x+1-25=0[/tex]

[tex]\\ \rm\Rrightarrow 2x^2-6x-24=0[/tex]

[tex]\\ \rm\Rrightarrow x^2-3x-12=0[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{3\pm\sqrt{9+48}}{2}[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{3\pm \sqrt{57}}{2}[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{3\pm 7.2}{2}[/tex]

[tex]\\ \rm\Rrightarrow x=1.5\pm 3.6[/tex]

Take positive

[tex]\\ \rm\Rrightarrow x=1.5+3.6=5.1[/tex]

  • P=5.1
  • B=10.1
  • H=11.2
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