Find detailed and accurate answers to your questions on IDNLearn.com. Join our interactive community and get comprehensive, reliable answers to all your questions.
Sagot :
The length of one leg is x inches.
The sum of the legs is 49 inches, so the length of the other leg is 49-x inches.
The length of the hypotenuse is 41 inches.
Use the Pythagorean theorem:
[tex](\hbox{one leg})^2 + (\hbox{the other leg})^2=(\hbox{hypotenuse})^2 \\ x^2+(49-x)^2=41^2 \\ x^2+2401-98x+x^2=1681 \\ 2x^2-98x+2401=1681 \ \ \ |-1681 \\ 2x^2-98x+720=0 \\ \\ a=2 \\ b=-98 \\ c=720 \\ b^2-4ac=(-98)^2-4 \times 2 \times 720=9604-5760=3844 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-98) \pm \sqrt{3844}}{2 \times 2}=\frac{98 \pm 62}{4} \\ x=\frac{98-62}{4} \ \lor \ x=\frac{98+62}{4} \\ x=\frac{36}{4} \ \lor \ x=\frac{160}{4} \\ x=9 \ \lor \ x=40[/tex]
[tex] 49-x=49-9 \ \lor \ 49-x=49-40 \\ 49-x=40 \ \lor \ 49-x=9[/tex]
The lengths of the legs are 9 inches and 40 inches.
The sum of the legs is 49 inches, so the length of the other leg is 49-x inches.
The length of the hypotenuse is 41 inches.
Use the Pythagorean theorem:
[tex](\hbox{one leg})^2 + (\hbox{the other leg})^2=(\hbox{hypotenuse})^2 \\ x^2+(49-x)^2=41^2 \\ x^2+2401-98x+x^2=1681 \\ 2x^2-98x+2401=1681 \ \ \ |-1681 \\ 2x^2-98x+720=0 \\ \\ a=2 \\ b=-98 \\ c=720 \\ b^2-4ac=(-98)^2-4 \times 2 \times 720=9604-5760=3844 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-98) \pm \sqrt{3844}}{2 \times 2}=\frac{98 \pm 62}{4} \\ x=\frac{98-62}{4} \ \lor \ x=\frac{98+62}{4} \\ x=\frac{36}{4} \ \lor \ x=\frac{160}{4} \\ x=9 \ \lor \ x=40[/tex]
[tex] 49-x=49-9 \ \lor \ 49-x=49-40 \\ 49-x=40 \ \lor \ 49-x=9[/tex]
The lengths of the legs are 9 inches and 40 inches.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.