Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
From the text of the task we have:
smaller leg: x
longer leg: x+3
hypotenuse: x + 6
Using the Pythagorean Theorema:
[tex](x+6)^2=x^2+(x+3)^2\\ \\ x^2+12x+36=x^2+x^2+6x+9\\ \\ x^2-6x-27=0\\ \\ \Delta=(-6)^2-4.1.(-27)=36+108=144\\ \\ x=\frac{6 \pm \sqrt{144}}{2}=\frac{6 \pm 12}{2}\\ \\ x_1=-3\\ \\ x_2=9[/tex]
Obviously -3 is not possible, so, smaller leg is 9 cm
smaller leg: x
longer leg: x+3
hypotenuse: x + 6
Using the Pythagorean Theorema:
[tex](x+6)^2=x^2+(x+3)^2\\ \\ x^2+12x+36=x^2+x^2+6x+9\\ \\ x^2-6x-27=0\\ \\ \Delta=(-6)^2-4.1.(-27)=36+108=144\\ \\ x=\frac{6 \pm \sqrt{144}}{2}=\frac{6 \pm 12}{2}\\ \\ x_1=-3\\ \\ x_2=9[/tex]
Obviously -3 is not possible, so, smaller leg is 9 cm
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.