The hypotenuse C and one side A of right triangle are consecutive integers,then find the square of the third in terms of A and C.
Answers
Answered by
1
Answer:
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer.
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer.
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer. c=n+2
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer. c=n+2b=n+1
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer. c=n+2b=n+1a=n
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer. c=n+2b=n+1a=n
I will solve for consecutive integers. The process for solving for consecutive even will be very similar. Let's define the three sides of the triangle from largest to smallest by using equations. We will have the smallest side be equal to "n," an integer. c=n+2b=n+1a=n Now, let's substitute these three equations into the Pythagorean Theorem, or a2+b=c2
=c2
=c2 (n+2)2=n2+(n+1)2
=c2 (n+2)2=n2+(n+1)2
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying.
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying.
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term n2+4n+4=2n2+2n+1 <-- Combined Like Terms
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term n2+4n+4=2n2+2n+1 <-- Combined Like Terms
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term n2+4n+4=2n2+2n+1 <-- Combined Like Terms Now, let's move all of the terms onto one side to set it equal to 0.
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term n2+4n+4=2n2+2n+1 <-- Combined Like Terms Now, let's move all of the terms onto one side to set it equal to 0.
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term n2+4n+4=2n2+2n+1 <-- Combined Like Terms Now, let's move all of the terms onto one side to set it equal to 0. n2-2n+-3=0
=c2 (n+2)2=n2+(n+1)2 Now, we can solve for "n" by FOILing out each squared factor and simplifying. n2+4n+4=n2+n2+2n+1 <-- I foiled each squared term n2+4n+4=2n2+2n+1 <-- Combined Like Terms Now, let's move all of the terms onto one side to set it equal to 0. n2-2n+-3=0
Similar questions