Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns | Nagwa Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns | Nagwa

Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns Mathematics • Third Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

In a right triangle, the difference between the lengths of the perpendicular sides is 7 cm. If the hypotenuse is 35 cm, what is the perimeter of the triangle?

04:35

Video Transcript

In a right triangle, the difference between the lengths of the perpendicular sides is seven centimeters. If the hypotenuse is 35 centimeters, what is the perimeter of the triangle?

We will begin by sketching the right triangle as shown. We are told that the difference between the lengths of the perpendicular sides is seven centimeters. If we let the length of the shorter side be 𝑥 centimeters, this means that the length of the longer side is 𝑥 plus seven centimeters. We are also told that the length of the hypotenuse is 35 centimeters.

We recall that the Pythagorean theorem states that 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the hypotenuse and 𝑎 and 𝑏 are the lengths of the shorter sides. Substituting in the values from this question, we have 𝑥 plus seven all squared plus 𝑥 squared is equal to 35 squared. In order to square 𝑥 plus seven, we can write the parentheses out twice. Using the FOIL method, this simplifies to 𝑥 squared plus seven 𝑥 plus seven 𝑥 plus 49. And collecting like terms, this is equal to 𝑥 squared plus 14𝑥 plus 49. As 35 squared is 1225, we have 𝑥 squared plus 14𝑥 plus 49 plus 𝑥 squared is equal to 1225. We can then subtract 1225 from both sides and collect like terms, giving us two 𝑥 squared plus 14𝑥 minus 1176 equals zero.

We could try to factor the expression on the left-hand side. However, we notice that each of the terms is divisible by two. The quadratic therefore simplifies to 𝑥 squared plus seven 𝑥 minus 588 equals zero. This can be factored into two sets of parentheses, where the first term is 𝑥. The second terms in our parentheses will have a product of negative 588. This is equal to the constant term in our quadratic equation. The sum of these terms will be positive seven, which is equal to the coefficient of 𝑥 in our quadratic.

One way of finding these values is by first writing the factor pairs of 588. One such pair is 21 multiplied by 28. This means that negative 21 multiplied by 28 is equal to negative 588. Negative 21 plus 28 is equal to seven, which means that our two parentheses are 𝑥 minus 21 and 𝑥 plus 28. As the product of these equals zero, either 𝑥 minus 21 equals zero or 𝑥 plus 28 equals zero. This means that either 𝑥 is 21 or negative 28.

Since we are dealing with lengths of a triangle, we know that 𝑥 cannot be negative. The value of 𝑥 is therefore 21. So the side length is 21 centimeters. This is the length of the shortest side of our triangle. The side perpendicular to this is seven centimeters longer. So this is equal to 28 centimeters.

We now have a right triangle with side lengths 21, 28, and 35 centimeters. And we can calculate the perimeter by finding the sum of these. 21 plus 28 plus 35 is equal to 84. The perimeter of the right triangle is 84 centimeters.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy