Math / Geometry

QuestionThe length of the longer leg of a right triangle is 19 cm more than five times the length of the shorter leg. The length of the hypotenuse is 20 cm more than five times the length of the shorter leg. Find the side lengths of the triangle.
Length of the shorter leg: $\square$ cm
Length of the longer leg: $\square$ cm Length of the $\square$ cm hypotenuse: 5

Studdy Solution

STEP 1

1. We have a right triangle with sides: shorter leg, longer leg, and hypotenuse.
2. The length of the longer leg is 19 cm more than five times the length of the shorter leg.
3. The length of the hypotenuse is 20 cm more than five times the length of the shorter leg.
4. We will use the Pythagorean Theorem to solve for the side lengths.

STEP 2

1. Define variables
2. Set up equations based on the problem statement
3. Use the Pythagorean Theorem to form an equation
4. Solve the system of equations
5. Verify the solution

STEP 3

Define the variable $x$ as the length of the shorter leg of the triangle.

STEP 4

Express the lengths of the other sides in terms of $x$ : - The longer leg is $5x + 19$ cm. - The hypotenuse is $5x + 20$ cm.

STEP 5

Apply the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$(5x + 20)^2 = x^2 + (5x + 19)^2$

STEP 6

Expand the squares in the equation:
$(5x + 20)^2 = (5x + 20)(5x + 20) = 25x^2 + 200x + 400$
$(5x + 19)^2 = (5x + 19)(5x + 19) = 25x^2 + 190x + 361$

STEP 7

Substitute these expanded forms back into the Pythagorean equation:
$25x^2 + 200x + 400 = x^2 + 25x^2 + 190x + 361$

STEP 8

Simplify the equation:
$25x^2 + 200x + 400 = 26x^2 + 190x + 361$

STEP 9

Rearrange the equation to bring all terms to one side:
$0 = 26x^2 + 190x + 361 - 25x^2 - 200x - 400$
$0 = x^2 - 10x - 39$

STEP 10

Solve the quadratic equation $x^2 - 10x - 39 = 0$ using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -10$ , and $c = -39$ .

STEP 11

Calculate the discriminant:
$b^2 - 4ac = (-10)^2 - 4(1)(-39) = 100 + 156 = 256$

STEP 12

Substitute the values into the quadratic formula:
$x = \frac{-(-10) \pm \sqrt{256}}{2(1)}$
$x = \frac{10 \pm 16}{2}$

STEP 13

Calculate the two possible solutions for $x$ :
$x = \frac{10 + 16}{2} = 13$
$x = \frac{10 - 16}{2} = -3$
Since a side length cannot be negative, $x = 13$ .

STEP 14

Verify the solution by calculating the other sides:
- Longer leg: $5x + 19 = 5(13) + 19 = 65 + 19 = 84$ cm - Hypotenuse: $5x + 20 = 5(13) + 20 = 65 + 20 = 85$ cm
Check using the Pythagorean Theorem:
$85^2 = 13^2 + 84^2$
$7225 = 169 + 7056$
$7225 = 7225$
The solution is verified.
The side lengths of the triangle are: - Length of the shorter leg: $\boxed{13}$ cm - Length of the longer leg: $\boxed{84}$ cm - Length of the hypotenuse: $\boxed{85}$ cm

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Studdy Solution

STEP 1

STEP 2

1. Define variables
2. Set up equations based on the problem statement
3. Use the Pythagorean Theorem to form an equation
4. Solve the system of equations
5. Verify the solution

STEP 3

Define the variable $x$ as the length of the shorter leg of the triangle.

STEP 4

Express the lengths of the other sides in terms of $x$ : - The longer leg is $5x + 19$ cm. - The hypotenuse is $5x + 20$ cm.

STEP 5

Apply the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$(5x + 20)^2 = x^2 + (5x + 19)^2$

STEP 6

Expand the squares in the equation:
$(5x + 20)^2 = (5x + 20)(5x + 20) = 25x^2 + 200x + 400$
$(5x + 19)^2 = (5x + 19)(5x + 19) = 25x^2 + 190x + 361$

STEP 7

Substitute these expanded forms back into the Pythagorean equation:
$25x^2 + 200x + 400 = x^2 + 25x^2 + 190x + 361$

STEP 8

Simplify the equation:
$25x^2 + 200x + 400 = 26x^2 + 190x + 361$

STEP 9

Rearrange the equation to bring all terms to one side:
$0 = 26x^2 + 190x + 361 - 25x^2 - 200x - 400$
$0 = x^2 - 10x - 39$

STEP 10

Solve the quadratic equation $x^2 - 10x - 39 = 0$ using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -10$ , and $c = -39$ .

STEP 11

Calculate the discriminant:
$b^2 - 4ac = (-10)^2 - 4(1)(-39) = 100 + 156 = 256$

STEP 12

Substitute the values into the quadratic formula:
$x = \frac{-(-10) \pm \sqrt{256}}{2(1)}$
$x = \frac{10 \pm 16}{2}$

STEP 13

Calculate the two possible solutions for $x$ :
$x = \frac{10 + 16}{2} = 13$
$x = \frac{10 - 16}{2} = -3$
Since a side length cannot be negative, $x = 13$ .

STEP 14

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Studdy Solution

STEP 1

STEP 2

1. Define variables2. Set up equations based on the problem statement3. Use the Pythagorean Theorem to form an equation4. Solve the system of equations5. Verify the solution

STEP 3

Define the variable x x x as the length of the shorter leg of the triangle.

STEP 4

Express the lengths of the other sides in terms of x x x: - The longer leg is 5x+19 5x + 19 5x+19 cm. - The hypotenuse is 5x+20 5x + 20 5x+20 cm.

STEP 5

Apply the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:(5x+20)2=x2+(5x+19)2 (5x + 20)^2 = x^2 + (5x + 19)^2 (5x+20)2=x2+(5x+19)2

STEP 6

Expand the squares in the equation:(5x+20)2=(5x+20)(5x+20)=25x2+200x+400 (5x + 20)^2 = (5x + 20)(5x + 20) = 25x^2 + 200x + 400 (5x+20)2=(5x+20)(5x+20)=25x2+200x+400(5x+19)2=(5x+19)(5x+19)=25x2+190x+361 (5x + 19)^2 = (5x + 19)(5x + 19) = 25x^2 + 190x + 361 (5x+19)2=(5x+19)(5x+19)=25x2+190x+361

STEP 7

Substitute these expanded forms back into the Pythagorean equation:25x2+200x+400=x2+25x2+190x+361 25x^2 + 200x + 400 = x^2 + 25x^2 + 190x + 361 25x2+200x+400=x2+25x2+190x+361

STEP 8

Simplify the equation:25x2+200x+400=26x2+190x+361 25x^2 + 200x + 400 = 26x^2 + 190x + 361 25x2+200x+400=26x2+190x+361

STEP 9

Rearrange the equation to bring all terms to one side:0=26x2+190x+361−25x2−200x−400 0 = 26x^2 + 190x + 361 - 25x^2 - 200x - 400 0=26x2+190x+361−25x2−200x−4000=x2−10x−39 0 = x^2 - 10x - 39 0=x2−10x−39

STEP 10

Solve the quadratic equation x2−10x−39=0 x^2 - 10x - 39 = 0 x2−10x−39=0 using the quadratic formula:x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x=2a−b±b2−4ac​​where a=1 a = 1 a=1, b=−10 b = -10 b=−10, and c=−39 c = -39 c=−39.

STEP 11

Calculate the discriminant:b2−4ac=(−10)2−4(1)(−39)=100+156=256 b^2 - 4ac = (-10)^2 - 4(1)(-39) = 100 + 156 = 256 b2−4ac=(−10)2−4(1)(−39)=100+156=256

STEP 12

Substitute the values into the quadratic formula:x=−(−10)±2562(1) x = \frac{-(-10) \pm \sqrt{256}}{2(1)} x=2(1)−(−10)±256​​x=10±162 x = \frac{10 \pm 16}{2} x=210±16​

STEP 13

Calculate the two possible solutions for x x x:x=10+162=13 x = \frac{10 + 16}{2} = 13 x=210+16​=13x=10−162=−3 x = \frac{10 - 16}{2} = -3 x=210−16​=−3Since a side length cannot be negative, x=13 x = 13 x=13.

STEP 14

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1. Define variables
2. Set up equations based on the problem statement
3. Use the Pythagorean Theorem to form an equation
4. Solve the system of equations
5. Verify the solution

Define the variable $x$ as the length of the shorter leg of the triangle.

Express the lengths of the other sides in terms of $x$ : - The longer leg is $5x + 19$ cm. - The hypotenuse is $5x + 20$ cm.

Apply the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$(5x + 20)^2 = x^2 + (5x + 19)^2$

Expand the squares in the equation:
$(5x + 20)^2 = (5x + 20)(5x + 20) = 25x^2 + 200x + 400$
$(5x + 19)^2 = (5x + 19)(5x + 19) = 25x^2 + 190x + 361$

Substitute these expanded forms back into the Pythagorean equation:
$25x^2 + 200x + 400 = x^2 + 25x^2 + 190x + 361$

Simplify the equation:
$25x^2 + 200x + 400 = 26x^2 + 190x + 361$

Rearrange the equation to bring all terms to one side:
$0 = 26x^2 + 190x + 361 - 25x^2 - 200x - 400$
$0 = x^2 - 10x - 39$

Solve the quadratic equation $x^2 - 10x - 39 = 0$ using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -10$ , and $c = -39$ .

Calculate the discriminant:
$b^2 - 4ac = (-10)^2 - 4(1)(-39) = 100 + 156 = 256$

Substitute the values into the quadratic formula:
$x = \frac{-(-10) \pm \sqrt{256}}{2(1)}$
$x = \frac{10 \pm 16}{2}$

Calculate the two possible solutions for $x$ :
$x = \frac{10 + 16}{2} = 13$
$x = \frac{10 - 16}{2} = -3$
Since a side length cannot be negative, $x = 13$ .