Math / Geometry

QuestionThe length of one leg of a right triangle is 7 cm more than that of the other leg. The length of the hypotenuse is 3 cm more than double that of the shorter leg. Find the lengths of each of the three sides.

Studdy Solution

STEP 1

1. Let the length of the shorter leg be $x$ cm.
2. The length of the longer leg is $x + 7$ cm.
3. The length of the hypotenuse is $2x + 3$ cm.
4. Use the Pythagorean theorem to relate the sides of the triangle.

STEP 2

1. Set up the equation using the Pythagorean theorem.
2. Solve the quadratic equation for $x$ .
3. Calculate the lengths of the legs and the hypotenuse.
4. Verify the solution.

STEP 3

Using 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, set up the equation:
$(x)^2 + (x + 7)^2 = (2x + 3)^2$

STEP 4

Expand each term in the equation:
$x^2 + (x + 7)^2 = (2x + 3)^2$ $x^2 + (x^2 + 14x + 49) = (4x^2 + 12x + 9)$

STEP 5

Combine like terms:
$2x^2 + 14x + 49 = 4x^2 + 12x + 9$

STEP 6

Rearrange the equation to bring all terms to one side:
$0 = 4x^2 + 12x + 9 - 2x^2 - 14x - 49$ $0 = 2x^2 - 2x - 40$

STEP 7

Simplify the equation by dividing all terms by 2:
$0 = x^2 - x - 20$

STEP 8

Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -1$ , and $c = -20$ :
$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1}$ $x = \frac{1 \pm \sqrt{1 + 80}}{2}$ $x = \frac{1 \pm \sqrt{81}}{2}$

STEP 9

Calculate the roots:
$x = \frac{1 \pm 9}{2}$
This gives two possible solutions:
$x = \frac{10}{2} = 5$ $x = \frac{-8}{2} = -4$
Since a length cannot be negative, we take $x = 5$ .

STEP 10

Calculate the lengths of the other sides using $x = 5$ :
Shorter leg: $x = 5$ cm
Longer leg: $x + 7 = 5 + 7 = 12$ cm
Hypotenuse: $2x + 3 = 2(5) + 3 = 10 + 3 = 13$ cm

STEP 11

Verify the solution by checking the Pythagorean theorem:
$5^2 + 12^2 = 13^2$ $25 + 144 = 169$ $169 = 169$
The solution satisfies the Pythagorean theorem.
The lengths of the sides of the triangle are:
Shorter leg: $5$ cm
Longer leg: $12$ cm
Hypotenuse: $13$ cm

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QuestionThe length of one leg of a right triangle is 7 cm more than that of the other leg. The length of the hypotenuse is 3 cm more than double that of the shorter leg. Find the lengths of each of the three sides.

Studdy Solution

STEP 1

1. Let the length of the shorter leg be $x$ cm.
2. The length of the longer leg is $x + 7$ cm.
3. The length of the hypotenuse is $2x + 3$ cm.
4. Use the Pythagorean theorem to relate the sides of the triangle.

STEP 2

1. Set up the equation using the Pythagorean theorem.
2. Solve the quadratic equation for $x$ .
3. Calculate the lengths of the legs and the hypotenuse.
4. Verify the solution.

STEP 3

Using 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, set up the equation:
$(x)^2 + (x + 7)^2 = (2x + 3)^2$

STEP 4

Expand each term in the equation:
$x^2 + (x + 7)^2 = (2x + 3)^2$ $x^2 + (x^2 + 14x + 49) = (4x^2 + 12x + 9)$

STEP 5

Combine like terms:
$2x^2 + 14x + 49 = 4x^2 + 12x + 9$

STEP 6

Rearrange the equation to bring all terms to one side:
$0 = 4x^2 + 12x + 9 - 2x^2 - 14x - 49$ $0 = 2x^2 - 2x - 40$

STEP 7

Simplify the equation by dividing all terms by 2:
$0 = x^2 - x - 20$

STEP 8

Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -1$ , and $c = -20$ :
$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1}$ $x = \frac{1 \pm \sqrt{1 + 80}}{2}$ $x = \frac{1 \pm \sqrt{81}}{2}$

STEP 9

Calculate the roots:
$x = \frac{1 \pm 9}{2}$
This gives two possible solutions:
$x = \frac{10}{2} = 5$ $x = \frac{-8}{2} = -4$
Since a length cannot be negative, we take $x = 5$ .

STEP 10

Calculate the lengths of the other sides using $x = 5$ :
Shorter leg: $x = 5$ cm
Longer leg: $x + 7 = 5 + 7 = 12$ cm
Hypotenuse: $2x + 3 = 2(5) + 3 = 10 + 3 = 13$ cm

STEP 11

Verify the solution by checking the Pythagorean theorem:
$5^2 + 12^2 = 13^2$ $25 + 144 = 169$ $169 = 169$
The solution satisfies the Pythagorean theorem.
The lengths of the sides of the triangle are:
Shorter leg: $5$ cm
Longer leg: $12$ cm
Hypotenuse: $13$ cm

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QuestionThe length of one leg of a right triangle is 7 cm more than that of the other leg. The length of the hypotenuse is 3 cm more than double that of the shorter leg. Find the lengths of each of the three sides.

Studdy Solution

STEP 1

1. Let the length of the shorter leg be x x x cm.2. The length of the longer leg is x+7 x + 7 x+7 cm.3. The length of the hypotenuse is 2x+3 2x + 3 2x+3 cm.4. Use the Pythagorean theorem to relate the sides of the triangle.

STEP 2

1. Set up the equation using the Pythagorean theorem.2. Solve the quadratic equation for x x x.3. Calculate the lengths of the legs and the hypotenuse.4. Verify the solution.

STEP 3

Using 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, set up the equation:(x)2+(x+7)2=(2x+3)2 (x)^2 + (x + 7)^2 = (2x + 3)^2 (x)2+(x+7)2=(2x+3)2

STEP 4

Expand each term in the equation:x2+(x+7)2=(2x+3)2 x^2 + (x + 7)^2 = (2x + 3)^2 x2+(x+7)2=(2x+3)2 x2+(x2+14x+49)=(4x2+12x+9) x^2 + (x^2 + 14x + 49) = (4x^2 + 12x + 9) x2+(x2+14x+49)=(4x2+12x+9)

STEP 5

Combine like terms:2x2+14x+49=4x2+12x+9 2x^2 + 14x + 49 = 4x^2 + 12x + 9 2x2+14x+49=4x2+12x+9

STEP 6

Rearrange the equation to bring all terms to one side:0=4x2+12x+9−2x2−14x−49 0 = 4x^2 + 12x + 9 - 2x^2 - 14x - 49 0=4x2+12x+9−2x2−14x−49 0=2x2−2x−40 0 = 2x^2 - 2x - 40 0=2x2−2x−40

STEP 7

Simplify the equation by dividing all terms by 2:0=x2−x−20 0 = x^2 - x - 20 0=x2−x−20

STEP 8

STEP 9

Calculate the roots:x=1±92 x = \frac{1 \pm 9}{2} x=21±9​This gives two possible solutions:x=102=5 x = \frac{10}{2} = 5 x=210​=5 x=−82=−4 x = \frac{-8}{2} = -4 x=2−8​=−4Since a length cannot be negative, we take x=5 x = 5 x=5.

STEP 10

Calculate the lengths of the other sides using x=5 x = 5 x=5:Shorter leg: x=5 x = 5 x=5 cmLonger leg: x+7=5+7=12 x + 7 = 5 + 7 = 12 x+7=5+7=12 cmHypotenuse: 2x+3=2(5)+3=10+3=13 2x + 3 = 2(5) + 3 = 10 + 3 = 13 2x+3=2(5)+3=10+3=13 cm

STEP 11

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1. Let the length of the shorter leg be $x$ cm.
2. The length of the longer leg is $x + 7$ cm.
3. The length of the hypotenuse is $2x + 3$ cm.
4. Use the Pythagorean theorem to relate the sides of the triangle.

1. Set up the equation using the Pythagorean theorem.
2. Solve the quadratic equation for $x$ .
3. Calculate the lengths of the legs and the hypotenuse.
4. Verify the solution.

Using 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, set up the equation:
$(x)^2 + (x + 7)^2 = (2x + 3)^2$

Expand each term in the equation:
$x^2 + (x + 7)^2 = (2x + 3)^2$ $x^2 + (x^2 + 14x + 49) = (4x^2 + 12x + 9)$

Combine like terms:
$2x^2 + 14x + 49 = 4x^2 + 12x + 9$

Rearrange the equation to bring all terms to one side:
$0 = 4x^2 + 12x + 9 - 2x^2 - 14x - 49$ $0 = 2x^2 - 2x - 40$

Simplify the equation by dividing all terms by 2:
$0 = x^2 - x - 20$

Calculate the roots:
$x = \frac{1 \pm 9}{2}$
This gives two possible solutions:
$x = \frac{10}{2} = 5$ $x = \frac{-8}{2} = -4$
Since a length cannot be negative, we take $x = 5$ .

Calculate the lengths of the other sides using $x = 5$ :
Shorter leg: $x = 5$ cm
Longer leg: $x + 7 = 5 + 7 = 12$ cm
Hypotenuse: $2x + 3 = 2(5) + 3 = 10 + 3 = 13$ cm