the hypotenuse of a right angle triangle is 1 more than twice the base, the altitude is 7 more than the base. find out it's sides
Answers
Answer:
Let the altitude of the right angled triangle be denoted by the variable x cm.
Then, the hypotenuse will be: 2x + 1 cm.
And the base of the right angled triangle is: x + 7 cm.
Using Pythagoras Theorem to find the value of x: (2x+1)² = x² + (x+7)².
= 4x² + 4x + 1 = x² + x² + 14x + 49
= 4x² + 4x + 1 = 2x² + 14x + 49
= 2x² - 10x - 48 = 0
Dividing the equation by (2): x² - 5x - 24 = 0.
The above equation is of the form ax² + bx + c = 0, where a = 1, b = - 5 and c = - 24, which is a quadratic equation.
Using quadratic formula to find the possible value of x:
[refer the attachment]
∵ x is the length of the altitude of the right angled triangle it should be positive.
So, we consider only the addition of the numbers given in the numerator since the 2nd number is larger than the fit and subtraction will give a negative number.
x = 16/ 2
x = 8 cm
2x + 1 = 2(8) + 1 = 16 + 1 = 17 cm
x + 7 = 8 + 7 = 15 cm
Therefore: Altitude = 8 cm
Base = 15 cm
Hypotenuse = 17 cm
Given :
- The hypotenuse of a right angle triangle is 1 more than twice the base.
- The altitude is 7 more than the base.
To Find :
- Sides of the triangle
Solution :
Let,
- the base of the triangle be x
According to the question :
- Hypotenuse will be (2x + 1)
- Altitude(Perpendicular) will be (x + 7)
Using Pythagoras Theorem : H² = P² + B²
⟶ (2x +1)² = (x + 7)² + x²
⟶ 4x² + 1 + 4x = x² + 49 + 14x + x²
⟶ 4x² + 1 + 4x - 2x² - 14x - 49 = 0
⟶ 2x² - 10x - 48 = 0
⟶ 2(x² - 5x - 24) = 0
⟶ x² - 5x - 24 = 0/2
Splitting Middle term
⟶ x² - (8 - 3)x - 24 = 0
⟶ x² - 8x + 3x - 24 = 0
⟶ x(x - 8) + 3(x - 8) = 0
⟶ (x - 8) (x + 3) = 0
⟶ x = 8, x = -3
- Hence, the base of a triangle can't be negative So, the base = 8.
We have,
- Hypotenuse = 2x + 1
⟶ Hypotenuse = 2(8) + 1
⟶ Hypotenuse = 17.
- Altitude = x + 7
⟶ Altitude = 8 + 7
⟶ Altitude = 15.
Hence, We got the sides of the triangle :
- Base = 8
- Hypotenuse = 17
- Altitude = 15