The area of the right angled triangle abc (x+4)(x+3) is twice the area of a square with sides x cm.Write a quadratic equation in terms of x.
To solve this problem, we need to understand the formulas for the areas of a right-angled triangle and a square.
The area of a right-angled triangle is given by the formula: Area = (1/2) * base * height.
Given the sides of the triangle as (x + 4) and (x + 3), the base and height can be taken as (x + 4) and (x + 3) respectively.
The area of a square is given by the formula: Area = side^2.
Given that the side of the square is x cm, the area of the square can be written as x^2.
Now, according to the problem statement, the area of the triangle is twice the area of the square:
(1/2) * (x + 4) * (x + 3) = 2 * x^2
To eliminate the fraction, multiply both sides of the equation by 2:
(x + 4)(x + 3) = 4x^2
Now, we can expand the expression on the left-hand side:
x * x + x * 3 + 4 * x + 4 * 3 = 4x^2
Simplifying further:
x^2 + 3x + 4x + 12 = 4x^2
x^2 + 7x + 12 = 4x^2
Moving all terms to the left-hand side:
4x^2 - x^2 - 7x - 12 = 0
Combining like terms:
3x^2 - 7x - 12 = 0
Therefore, the quadratic equation in terms of x is: 3x^2 - 7x - 12 = 0.