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lesson 2.4 pt 4 assignment geometry write and solve an equation to find…

Question

lesson 2.4 pt 4 assignment
geometry write and solve an equation to find the value of x so that the figures have the same area
area of rectangle = length(width)
area of triangle = 1/2(base)(height)
3(x + 6) = 2(8)(x + 4) - 24
3(x + 6) = 1/2(8)(x + 4) 2
3(x + 6) = 1/2(8)(x + 6) - 72
3(x + 4) = 1/2(8)(x + 6) - 72
3(x + 6) = 1/2(x + 4) - 28

Explanation:

Step1: Find the area of the triangle

The formula for the area of a triangle is \(A=\frac{1}{2}\times base\times height\). Here, the base is \(8\) and the height is \(x + 2\). So the area of the triangle \(A_{triangle}=\frac{1}{2}\times8\times(x + 2)=4(x + 2)=4x+8\).

Step2: Find the area of the rectangle

The formula for the area of a rectangle is \(A = length\times width\). Here, the length is \(x + 6\) and the width is \(2\). So the area of the rectangle \(A_{rectangle}=2\times(x + 6)=2x+12\).

Step3: Set the two - areas equal

Since the areas are equal, we have the equation \(4x + 8=2x+12\).
Subtract \(2x\) from both sides: \(4x-2x + 8=2x-2x+12\), which simplifies to \(2x+8 = 12\).
Subtract \(8\) from both sides: \(2x+8 - 8=12 - 8\), so \(2x=4\).
Divide both sides by \(2\): \(x = 2\).

Answer:

\(x = 2\)