Find the area of the middle region for the shape.

Question

One corner angle of a square of side 4 inches is trisected. Find the area of the middle region.

Answer 1

Let A be the area of the square, A₁ is the upper region, A₃ is the lower region and A₂ is the middle region.

A = 4² = 16 sq. inch

A₁ = A₃

A₃ is a 30-60-90 triangle. To get the length of the shorter leg (y) it will be:

y√3 =4

y =4/√3

y = (4√3) / 3.

Therefore the slope of the line f(x) is m = ((4√3) / 3)/4 = (√3) / 3 with a y-intercept of 0.

Using the slope-intercept form of linear equation we have:

f(x) = (√3 / 3)x

Using definite integral:

      ₄

A₃ = ∫ (√3 / 3)x dx

       ⁰

A₂ = A - (A₁+A₃). Since A₁ = A₃

A₂ = A - (A₃+A₃)

A₂ = A - 2A₃

                      ₄

A₂ = 16 - 2 ∫ (√3 / 3)x dx 

                ⁰

                                         ₄

A₂ = 16 - 2 [(√3 / 6)x²] 

                            ⁰

A₂ = 16 - 2 [(16√3) / 6 - 0]

A₂ = 16 - (16√3) / 3

A₂ = 16 - (16√3) / 3

A₂ = (48-16√3) /3 ≈ 6.762 sq. in.

Answer 2

When you trisect the corner angle, each angle will be 30°

The 2 "side" triangles will be congruent 30-60 sight triangles

The middle area will be the area of the square less the area of the 2 congruent triangles.

The short side of the right triangle will be (1/3)s√3.

The area will be (1/2)s * (1/3)s√3.

The are of the 2 "side" triangles will then b s*(1/3)s√3 and the area of the middle section will be s²-(1/3)s²√3.

I will leave it to you to plug in the numbers and do the arithmetic.