Question

A rectangular piece of paper ABCD measuring 4 cm x 16 cm is folded along the line MN so that

vertex C coincides with vertex A, as shown in the picture. What is the area of the pentagon ABNMD' ?​

Answer 1

Answer: First, we need to find the length of the line segment MN. Since MN is a fold line, it divides the rectangle into two congruent right triangles ABC and CDM. We can use the Pythagorean theorem to find the length of MN:

AC² + CM² = AM²

Since AC = 16 cm and CM = 2 cm (half the width of the rectangle), we have:

16² + 2² = AM²

256 + 4 = AM²

260 = AM²

AM = sqrt(260) = 2sqrt(65) cm

Since MN is the hypotenuse of right triangle ACM, we have:

MN = 2AM = 4sqrt(65) cm

Now, let's draw a line segment from B to MN, perpendicular to MN, and let the intersection point be E. Since triangle ABN is similar to triangle CDM, we have:

BN/DM = AB/CD

BN/2 = 4/16

BN = 1 cm

Since triangle BEN is a right triangle, we can use the Pythagorean theorem to find the length of BE:

BE² + EN² = BN²

BE² + (MN - DM)² = 1²

BE² + (4sqrt(65) - 2)² = 1

BE² + 64*5 - 16sqrt(65) + 4 = 1

BE² = -64*5 + 16sqrt(65) - 3

BE = sqrt(-64*5 + 16sqrt(65) - 3)

Note that BE is an imaginary number, which means that point E is actually below line segment MN. Therefore, the area of pentagon ABNMD' is zero.

Step-by-step explanation:

Answer 2

Answer: 47 square cm

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Step-by-step explanation:

Grab some graph paper or use GeoGebra.

Place point A at the origin (0,0). Move 16 units to the right to plot B at (16,0).

Then move 4 units up to get to C(16,4). Then move 16 units left to arrive at D(0,4)

Here are the four points so far:

• A = (0,0)
• B = (16,0)
• C = (16,4)
• D = (0,4)

Next draw a line through A and C.

The equation of line AC is y = 0.25x; I'll skip the steps showing how I got that equation. But let me know if you need to see those steps.

The perpendicular bisector of segment AC is the equation y = -4x+34. Use the fact that the perpendicular line has a negative reciprocal slope. Meaning the slope 0.25 has the negative reciprocal -4. Also, use the center point (8,2) to help determine this perpendicular bisector equation.

Why is the perpendicular bisector so important? It's the mirror line. We'll reflect C over this line to land on A.

All points to the right of the mirror line will also reflect over to land somewhere to the left of the mirror. This will form the pentagon ABNMD where segment NM is the mirror line.

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If we were to intersect the mirror line y = -4x+34 with the horizontal line y = 4, then we'll find the intersection point is (7.5,0) which is the location of point M in the diagram below.

Intersect y = 0 (aka the x axis) with y = -4x+34 to find the location of point N(8.5, 0)

So we should have

M = (7.5, 0)

N = (8.5, 0)

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Pentagon ABNMD is composed of the following triangles

• Triangle ADM
• Triangle MNA
• Triangle ABN

But notice carefully that triangle NPQ has folded over mirror line MN to land exactly on top of triangle ABN. This means triangle ABN is congruent to triangle NPQ due to reflectional symmetry.

Also due to the symmetry of the fold, triangle ADM = triangle NPQ

Because of symmetry we have:

• triangle ADM = triangle NPQ
• triangle NPQ = triangle ABN

Apply the transitive property to find triangle ADM = triangle ABN

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area of triangle ADM = 0.5*base*height

area of triangle ADM = 0.5*MD*AD

area of triangle ADM = 0.5*7.5*4

area of triangle ADM = 15

Therefore, triangle ABN is also 15 square cm as well.

area of triangle MNA = 0.5*base*height

area of triangle MNA = 0.5*AN*4

area of triangle MNA = 0.5*8.5*4

area of triangle MNA = 17

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Summary of the triangle areas:

• area of triangle ADM = 15
• area of triangle MNA = 17
• area of triangle ABN = 15

Therefore,

pentagon ABNMD = (triangle ADM)+(triangle MNA)+(triangle ABN)

pentagon ABNMD = (15)+(17)+(15)

pentagon ABNMD = 47 square cm is the final answer.

The diagram is shown below. I used GeoGebra to make it. The diagram is to scale.

Notes:

• P is the old location of point B (where it used to be before the paper folded)
• Q is the old location of point C (where it used to be before the paper folded)