Question

Please help, thank you!!

Answer 1

Answer as a fraction = 1/7

Answer in decimal form = 0.1428571

The decimal value is approximate.

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Explanation:

We'll be adding and subtracting areas, before we can divide the areas mentioned.

When I write something like "area(ACE)", I'm referring to the area of triangle ACE. It's to save time.

Any time there are 3 letters in the parenthesis for the area function, I'm talking about a triangle. Four letters refers to a quadrilateral. The order of the letters doesn't matter.

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Let's focus on triangle ACE. If we rotate it around so that AC is horizontal, then x is the base. Let h be the height. We can then say:

area(ACE) = 0.5*base*height = 0.5xh

since triangle AEG is the same area, we have

area(AEG) = 0.5xh

and furthermore,

area(ACG) = area(ACE) + area(AEG)

area(ACG) = 0.5xh+0.5xh

area(ACG) = xh

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Let's list out three facts:

1. Segment AC is parallel to segment DB. By the alternate interior angle theorem, this means angle ACE = angle HDG
2. Point G is the midpoint of segment CD, so we know CG = GD
3. By the vertical angles theorem, angle AGC = angle HGD

In short,

1. angle ACE = angle HDG
2. segment CG = segment GD
3. angle AGC = angle HGD

Those observations allow us to use the angle side angle (ASA) congruence theorem to prove that triangle ACG is congruent to triangle HDG. Statements 1 and 3 refer to the "angle" part of "ASA", while statement 2 is the side sandwiched in between said angles.

The key takeaway is that triangles ACG and HDG have the same area. Congruent triangles are identical clones of each other.

So,

area(HDG) = area(ACG)

area(HDG) = xh

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Now let's focus on triangle DEB

This triangle has side lengths 3 times larger compared to the smaller counterpart triangle ACE. These two triangles are similar (we can prove that by the Angle Angle Similarity Theorem)

Since the side lengths are 3 times larger, this makes the area 3^2 = 9 times larger

area(DEB) = 9*area(ACE)

area(DEB) = 9*(0.5xh)

area(DEB) = 4.5xh

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Next we'll use these two facts

• area(DEB) = 4.5xh
• area(HDG) = xh

to find that...

area(EGHB) = area(DEB) - area(HDG)

area(EGHB) = 4.5xh - xh

area(EGHB) = 3.5xh

Another way to find this area is to compute area(HAB) and subtract off the value of area(AEG). You should get the same result as above.

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Hopefully everything makes sense so far. If not, then review the earlier sections or feel free to ask about any step mentioned. There's a lot to keep track of.

At this point we have the areas we need.

• area(AEG) = 0.5xh
• area(EGHB) = 3.5xh

which I'll call m and n respectively

Divide m over n to get:

m/n = (0.5xh)/(3.5xh)

m/n = (0.5)/(3.5)

m/n = 5/35

m/n = 1/7 which is the exact fraction form

m/n = 0.1428571 which is the approximate decimal form

As you can see, the values of x and h don't matter because they cancel out at the end.

Answer 2

Answer:

$\sf \dfrac{Area\:of\: \triangle\: AEG}{Area \: of \: quadrilateral\:EGBH}=\dfrac{1}{7}$

Step-by-step explanation:

If G is the midpoint of CD, and AC is parallel to DB, then AC = DH.

Therefore, G is the midpoint of AH and ΔACE is similar to ΔDBE.

As AC : DB = 1 : 3

⇒ Area of ΔACE : Area of ΔDBE = 1² : 3² = 1 : 9

We are told that Area ΔACE = Area ΔAEG.

⇒ Area ΔACG = 2 × Area ΔACE

As AC = DH, and G is the midpoint of CD:

⇒ ΔACG ≅ ΔHDG

⇒ Area ΔHDG = 2 × Area ΔACE

Area of quadrilateral EGHB = Area of ΔDBE - Area ΔHDG

                                              = Area of ΔDBE - 2 × Area of ΔACE

Therefore:

$\sf \implies \dfrac{Area\:of\: \triangle\: AEG}{Area \: of \: quadrilateral\:EGBH}$

$\sf \implies \dfrac{Area\:of\: \triangle\: ACE}{Area \: of \: \triangle\:DBE - 2 \times Area\:of\: \triangle ACE}$

Using the ratio of Area ΔACE : Area ΔDBE =  1 : 9

$\implies \sf \dfrac{1}{9-2}$

$\implies \sf \dfrac{1}{7}$