Geometry
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These are similar triangles so the sides of CDE are 1/3 of the sides of ADG. This makes the area of CDE 1/9 of the area of ADG, making ADG 9 x 42 = 378
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1. We are given the area of the top triangle as 42. Use the Area of a triagle formula to determine the Base and Height.
A = 1/2 of the base x the height (A = 1/2 bh)
42 =1/2 base x height
42 x 2 = 1/ 2 x 2 x base x height
84 = base x height
84 = 12 x 7
Next, use the Pythagorean theorum to solve for the hypotenuse.
12 **2 + 7 **2 = c **2
Simplify this to 19.
The segments AB BC and CD are equal, and since you know them to be 19 a piece, you now have enough information to determine the lengths of the base and long side because the triangles are "similar" by a factor of three  3. So you can multiply the sides by three to arrive at the larger scaled triangle.
7 * 3 = 21 (bottom / base)
AND
12 * 3 = 36 ( long side / height)
Now, plug this into the Area of a Triangle = 1/2 base x height.
Area = 1/2 x 21 * 36
Area of ADG is 378
A = 1/2 of the base x the height (A = 1/2 bh)
42 =1/2 base x height
42 x 2 = 1/ 2 x 2 x base x height
84 = base x height
84 = 12 x 7
Next, use the Pythagorean theorum to solve for the hypotenuse.
12 **2 + 7 **2 = c **2
Simplify this to 19.
The segments AB BC and CD are equal, and since you know them to be 19 a piece, you now have enough information to determine the lengths of the base and long side because the triangles are "similar" by a factor of three  3. So you can multiply the sides by three to arrive at the larger scaled triangle.
7 * 3 = 21 (bottom / base)
AND
12 * 3 = 36 ( long side / height)
Now, plug this into the Area of a Triangle = 1/2 base x height.
Area = 1/2 x 21 * 36
Area of ADG is 378

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Since you know AG = 3CE and DG = 3DE ...eq(1),
Now the area of the top most triangle(CDE)
=1/2(EC x DE) = 42 ...eq(2) (given)
And Now, consider the area of the entire triangle(ADG) = 1/2(AG x DG)
= 1/2(3EC x 3DE) replace the values using (eq(1))
= 9/2(EC x DE)
= 9(1/2(EC x DE)) eq(3) in this 1/2(EC x DE) is equal to 42(given) i.e. eq(2) replace it in the eq(3) with the value 42.
= 9 (42) = 378.
I hope this is helpful.
MYBKExperience
Now the area of the top most triangle(CDE)
=1/2(EC x DE) = 42 ...eq(2) (given)
And Now, consider the area of the entire triangle(ADG) = 1/2(AG x DG)
= 1/2(3EC x 3DE) replace the values using (eq(1))
= 9/2(EC x DE)
= 9(1/2(EC x DE)) eq(3) in this 1/2(EC x DE) is equal to 42(given) i.e. eq(2) replace it in the eq(3) with the value 42.
= 9 (42) = 378.
I hope this is helpful.
MYBKExperience
In the figure below AB=BC=CD. If the area of the Traingle CDE is 42,what is the area of the triangle ADG?

similar triangle sides of CDE are 1/3 of the Side of ADG.
9 X Area of triangle CDE = area of Triangle ADG
Area of Triangle ADG =9 X 42
Area of Triangle ADG =378
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similar triangle sides of CDE are 1/3 of the Side of ADG.
9 X Area of triangle CDE = area of Triangle ADG
Area of Triangle ADG =9 X 42
Area of Triangle ADG =378
<a href="https://www.andlearning.org/mathformula/">Math Formulas</a>

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thank you very much for taking out your time and helping with thisFerguson259 wrote: ↑Wed Sep 19, 2018 1:16 amSince you know AG = 3CE and DG = 3DE ...eq(1),
Now the area of the top most triangle(CDE)
=1/2(EC x DE) = 42 ...eq(2) (given)
And Now, consider the area of the entire triangle(ADG) = 1/2(AG x DG)
= 1/2(3EC x 3DE) replace the values using (eq(1))
= 9/2(EC x DE)
= 9(1/2(EC x DE)) eq(3) in this 1/2(EC x DE) is equal to 42(given) i.e. eq(2) replace it in the eq(3) with the value 42.
= 9 (42) = 378.
I hope this is helpful.