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Geometry

by VenugopalGurram » Wed Dec 12, 2012 1:43 am
In the figure below AB=BC=CD. If the area of the Traingle CDE is 42,what is the area of the triangle ADG?

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by [email protected] » Wed Dec 12, 2012 8:25 am
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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by cindyb » Sat Sep 21, 2013 10:38 am
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

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Geometry

by Ferguson259 » Wed Sep 19, 2018 1:16 am
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.

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by manchun » Tue Nov 27, 2018 4:00 am
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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Re: Geometry

by harrycharlesop » Fri Oct 09, 2020 9:46 pm
Ferguson259 wrote:
Wed Sep 19, 2018 1:16 am
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.
thank you very much for taking out your time and helping with this