MHB What is the multiplier for finding the area of triangle ADG?

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To find the area of triangle ADG, the area formula A = (bh)/2 is applied, where b represents the base and h the height. The area of triangle CDE is expressed in terms of segments AG and DG, leading to the equation (1/3 AG * 1/3 DG)/2 = 42. To determine the area of triangle ADG, a multiplier must be identified for both sides of the equation. The discussion highlights the challenge of interpreting a small image related to the problem. Ultimately, the focus is on deriving the correct multiplier for calculating the area of triangle ADG.
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Not sure how to use a single area and line segments that are same to calculate the areas and line segments for the areas.
 

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Using the formula for the area, $A$, of a triangle

$$A=\frac{bh}{2}$$

where $b$ is the base of the triangle and $h$ is the height of the triangle, can you state the area of triangle $CDE$ in terms of $AG$ and $DG$?
 
(3)42 = 84/2 so b times h equals 84
 
$$\frac{\frac13AG\cdot\frac13DG}{2}=42$$

Now, what number can we multiply both sides of the above equation by to find $\triangle{ADG}$?
 
greg1313 said:
$$\frac{\frac13AG\cdot\frac13DG}{2}=42$$

Now, what number can we multiply both sides of the above equation by to find $\triangle{ADG}$?

Greg, I'm glad you can read that tiny sideways image, because I sure can't. (Bandit)
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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