An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF, and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?
Good Question +1.
Let's discuss the step-by-step approach to solve this Question.
Step 1: Draw the figure w.r.t to that data given in the question. The naming of the figure should be as per the info given. Please refer to Fig. 1 in the attachment.
Attachment:
Triangle_GC.JPG [ 54.41 KiB | Viewed 1027 times ]
Step 2: Use the information given and try to figure out the unknown sides.
ADEF is a square. So, let's say the side
AD = AF = FE = ED = a .
Since △ ABC is an equilateral triangle, AB = BC = AC
If you analyze the △ABD and △AFC i.e △1 and △2 in the fig.2, they are 2 right-angled triangles with the same hypotenuse ( AB = AC) and one of the sides is also equal (AD = AF = a).
If the hypotenuse and a side of a right-angled triangle are equivalent to the hypotenuse and a side of the second right-angled triangle, then the two right triangles are said to be congruent. So, we can conclude that the other side of the triangles is also equal. i.e
DB = FC .
Let's say DB = FC = b. Then,
BE = a - b , also
CE =a-b. Please refer the figure 2 for better understanding.
Step 3: We are asked to find the ratio of the area of △ BEC to that of △ADB?
Area of a triangle = 1/2 * base * height
Area of △ BEC = 1/2 * (a-b) * (a-b)
Area of △ ADB= 1/2 * a * b
Area of △ BEC/Area of △ ADB = (1/2 * (a-b) * (a-b))/(1/2 * a * b) =
\((a-b)^2/ab\)Since there are 2 variables in this ratio, we need to convert them to a single variable in order to get the value for this ratio.
For that, we need to find a relationship between a and b.
Analyzing △ 1 and △ 3 in figure 3, we can see that their hypotenuse are equal (△ ABC is equilateral, AB =BC )
Can we equate the hypotenuse in both triangles to find the connection between a and b? Absolutely yes.
\(AB^2 = BC^2\)
In △ 1, \(AB^2\) = \(a^2 + b^2\) and in △ 2, \(BC^2 = (a-b)^2 + (a-b)^2\)
Equating both,
\( a^2 + b^2\) = \((a-b)^2 + (a-b)^2\)\(a^2 + b^2 = 2(a-b)^2\)
In the above equation, we can replace \(a^2 + b^2\) by \((a-b)^2 + 2ab\) ( Using algebraic identities)
\((a-b)^2 + 2ab = 2(a-b)^2\)
\((a-b)^2 = 2ab\)
Area of △ BEC/Area of △ ADB = \((a-b)^2\)/ab. =
2ab/ab = 2Option C is the answer.Thanks,
Clifin J Francis,
GMAT Quant Mentor