An Equilateral Triangle ABC is Inscribed in Square ADEF GMAT Problem Solving

Question: 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?

1. \(\frac{4}{3} \)
2. \(\sqrt{3}\)
3. \(\sqrt{2} \)
4. \(\frac{5}{2}\)
5. \(\sqrt{5} \)

“An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF, and BEC.” - this is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. In the GMAT Problem Solving section, examiners measure how well the candidates make analytical and logical approaches to solve numerical problems. In this section, candidates have to evaluate and interpret data from a given graphical representation. In this section, mostly one finds out mathematical questions. Five answer choices are given for each GMAT Problem solving question.

Solution and Explanation:

Approach Solution:

This question has only one approach.

Explanation:

Given to us that an equilateral triangle is inscribed in a square ADEF. It forms three right triangles: ADB, ACF, and BEC. It is asked to find out the ratio of the area of triangle BEC to that of the triangle ADB.
A triangle is inscribed in the square means all the sides of the triangle are on the sides of the square and the triangle lies inside the square.
Firstly some properties of triangles and squares should be known to you

• All the sides in an equilateral triangle are equal.
• All the sides make an angle of 60 degrees with each other.
• All the angles of a square are 90 degrees.
• All sides of the square are equal and perpendicular to each other.

Consider the square ADEF,

Let the equilateral triangle be ABC.
Let point B of the triangle divide the side DE into two ratios y:x respectively
Total length of the side of the square will become = x+y.
EC = BE = x
CF = DB = y
From the given diagram we can see that
In the triangle ADB and BEC
It is clearly visible that the hypotenuse AB and BC are the same because the triangle ABC is equilateral and has all sides equal.

So we can write,
In triangle ADB and BEC

\(AD^2+DB^2=BE^2+EC^2\)

=> \(y^2 +(x+y)^2=x^2+x^2\)
=> \(y^2+x^2+y^2+2xy=2x^2\)
=> \(2y^2+2xy=x^2\)
=> x = \(\sqrt{2y^2 + 2xy}\)

Now firstly it is important to find the area of both the triangles to get the ratio.
Let A1 be the area of triangle BEC and A2 be the area of triangle BEC
A1 = \(\frac{1}{2}\)* b* h = \(\frac{1}{2}\) * x * x
Putting the value of x we get,

A1 = \(\frac{1}{2}\) * \(\sqrt{2y^2 + 2xy}\) * \(\sqrt{2y^2 + 2xy}\)

A1 = \(\frac{1}{2}\) * (\(2y^2 + 2xy\)) = \(y^2\)+ xy
Area of triangle ADB = A2 = \(\frac{1}{2}\)*b*h
A2 = \(\frac{1}{2}\) (y * (x+y))

Ratio = area of BEC / area of ADB = A1 /A2
=\(y^2\) + xy / (\(\frac{1}{2}\)(\(y^2\) + xy)
=1/(\(\frac{1}{2}\))
= 2
Therefore the correct answer is option C.

Correct Answer: C

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