One side length of a left square is m and one side length of a right

Did not follow your explanation.You have made some mistakes which I have highlighted and also I couldn't understand as to whyCG=GF=m*DG=n*GH

Can you please explain how A is the answer.

How can CG = CF. CG is side of a Square and CF is diagonal. They can never be equal. Can you please explain.

Thank you for the question! Please refer to the attachment.

There is no way I could have solved this in 2 mins

This question can actually become very easy.
The trick on these geometry questions is that you can assume things to your convenience, as long as it does not violate any of the conditions on the question.
We can easily assume m=n, we don't violate any conditions on the qus. The answer is pretty straightforward then.

One side length of a left square is m and one side length of a right square is n. What is the area of a middle square, in terms of m and n?

Please refer the attached diagram.
Given that \(BC = m\) and \(EH = n\)

To solve this question use the fact that middle figure is also square. => BJ = JH => \(CJ = n\) and \(JE = m\).

Hence, \(BJ = \sqrt{m^{2} + n^{2}}\) and area of middle square = \({m^{2} + n^{2}\)

The simplest way to solve this question is to assume the base angle in left right angled triangle as x. Since angle in square is 90deg. So the base angle in the right triangle is 90-x. Now(for left triangle) sin(90-x)=cosx = m/a and (for right triangle) sin x=n/a. Now use cos^2 x + sin^2 x =1. And you have result without Pythagoras theorem.

Hi,

Can someone please explain how did we go from BJ = CJ to CJ = n and JE = m ?

is that because the two triangles are congruent ?

I don't understand how CG=CF?

Can anybody share a simple solution to this problem

Based on the diagram- m^2 < n^2. Hence eliminate option B.
In all the other options the dimensions are not correct- meaning- area has to be of the form (length)^2. None of the other options are of that form (you can't do m + n^2 or n + m^2, it's like adding 1 metre and 1 sq.metre)

Hence answer has to be Option A.

Using this method you can solve this in 20 seconds max.

My approach.
Name LHS Triangle as ABC with B as 90 deg and RHS Triangle as EDC with angle D as 90 deg.
In LHS triangle ABC
Angle B= 90 deg, Assume angle ACB=x, then angle BAC=90-x. -------------(1)
Angle ACE = 90 deg which means angle ECD = 90-x (Angles ACB, ACE and ECD add up to 180 deg). -------------(2)
Now RHS triangle EDC,
Angle D = 90 deg, angle ECD = 90-x and angle CED = x -------------------(3)
From (1) & (2), triangles ABC and EDC are similar triangles.
AC/BC = CE/ED
a/BC=a/n, BC = n.
Also, AC^2 = AB^2+BC^2, a^2 = m^2+n^2. And we know that a^2 is the area of the middle square. So Area of the middle square is m^2+n^2.

Brilliant solution, i tried using Pytha and got lost... then tried to make it easier by assuming M = N and viola got the answer..

Have to use the rule that:

The sum of 2 remote interior angles are equal to the exterior angle

If you use that rule and label the angles

Then realize that the hypotenuse of each right triangle is equal length because it is one of the sides of the same tilted square

you will see that the 2 right triangles are Congruent Right Triangles:

Labeling the Tilted Square’s Side = H = hypotenuse of 2 right triangles

The Right triangle to the LEFT of the titled square will have the horizontal leg = N

And the Right triangle to the RIGHT of the tilted square will have the horizontal leg = M


Area of the tilted square = (H)(H) = (H)^2

And you can use Pythagorean Theorem with either triangle to determine:

(M)^2 + (N)^2 = (H)^2


Answer
(A)

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Look at the picture. Make up some numbers. Check the answer choices. Get the question right in 30 seconds without messing around with variables. Move on!!

Two ways.

First, the really easy way if you pay attention to the opportunity to manipulate the drawing (which you should always look to do!!). Make the two side squares the same size and m=n=2. Now the side of the middle square is \(2\sqrt{2}\), so the area is 8.

A. 4+4 Keep it.
B. 4-4 Nope.
C. 2+4 Nope
D. 4+2 Nope.
E. 2+2 Nope.

Answer choice A.


Second, use the drawing as it appears and estimate. Let's make up values that are reasonably close to what we see on the screen. How about m=2 and n=3? Left square is area 4. Right square is area 9. Middle square is area of what? More than 9, but less than 2*9? Fine.
A. 4+9 Keep it.
B. 4-9 Nope.
C. 2+9 Keep it.
D. 4+3 Nope.
E. 2+3 Nope.

Down to A and C. Is the middle square really only the right one plus half the left one? That feels too small. Plus, the relationship between the middle square and each of the side ones is the same, so why wouldn't m and n both be treated the same way in the answer? I don't like C.

Answer choice A.