If the shaded region in the diagram above is a square, then what fract

Please find the video solution as attached here.



Answer: Option B

That solution is not correct. The method is fine -- the smallest triangles in the picture are similar to the largest ones, so their legs are in a 1 to 3 ratio -- but in the smallest right triangle, with sides x, 3x and 1, you seem to have concluded that x = 1/2. That would make 3x longer than the hypotenuse, so that is clearly wrong. To fix the solution, you need to square 3x correctly:

x^2 + (3x)^2 = 1
x^2 + 9x^2 = 1
x^2 = 1/10
x = √10/10

Then by noticing that the long diagonal lines of length √10 are made up of two lengths x and 3x, along with a side of the square, we find the side of the square is √10 - 4x = √10 - 4(√10/10) = √10 (1 - 2/5) = (3/5)√10, and the area of the square is the square of this, (9/25)(10) = 18/5. To find what fraction this represents of the big square of area 9, we divide this by 9 to get 2/5, so the answer is B.

Another way to do the problem: look at the triangle you get by connecting points A, B and the point on the square closest to A, which I'll call P. That's a right triangle (because we have a 90 degree angle in the square, all of the angles around the square's vertices are 90 degrees) with a hypotenuse of 3. Notice that inside the big square, we have the blue square, and four triangles identical to ABP. So if we could find the area of ABP, we could find the area of the blue square.

Inside triangle ABP, we have a smaller right triangle, on the right of the diagram, with hypotenuse 1. This triangle has the same angles as ABP, so is similar to ABP, but it's hypotenuse is 1/3 as long. If two shapes are similar, and the lengths in one are 1/3 the lengths in the other, the area of the smaller shape will be (1/3)^2 = 1/9 as big. So if the area of the small triangle is x, the area of ABP is 9x. If you now look at the slightly larger right triangle containing AB and the top portion of AD (the part of length 1), that is made up of ABP and another of the small triangles. So its area is 10x. But we know its dimensions; its area is also 3/2. So 10x = 3/2, and x = 3/20. Since the area of ABP is 9x, the area of ABP is 27/20, and the area of four copies of triangle ABP is 27/5. Subtracting that from 9, the area of the big square, we find the area of the blue square is 18/5. We need to divide that by 9 to answer the question, so the answer is 2/5.

From right triangle ADG,

\(AG^2=AD^2+DG^2=9+1\)

\(AG=\sqrt{10}\)

Property: In a right triangle, when a perpendicular is drawn from the vertex with the right angle to the hypotenuse, the right triangle is divided into two similar right triangles each of which is similar to the original right triangle

In triangle ADG, DR is perpendicular to the hypotenuse AG

So triangles ADG & DRG are similar

\(\frac{AG}{DG} = \frac{AD}{DR}\)

\(\frac{\sqrt{10}}{1} = \frac{3}{DR}\)

\(DR=\frac{3}{\sqrt{10}}\)

Similarly, \(\frac{AG}{DG}=\frac{DG}{GR}\)

\(\frac{\sqrt{10}}{1}=\frac{1}{GR}\)

\(GR=\frac{1}{\sqrt{10}}\)

We have \(PR=AG-AP-GR\)

By symmetry, \(AP=DR\)

So, \(PR=AG-DR-GR=\sqrt{10}-\frac{3}{\sqrt{10}}-\frac{1}{\sqrt{10}}\)

\(PR=\frac{6}{\sqrt{10}}\)

So the ratio of area of PQRS to area of ABCD \(= \frac{(\frac{6}{\sqrt{10}})^2}{3^2}\) \(= \frac{\frac{36}{10}}{9}\) \(= \frac{2}{5}\)

Answer is (B)

Please refer to the image.

So there are 4 right angle triangles. Let me take AOB. The area of it is 1/2*3*1=3/2 sq. units. BO is the hypotenuse \sqrt{ \(3^2\)+\(1^2\) }= \sqrt{10}.

We can see AP is the height of the triangle AOB so 3/2=1/2 * AP * \sqrt{10} or, AP=3/ \sqrt{10} .

Now take the right angle triangle AOP, we know AO and AP. So, OP= \sqrt{ \(1^2\)-\((3/ [square_root]10} )^2\) [/square_root] or, OP=1/ \sqrt{10}.

Therefore, each side of the shaded square is \sqrt{10}-3/\sqrt{10}-1/\sqrt{10}=6/ \sqrt{10}

The area of the shaded square becomes 36/10=3.6. The area of the larger square is 3*3=9.

3.6/9=2/5. Option (E) is correct.­