In the 7-inch square above, another square is inscribed. What fraction

So every corner makes a right angled triangle with sides 3-4-5...
And hypotenuse of this triangle is the side of inner square, so 5..

Area of shaded region = \(7^2-5^2=49-25=24\)
Total area =7^2=49

Shaded region as a fraction of total =\(\frac{24}{49}\)

B

_________________

The area of the larger square is 7^2 = 49. We see that each of the shaded regions is a right triangle that is a 3-4-5 right triangle. Thus, each of these right triangles has an area of ½ x 3 x 4 = 6, and the total area of the four triangles is 6 x 4 = 24. So the fraction of the larger square is shaded is 24/49.

Answer: B
_________________

chetan2u ScottTargetTestPrep
I have question ; we are given that a square is inscribed in square ; we can notice that the digonal of the inside square = 7 so side of small square ; s=7/√2
area = 49/2
so shaded region ; 49-49/2 = 49/2
ratio ; 49/2 /49 = 1/2
IMO C

why is this approach and answer wrong?

The highlighted part is not correct, you can not calculate the diagonal of the square that way, chetan2u s approach is IMHO the most elegant way to solve the question...
_________________

Abhishek009
Could you please share reason why won't the diagonal of small square be 7 ? Is it because figure is not drawn to scale or some other reason?

Posted from my mobile device

Bro , the sides of the square inscribed inside the square is 5 units and the sides of the bigger square is 7 units....

Thus the diagonal of the bigger square will be 7√2 and the diagonal of the smaller square will be 5√2

Hope this helps..
_________________

Hi Archit3110 , the diagonal of the inner square would have been 7 units if in the figure, all the points, where the inner square is touching the outer square are the midpoints of the sides of outer square

Posted from my mobile device

Yes, your assumption is wrong. From the picture, you can actually see that the diagonal of the inscribed square is slanted, that is, it’s actually longer than a side of the circumscribed square. The only way the diagonal of the inscribed square is equal to the side length of the circumscribed square is when the vertices of the inscribed square are the midpoints of the sides of the circumscribed square. However, that is not the case here. You can see that the vertices of the inscribed square are a little off from the midpoints of the sides of the circumscribed square.
_________________

How did we arrive at that its a 3-4-5 triangle?

how do we know for sure that it is a 30-60-90 triangle ? what part of question gave u that clue? Please explain how u proved congruency of triangle if so ?
Proof will be appreciated.

VeritasKarishma Bunuel chetan2u ScottTargetTestPrep can u help ?

A 3-4-5 triangle is not the same as 30-60-90 triangle.
30-60-90 has a ratio of 1:sqrt(3):2 sides, not 3:4:5.



As for how do we know that the triangles are congruent, note that a + b = 90 and b + c = 90. So a = c.
So the two triangles have 90 degree angles and a = c. By AA, the triangles are congruent. Since their hypotenuse sides are equal (sides of a square), all other sides would be equal too.
_________________

Discussion:

The triangles in the figure are 3-4-5 triangles, not 30-60-90 triangles. A 30-60-90 triangle has side lengths in the ratio of x : x√3 : 2x and the interior angles of a 3-4-5 right triangle are roughly 53.1 and 36.9.

In order to prove that all the triangles in the figure are congruent, first we show that they are all similar. Call any interior angle of any triangle x. The other interior angle of the same triangle is 90 - x. The interior angle of the triangle which touches the 90 - x degree angle has measure 180 - (90 + 90 - x) = x. Continuing this way, we see that the interior angles of every triangle in the figure are x and 90 - x. Thus, all the triangles are similar.

Once we establish that all the shaded triangles are similar, it just remains to notice that the hypotenuses of all these triangles have the same length; so the triangles are not just similar but also congruent.
_________________

I am not sure if this is correct but I used a quicker method.
I looked at the distance of the inner square's vertex and gauged how close it was to the midpoint of the outer square.
If it were to fall right on the midpoint the ratio would have been 1/2, which is not, so I can eliminate C.
1/2 would be where the ratio would be at its maximum point, so from there it can only decrease, and since D) and E) are over 1/2, they also can be eliminated.
Between A) and B), B) is a lot closer to 1/2 than A) so I picked B). For A) to be correct the vertex of the inner square would have to be a lot closer to the corners of the outer square.

Hope this helps.
Thanks!