In the diagram above, triangle ABC is equilateral, figure SQRE is a sq

its C.

side of triangle is 2.draw a perpendicular from A to the bottom opf the square.Take it as point Z.

since AC =2,AE=sqrt3(30,60,90 trianghoe)


now AQ is sqrt3/2 (as it half side of sqare)

now AQY is 30,60,90 triangle too. with AQ=sqrt3/2

so QY=3/2(sqrt3 *sqrt3/2)

yr=qr-qy=sqrt3/2-3/2

option C

find side of triangle: 6/3 = 2 inch per side
find area of equilateral triangle: 2²* sqrt(3) / 4 = sqrt (3)
find height: area of triangle = sqrt(3) = 1/2 * base * height -> solve for height: sqrt (3) = 1/2 * 2 * height -> height= sqrt (3)
find third side: its a 90° triangle -> ratio is 1:sqrt(3):2 -> third site = 1 inch.
find RB: (2 - sqrt(3)) / 2
find RY: use Thales' theorem -> 1 / sqrt(3) = RB / RY -> 1 / sqrt(3) = ((2 - sqrt(3)) / 2) / RY -> solve for RY -> RY = sqrt(3) - 1.5

C

hsbinfy :
Can you please explain how did you get QY=3/2.I am getting it as 1/2.
Please explain how did you apply 30-6-90 to that triangle and how did you got tht value.

Thanks

See the attachment.

For 30-60-90 triangle, sides are in the ratio 1 : √3 : 2.
You just have to apply this rule to the triangles in the figure.

You will see that RY = √3 - 1.5

Answer C

Here's the step-by-step guide to thinking through this question.

Let the length of RY be x inches.



We are given that Triangle ABC is equilateral. This means, Angle ABC = 60 degrees

In right triangle BRY,

\(\frac{RY}{BR} = tan60 =\sqrt{3}\)

That is, \(\frac{x}{BR} = \sqrt{3}\) . . . (1)

If we can find the value of BR, we will be able to find the value of x.

So, let's try to find more about BR now.

We are given that the perimeter of the equilateral triangle ABC = 6 inches

This means, each side of triangle ABC = 6/3 = 2 inches

Now, let each side of square QRES be 2a units.

Since we are given that A is the mid-point of side QS, this means that equilateral triangle ABC is placed symmetrically about the square QRES.

Therefore, \(CE = BR = \frac{(2 - 2a)}{2} = 1 - a\) . . . (2)

Substituting (2) in (1), we get:

\(\frac{x}{(1-a)} = \sqrt{3}\) . . . (3)

Equation 3 contains two unknowns: x and a. So, to find a unique value of x, we now know that we should try to find another relation involving x and/or a.

We get it by dropping a perpendicular from A on side BC.

In right triangle APB,

\(\frac{AP}{BP} = tan60 = \sqrt{3}\)

That is, \(\frac{2a}{1} = \sqrt{3}\)

That is, \(a = \frac{(sqrt3)}{2}\) . . . (4)

By solving (3) and (4), we get \(x = \sqrt{3} - 1.5\)

Hope this helped!

Best Regards

Japinder

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I can not understand how we get 1.5 i understand how we got Square root 3 please expert help. thank you in advance.

All of you are doing this problem incorrectly if you even begin to do the algebra or geometry.

If you know that the side of the triangle is 2 inches, then you know the length of RY has to be significantly less than that, so take a look at the answer choices.

A) 1/2 - way too big.
B) 1.5 - way too big.
C) .2 - close, keep this in mind
D) .3 - close, but still too big.
E) .85 - way too big.

The answer is between C and D, and because that segment looks a lot closer to 1/5 than to 1/3, go with answer C.

Time ~20 seconds

Here is my solution. Since, this one had a diagram and lot of rules, I posted an image. Hope this helps.

Different approach to solve this question.

In the diagram above, triangle ABC is equilateral and its perimeter is 6 inches => side of triangle is 2 inch
Also, figure SQRE is a square, and A is the midpoint of SQ
Clearly, RQ (side of square) < Side of triangle = 2 inch

We are required to find length of segment RY which is part of Side of a square.

If we see the options A, B and E can be immediately ruled out as they are quite high a value for line segment RY.
Option D is also incorrect as it is the difference between 2 and sqrt3 and we know that side of a square is less than 2 inch
so we are left with Option C.

PS: This approach is not a standard method to solve, however, it may come handy if you are short of time and helps you in making a calculative guess.

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