Points A,B,C and D are midpoints of the sides of the larger square. If the smaller square has area 60, what is the area of the bigger square?

Guest Jul 23, 2020

#1**0 **

*Points A,B,C and D are midpoints of the sides of the larger square. If the smaller square has area 60, what is the area of the bigger square?*

First, for reference, let's name one of the corners of the large square.

Any of the corners will do, so let's just choose the top left one. Call it **V**.

Since the area of the small square is 60, then each side is length sqrt(60).

DVA is an isoceles right triangle, with hypotenuse length sqrt(60).

Therefore, VA is length sqrt(60) divided by sqrt(2), and the entire side is twice that.

2 • sqrt(60)

The side of the large square is ——————

sqrt(2)

4 • 60

Square that side, and you have the area of the large square ———— thus **A _{largesquare} = 120**

2

_{.}

Guest Jul 23, 2020

#2**0 **

Hey, it's me again. I was looking at the picture and an idea fell out of the sky.

I don't know how to draw on this site, so I'll try to explain it in words.

You have the large square which is the outside one.

You have the small square which is the inside one.

Draw a line from point A to point C and

draw a line from point D to point B.

Now you have four tiny squares, each of which is cut into two equal triangles by a diagonal line.

The small square contains four of those triangles and

the large square contains eight of those triangles.

Thus, the large square contains twice the area of the small square.

The area of the small square is 60, so twice that is the area of the large square, i.e., **120**

Neat, huh? Even if I do say so myself. TTFN.

_{.}

Guest Jul 24, 2020