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How to find the area of a square ABCD in looking for Pythagoras?​

How to find the area of a square ABCD in looking for Pythagoras?​-example-1
User MDaniyal
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1 Answer

3 votes

Answer:

A = 25 units²

Explanation:

You need to visualize two triangles on the lower left and right that are OUTSIDE of the square.

One leg of the first triangle is formed by starting at Point A and going down 4 units. The other leg is formed by starting at point D and moving left 3 units. Line segment AD is the hypotenuse of this triangle.

The leg lengths are 3 and 4, so the hypotenuse is...

3² + 4² = c²

9 + 16 = c²

25 = c²

5 = c

So side AD of the square is 5 units long.

We can do the same concept using points C and D. We can make a triangle that has legs with length 3 and 4 with CD being the hypotenuse. The length of CD is therefore 5 as well.

The area of the square is

A = 5*5 = 25 units²

User Vettis
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