Math Is Fun Forum

CurlyBracket

Member

From: Your Maths Textbook

Registered: 2022-01-03

Posts: 155

Folding Papers

Here's the question:

And here's what I did:

AB = 4

BN = x

In Δ ABN,

(16-x)² = 4² + x²

So, x = 15/2

AN = (16 - x²)
     = 17/2

Now, area of ABNMD` = ABCD - ANM

                                = (4 * 16) - ((1/2) * (15/2 ) * 4)

                                = 64 - 15

                                = 49 cm²

Where am I mistaken?

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Filling the unforgiving minute with sixty seconds’ worth of distance run!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Folding Papers

I don't think that's right for ANM.

It's isosceles with AN = AM

You can calculate its height as it must be half the diagonal of the original rectangle. Then use pythag to compute NM.

The diagram makes it look like MN is parallel to AB. It isn't.

Try cutting out a paper rectangle and fold as instructed .

LATER ADDITION:

E is the middle of the original rectangle.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob