Carlie’s Doll Dresses

fig.g12.3Carlie’s Idea

Carlie was watching a fashion show on TV.  Suddenly, an idea came to her. She knew that many of her friends dressed up their Barbie dolls for playing. All of them wished that their dolls had beautiful dresses.  The stores charged a lot of money for fancy doll dresses.  Being a go getter, she started thinking about it. She could invest the money she had made from the lemonade stand to start making dresses for dolls.

Carlie talked to her mom who was supportive but laid down some rules like doing homework, sleeping on time.  Carlie agreed to the rules.

Carlie’s mom: Do you have enough money?

Carlie: Yes mom.  I want to go to a fabric shop to get one meter each of three different types of fabrics and some other things.

The cheapest fabrics are the remnants

Carlie went with her mom to the fabric store and looked at the remnants because they were very cheap.  These remnants were from 1.5 meter wide bolts.  Carlie picked up three pieces of fabric of different designs. The first piece was only 50 cents and it was half a meter long, the second cost $1.50 and it was 0.8 meters, and the third one was $4 but it was 1.5 meters long and beautiful.

She also spent a total of $9 to buy two scotch tapes sticky on both sides, two scotch tapes sticky only on one side, one thick sticky red patch of 15 cm x 20 cm, and a box of sticky rhinestones.  That was a total of $15.

Carlie’s Design

Carlie knew that it would be too much work for her to make doll dresses and sell them.  She talked to her friend Amelia and made her a junior partner in the business.

Amelia: How many dresses will we make?

Carlie:  As many as we can sell.

Amelia: How are we going to make the dresses?

Carlie:  I  have an idea. We’ll make blouses and skirts.  For a skirt, we will take a 30 cm x 30 cm square piece of fabric.  We will draw lines connecting the corners of the square in an “X” shape.

Then we will cut along these lines to get four pieces. This will make four identical right angle triangles.

fig.12g.1.jpgEach triangle will have a bottom of 30 cm. I guess its height will be about 15 cm.  We will fold it around to make a cone. Then we will cut of the top of the cone and put a heavy red adhesive patch to make a belt. This will give about 13 cm high dress with a large boundary of 30 cm at the bottom.

Oh, we will also have the blouse – just 3 – 4 cm long tube top so that the doll can show her waist off.  That’s the fashion these days.

Amelia:  You are a genius.   How are you going to stitch the sides of the triangle to make it into a cone?

Carlie:  Stitching will be a lot of work. We could use a two sided thin sticky tape instead.  You know we could also use a simple thin sticky tape at the bottom, make some folds in the helm.  That will make the bottom of the skirt look nice and heavy.

Amelia:  What kind of fabric are we going to use and how much do we need?

Carlie:  Here are the three pieces of fabric I bought. I also bought sticky rhinestones to make the dresses look pretty.

Carlie’s cut or Amelia’s cut?

They decided to start with the smallest piece.  It was half a meter (50 cm) by 1.5 meters (150 cm). Carlie said that she could cut it into five square pieces of 30 cm x 30 cm. That would make 20 dresses because each square made 4 dresses.  They would also have a sufficient amount of material left to make the matching tube tops.

Carlie was about to start cutting when Amelia stopped her and said, “You said that we could get only 20 dresses from the 0.5 meter x 1.5 meter piece.  I bet you I can get you more.”

Carlie: We will just get a 150 cm long strip and cut it into 5 pieces each 30 cm long.  Then we will make 4 skirts from each to get a total of 20 skirts.  Genius, how are you going to get more?

fig.g12.2Amelia showed a picture she had drawn and said:  This is how.  Your cuts are on the left. Each triangle you made had a base of 30 cm and each of the other two sides was 21.2 cm.

My picture is on the right.  If I make a 21.2 cm x 21.2 cm square piece and cut it only once at the connecting the corners,  I will get two skirts which would be the same as yours – a base 30 cm, sides of 21.2 cm  and a height 15 cm.

From the same 0.5 meters x 1.5 meters piece, I can get two rows of 21.2 cm (totalling 42.4 cm) and then cut each row into 7 pieces of 21.2 cm length – totaling 148.4 cm. Each square piece will give two skirts.  This means, we can get 2 x 7 x 2 or 28 skirts out instead of 20.  That will be 8 more skirts from the same piece.

Carlie:  Are you trying to show your boss off?

Amelia: Remember, we are partners.  I will get paid more if there is a larger profit.

Carlie:  That’s right.  Now, what about the other pieces.?

Amelia: I don’t think it will make a big difference. Besides, we also need fabric for the blouses.

Amelia visited Carlie the next day.

Amelia:  I talked to some of my friends about the doll dress. They loved the idea.  Some of them may phone you soon.

Carlie: We should be able to charge them $2 for the doll plus the tube top.  The stores will charge them $10 – $20 for a good doll dress.


Ever since he saw the first stop sign, Tim has a fascination for octagons.  He wants a 1 centimeter thick silver sheet of an octagon with a perimeter of 24 cm.  Because, silver is sold by weight, he wants to know what it would cost at the rate of $600 per kilogram of silver. Note that 1 cubic centimeter of silver is 10.5 g.

Solution: Draw the octagon ABCDEFGH. It has a perimeter of 24 cm and 8 equal sides.  Therefore, each side equals 24/8 or 3 cm. First, we need to determine the area of this octagon. Extend the sides AB and KJ to construct the quadrangle LIJK.

Now the area of the octagon = Area of the quadrangle LIJK – sum of the area of the triangles BIC, DJE,GKF, KLH

fig.g12.4First, we determine the area of the triangle BIC

Sum of all exterior angles of an octagon is 360° and hence each anterior angle is 45°.  Angle JBC = angle XCD =45°. Angle JCB is opposite to angle XCD and hence it must also equal 45°. Therefore, BIC the third angle of the triangle equals 180°- 45°- 45° = 90°.  From Pythagoras theorem BC2 = BI2 + CI2

Because angles CBI and BCI are equal, it is also an isosceles triangle with BI = CI.  Hence or  BI2  = BC2/ 2.  When BC = 3 cm, BI2 = 9/2 =4.5 or BI = 2.12 cm. Therefore, the area of the triangle BCI = base x height/2 = 2.12 x 2.12/2 = 2.25 cm2.

Area of the auadrangle LIJK

Using the same approach as for the angle BIC, one can show that all angles of this quadrilateral are 90°.  One can also show that LA is the same length as BI,  Therefore, side LI of the quadrilateral is 3 + 2.12 +2.12 or 7.24 cm.

Using the same approach  all four sides of the quadrilateral can be shown to be the same length.

Now determine the area of the octagon.

The area of the octagon = area of the quadrangle IJKL – sum of areas of triangles BIC, EJD, FKG and HLA = 7.242 – 4 x 2.25 = 52.42 – 9 = 43.42 cm2.

Volume of the octagon = area of the octagon (43.42 cm2) x thickness (1 cm) = 43.42 cm3.

Since 1 cubic centimeter of silver weighs 10.5 g, the weight of the octagon = 43.42 x 10.5 = 455.88 g or 0.45588 kg.

At $600/kg, it will cost $275.53.