# My kite is bigger than yours

Ms. Rania Ali taught students of a primary school located in a slum. Most students came there because they had to. The kids had at best marginal interest in learning. Ms. Ali was aware of her challenge but the great teacher found new ways to get attention of her students. This is the story of the day when two students were about to be expelled for fighting in the school but she saved them by giving them a geometry lesson.

In India and Pakistan, the end of winter is celebrated with a kite flying festival which falls on basant panchami. Basant is the spring season when children begin to play outside because soon the scorching hot days of summer would start. In the spirit of spring, Rajab and his friends had made a big kite which was 90 cm long and 30 cm wide. They were really proud of it until they ran into Rehan and his friends. This group had gotten together and made a kite which was only 80 cm long but 40 cm wide. An argument started as to whose kite was bigger. With the help of their inflamming friends, they began to yell louder and louder as if yelling would be the convincing factor rather than the size of the kite. No, they did not hit each other or call names but the nonstop shouting with increasing loudness in the school became unbearable. The principal heard it and called Rajab and Rehan into her office. She said that this behaviour was intolerable and that both would be expelled. Some of the other students told Ms. Ali who then went to the principal’s office. She told the principal that she had never before seen either of the two students misbehave. At the end, she convinced the principal that they should not be expelled if by the end of the day the two would come to her office and show her how they could have settled the argument logically rather than by quarreling.

Ms. Ali entered her class where both Rajab and Rehan were also present.

# Rajab and Rehan would be expelled unless….

Ms. Ali: You all know Rajab and Rehan would be expelled unless they can decide whose kite is bigger and explain it to the principal. How should we start?

Mehak: We could ask one of them to draw their kite.

Ms. Ali: Okay, Rajab. Draw a picture of your kite on the board.

Rajab took coloured chalks and proudly drew a picture of his kite. Ms. Ali asked him to label all corners of the kite and also the middle. Here is what the kite looked like.

Ms. Ali: Colourful kite, Rajab. Who wants to tell me if the kite is symmetrical?

Arisha; It has to be. Top of the kite above AC has to be exactly the same as the part below it.

Mehak: I have seen kites which are a little bit crooked but they don’t fly. So the triangle ABC has to be congruent to triangle ADC. That would also mean that all the four angles made by crossing of AC and BD have to be the same which is 90°.

Alia: Ms. Ali, I like it that you are making their fight into a geometry lesson,

Ms. Ali: Thanks Alia. Now tell me the area of the triangle ABE.

Alia: We learned last time that it would base x height/2. So, I would say AE x BE/2. Also the triangle BEC would have the area EC x BE/2.

# Kite as sum of four right angle triangles

Ms. Ali: Then what is the area of the kite, Rehan?

Rehan: First the area of the triangle ABC is the base AC times the height BE divided by two. Because, the kite is symmetrical, the triangle ADC would have the same area, So the area of the kite would be AC x BD/2 that means the length times width divided by two.

Ms. Ali: So how big was your kite Rehan?

Rehan: My kite was 80 cm long and 40 cm wide. Hence it had an area of 3200/2 cm^{2 }which is 1600 cm^{2}.

Ms. Ali: And Rajab?

Rajab: My kite was 90 cm long and 30 cm wide. This means its area was 90 x 30/2 or 1350 cm^{2}. Rehan, your kite with an area of 1600 cm^{2 }was bigger than mine.

Taheen: Ms. Ali, I want to draw something on the board.

Ms. Ali: Go ahead. Taheen.

# Kite as half a rectangle

Taheen: I kept the top half of his kite. Then I flipped the red triangle and pasted on the left side. Then I also flipped the blue triangle and pasted it on the right side. This way, I made a rectangle out of his kite – with the base AC and the height BE. The area of the kite is AC x BE which is the same as length x the original width (BD)/2 but I think this is more straight forward.

Ms. Ali: Very good Taheen. You helped us a lot. Rajab and Rehan, you can now tell the principal what you learned. Also, your homework would be to do the same thing to get the areas of a parallelogram and a trapezoid. I will draw the pictures for you and, in the next class, you have to explain how you got the areas.

Rajab: Do we have to?

Rehan: I think we have to because Ms. Ali just saved us from being expelled.

*Challenge*

Help Rajab and Rehan in finding the areas of a parallelogram and trapezoid.

Solution: ABCD is a parallelogram. Therefore, the side AB is parallel to CD and AC is parallel to BD. Draw a perpendicular CE from C to AB and another (BF) from B to CD. Triangles AEC and BDF are right angle triangles. Angle BAC equals angle BDF (opposite angles of a parallelogram are equal, you can also prove this independently using parallel line assumptions). Also side AC equals side BD. Therefore triangles AEC and BFD are congruent and hence AE = BF.

CEBF is a rectangle with a base EB and height EC. Therefore its area is EB x EC. Area of the triangle AEC = AE x CE/2 (base x height/2). Similarly the area of the triangle BFD = BF x CD/2. The area of the parallelogram = area of triangle AEC + area of triangle BFC + area of the rectangle CEBF = AB x CE (base x height).

For the trapezoid ABCD, also draw two similar verticles AE and BF.

The area of the trapezoid = area of the rectangle CDEF + area of the triangle ACE + area of the triangle DFB which is EF x CE + AE x CE/2 + FB x DF/2. DF equals CE because AB and CD are parallel. Therefore EF x CE + AE x CE/2 + FB x DF/2 =(AE/2 + EF+ FB/2) x CE = (AB + CD) x CE/2 which is height x (sum of the two parallel sides)/2.