# The tenth anniversary of the school

Ms. Rania Ali taught at a primary school located in a slum. 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 the students asked for help and she turned their request into a geometry lesson.

This was the tenth anniversary of the school. Yes, it was a big achievement that the school had survived here for 10 years despite every hurdle one could name. The current principal was proud that it happened under her leadership and wanted to celebrate this occasion. There were no funds available for such a celebration. Only one small local merchant came with a limited donation. This is why she could not refuse when some students asked if they could decorate the school premises. They wanted to hang triangular banners. She agreed with some conditions. The first condition was that they could make the banners but only the chowkidar (security guard) would hang them. She would supply them with 200 meters of string, glue and some colored banner paper. They could start during lunch hour one day and finish the job during rest of the school hours. The second condition was that they had to tell her that day itself how much banner paper they would need.

Ms. Ali entered the classroom and noticed that several students were huddled in one corner. She addressed the class and asked everyone to go to their own seats. They did so but then Mehak raised her hand. She said that the class had a math problem and wondered if she could help them instead of giving them a regular lesson. Ms. Ali asked about the problem and Mehak told her what they had to calculate.

Ms. Ali: Let us continue with the regular lesson. This lesson will help you in you problem. At the end we can also devote 10 minutes to it. Does anyone remember how to determine the area of a rectangle? You all learned it in your arithmetic class.

Rajab: The area of a rectangle is base multiplied by height. I am sure everyone remembers that.

# Ms. Ali: Now what is the area of triangle?

Arisha; Ms. Ali, you told us that there are many types of triangles. Which triangle are we talking about?

Ms. Ali: We are going to talk about an acute scalene triangle today but you will see that what you will learn today would apply equally to all types of triangles. Here is a scalene triangle ABC. Now I am going to draw a perpendicular BP from B onto AC. From A, I will draw a line AD which is parallel to BP and another line DB parallel to AC. Now, Taheen do you think that triangle ADB is similar to ABP?

Taheen: I remember this about parallel lines. DB and AP are parallel, Therefore, angles ABD and BAP will be the same because they are corresponding angles. Because BP is perpendicular to AC, it will also be a perpendicular to DB. Both these angles being 90° and angles ABD and BAP being the same, the third angle would also have to be the same. So the two triangles are similar. Ms. Ali, can I add something?

Ms. Ali: Yes. Taheen, what is it?

Taheen: The side AB is common to both the triangles. So the triangles are not only similar but they are also congruent.

Ms. Ali: Taheen, that was great. Does everyone agree?

# Congruent triangles have the same areas

Mehak: I think that mean that the triangles ABP and ADB have the same area.

No one questioned Mehak’s observation.

Ms. Ali: Alright then I will draw two more lines BE and EC on the right side, the same way I drew AD and DB on the left side. Now do you think that the triangle BEC will have the same area as BPC?

Taheen: Yes, they will and that means the triangle ABC is only half the area of the rectangle ADEC.

Rajab: Ms. Ali, I think you tricked us into learning that the area of a triangle is base x height /2.

Ms. Ali: I did not trick you. You needed to know this to solve your problem. Now, you know the size and the number of triangles you want for the decoration. What are they?

Mehak: We had decided that we would make isosceles triangles each with a base about 15 cm and a height of about 22 cm. I think that would like nice. Now, I know that each triangle would require 15 x 22/2 or 165 cm^{2 }of paper. Is this correct?

Ms. Ali: That’s correct. How many triangles would you want?

Arisha: I would like to see at least 3 triangles per meter and we have 200 meters of string. That means 600 triangles.

Rajab: So we need 165 cm^{2 }paper per triangle for 600 triangles. That’s a lot of paper 99,000 cm^{2}.

Rehan: It is a very large area if you talk that way. It comes out to 9.9 meters^{2}. So all we need to ask the principal is to give us 10 sheets of the paper because each piece is 1 meter^{2}.

Mehak: 10 sheets of different colours. Thanks for helping us out Ms. Ali.

So, Ms. Ali managed to get the students engaged in another geometry lesson. Can she do it again?

*Challenge*

Find the areas of the following acute, right angle and obtuse triangles. Given: length of AC in each triangle is 10 cm. The height is given by the perpendicular BD or BA to be 8 cm.

* Solution: *The area of a triangle is base x height/2, regardless of the shape of the triangle. In all these cases base = 10 cm, height = 8 cm. Therefore, area 10 x 8/2 = 40 cm^{2}.