Mon., March 5th, 2018 – Area of a Triangle

AREA OF A TRIANGLE
We first began to learn about the area of a triangle mid-February. Tomorrow we have a test! Students need to
  • calculate the area of a triangle when the base and height are given
  • find the base or height when the area and other dimension (height or base) are given
  • show an understanding of the relationship between the area of a triangle and a parallelogram or rectangle that has the same base and height
For practice, students can
  • review questions already completed (notebooks came home today)
  •  Calculate the area of any triangle they sketch (just make sure you have a base and height labelled)…and use the formula!
    • try to sketch all sorts of triangles
      • some with acute angles (height labelled with dotted lines inside the triangle)
      • some with one right angle (the height is one of the sides!)
      • some with one obtuse angle (you have to label the height with dotted lines outside the triangle)
  • Draw a variety of rectangles and/or parallelograms, label the height and base…calculate the area
    • …….and then cut them in half from one corner to another to show that the triangle with same height and base has half the area (prove it by calculating the area of the rectangle or parallelogram AND the area of the triangle)

 

BE ABLE TO ANSWER THIS:

What is the relationship between the area of a triangle, and the area of a parallelogram with the same base & height?

What is the relationship between the area of a parallelogram, and the area of a triangle with the same base & height?

 

Other Sample question from class:

How many triangles can you sketch that have an area of 12 cm2?

 

Incorrect – but a try!

 

Correct answer:

 

Are we able to calculate this area? — No….there is no height. The 12cm is a label for the adjacent side (beside the base….but NOT at right angles to the base….height must be at right angles to the base)

 

What if the height is 10cm? Does the triangle have an area of 12 cm2?  — No

A =   b   x h   ÷  2

=  1    x 10  ÷ 2

=  10   ÷   2

=  5 cm2     (NOT 12 cm2)

 

LEVEL 4 strategy for solving for a missing base or height (from previous blog post)

 

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