Triangle Area Demonstration This resource provides animated demonstrations

  • Slides: 25
Download presentation
Triangle – Area – Demonstration This resource provides animated demonstrations of the mathematical method.

Triangle – Area – Demonstration This resource provides animated demonstrations of the mathematical method. Check animations and delete slides not needed for your class.

(Hint: What is the total rectangular area? ) 2 cm 2 1 cm =

(Hint: What is the total rectangular area? ) 2 cm 2 1 cm = 1 cm 2 1 cm Find the area of each shape. 1 cm 2 4 cm 2 3 cm 2 4. 5 cm 2

What is the area of this square? 4 cm If we cut the square

What is the area of this square? 4 cm If we cut the square into two: what is the new shape? what is the area of the new shape?

What is the area of this rectangle? 3 cm 6 cm If we cut

What is the area of this rectangle? 3 cm 6 cm If we cut the rectangle into two: what is the new shape? what is the area of the new shape?

We start with a rectangle… 4 cm 6 cm If we cut the rectangle

We start with a rectangle… 4 cm 6 cm If we cut the rectangle into a triangle: what is the new shape? what is the area of the new shape?

We start with a rectangle… 4 cm Equal areas 6 cm If we cut

We start with a rectangle… 4 cm Equal areas 6 cm If we cut the rectangle into a triangle: what is the new shape? what is the area of the new shape?

What is the area of each triangle? 6 cm 4 cm 6 cm 8

What is the area of each triangle? 6 cm 4 cm 6 cm 8 cm 4 cm 5 cm Area = 16 cm 2 Area = 12 cm 2 Area = 15 cm 2 What is the formula to find the area of any triangle? Area of a triangle = half × base × height Height Base

6 cm 10 cm Which method would you use? 10 × 6 then divide

6 cm 10 cm Which method would you use? 10 × 6 then divide by 2 half the base then multiply by the height

6 cm 5 cm 4 cm What has Anne done wrong? We must use

6 cm 5 cm 4 cm What has Anne done wrong? We must use the perpendicular height from the base. The perpendicular height is at an angle of 90° from the base

Does the formula still work? 1) Doubling the triangle we make a parallelogram. 2)

Does the formula still work? 1) Doubling the triangle we make a parallelogram. 2) Area of a parallelogram = base × height 3) We want half that, so, half the base x height Height Base

Right or wrong? Would we use these measurements for base & height?

Right or wrong? Would we use these measurements for base & height?

For each triangle, which measurements can we use for the base, and for the

For each triangle, which measurements can we use for the base, and for the perpendicular height? 15 cm 2 7 cm 5 cm 16 4 cm cm 2 20 cm 2 7 cm 5 cm 10 cm 6 cm 8 cm 3 cm 10 cm 15 cm 2 Find the area of each triangle. 8 cm 4 cm 6 cm 11 cm 8 cm 10 cm 40 cm 2 4. 8 cm 10 cm 24 cm 2

Area of a triangle = half × base × height 4 cm = 4

Area of a triangle = half × base × height 4 cm = 4 cm 2

Area of a triangle = half × base × height 4 cm = 10

Area of a triangle = half × base × height 4 cm = 10 cm 2

Area of a triangle = half × base × height 7 cm 6 cm

Area of a triangle = half × base × height 7 cm 6 cm 5 cm = 15 cm 2

Area of a triangle = half × base × height 3 m 5 m

Area of a triangle = half × base × height 3 m 5 m 4 m = 6 m 2

Area of a triangle = half × base × height 7 cm 6 cm

Area of a triangle = half × base × height 7 cm 6 cm 8 cm = 24 cm 2

Area of a triangle = half × base × height 4 cm = 6

Area of a triangle = half × base × height 4 cm = 6 cm 2

Area of a triangle = half × base × height 6 m 7 m

Area of a triangle = half × base × height 6 m 7 m 5 m 10 m = 25 m 2

Area of a triangle = half × base × height 9 cm 6 cm

Area of a triangle = half × base × height 9 cm 6 cm 8 cm = 27 cm 2

Area of a triangle = half × base × height 8 cm = 32

Area of a triangle = half × base × height 8 cm = 32 cm 2

Area of a triangle = half × base × height 15 cm 12 cm

Area of a triangle = half × base × height 15 cm 12 cm 4 cm 6 cm = 30 cm 2

Area of a triangle = half × base × height 14 cm 8 cm

Area of a triangle = half × base × height 14 cm 8 cm 12 cm = 48 cm 2

Questions? Comments? Suggestions? …or have you found a mistake!? Any feedback would be appreciated

Questions? Comments? Suggestions? …or have you found a mistake!? Any feedback would be appreciated . Please feel free to email: tom@goteachmaths. co. uk