Area of triangle

### 1.1. Knowledge to remember

– For two congruent triangles (see figure)

– Take that one triangle, cut along the height to make two pieces of triangles 1 and 2.

– Match two pieces 1 and 2 to the remaining triangle to get a rectangle ABCD (see figure).

Based on the figure, we have:

Rectangle ABCD has length equal to base length DC of triangle EDC, width equals height EH of triangle EDC

The area of rectangle ABCD is 2 times the area of triangle EDC.

The area of rectangle ABCD is DC x AD = DC x EH.

So the area of triangle EDC is \(S = \frac{{DC.EH}}{2}\)

*To calculate the area of a triangle, multiply the base length by the height (same unit) and divide by 2.*

\(S = \frac{{a*h}}{2}\)

S is the area, a is the base length, h is the height

### 1.2. Solve textbook exercises page 88

**Lesson 1 of the textbook page 88:**

Calculate the area of a triangle with:

a) The base length is 8cm and the height is 6cm.

b) The base length is 2.3 dm and the height is 1.2 dm.

__Solution guide:__

a) The area of the triangle is:

\(\frac{{8 \times 6}}{2} = 24(c{m^2})\)

b) The area of the triangle is:

\(\frac{{2,3 \times 1,2}}{2} = 1.38\:(d{m^2})\)

Lesson 2 Textbook page 88:

Calculate the area of a triangle with:

a) The base length is 5m and the height is 24 dm

b) The base length is 42.5m and the height is 5.2m.

__Solution guide:__

a, Change 5m = 50dm

The area of that triangle is:

\(\frac{{50 \times 24}}{2} = 600(d{m^2})\)

b, The area of that triangle is:

\(\frac{{42,5 \times 5.2}}{2} = 110.5({m^2})\)

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