Practice calculating the area of a triangle

### 1.1. Solving exercises in the textbook Practice page 88

**Lesson 1 Textbook page 88**

Find the area of a triangle with base length aa and height hh:

a) a = 30.5dm and h = 12dm

b) a = 16dm and h = 5.3m

__Solution guide:__

a) The area of the triangle is:

\(\frac{{30.5 \times 12}}{2} = 183(d{m^2})\)

b) Change 5.3m = 53dm

The area of that triangle is:

\(\frac{{16 \times 53}}{2} = 424(d{m^2})\)

Lesson 2 Textbook page 88

Indicate the corresponding bottom and high line present in each of the triangles below:

__Solution guide:__

- Right triangle ABC:

– The bottom is AC and the corresponding high is BA.

– The bottom is AB and the corresponding high is CA.

- Right triangle DEG:

– The bottom is DE and the corresponding high is GD.

– The bottom is DG and the corresponding high is ED.

Lesson 3 Textbook page 88

a) Calculate the area of a right triangle ABC.

b) Calculate the area of the right triangle DEG.

__Solution guide:__

The area of triangle ABC is:

\(\frac{{3 \times 4}}{2} = 6\:(c{m^2})\)

The area of triangle DEG is:

\(\frac{{3 \times 5}}{2} = 7.5(c{m^2})\)

Lesson 4 Textbook page 88

a) Measure the lengths of the sides of rectangle ABCD and then calculate the area of triangle ABC.

b) Measure the lengths of the sides of the rectangle MNPQ and the side lengths ME.

Calculate:

– The sum of the area of triangle MQE and the area of triangle NEP.

– Area of triangle EQP.

__Solution guide:__

a) After measuring we have: AB = 4cm, BC = 3cm.

The area of triangle ABC is:

\(\frac{{4 \times 3}}{2} = 6\left( {c{m^2}} \right)\)

b, After measuring we have: MQ = NP = 3cm, MN = PQ = 4cm, ME = 1cm.

So, NE = MN – ME = 4 – 1 = 3 (cm)

The area of rectangle MNPQ is:

4 × 3 = 12(cm2)

The area of triangle MQE is:

\(\frac{{3 \times 1}}{2} = 1.5(c{m^2})\)

The area of triangle NEP is:

\(\frac{{3 \times 3}}{2} = 4.5(c{m^2})\)

The total area of the two triangles MQE and NEP is:

1.5 + 4.5 = 6(cm2)

The area of triangle EQP is:

12 − 6 = 6(cm2)

### 1.2. Solving exercises in textbooks General practice page 89

**Lesson 1 Textbook page 89**

The digit 3 in the decimal number 72,364 has a value of:

A. 3 B. \(\frac{3}{{10}}\) C. \(\frac{3}{{100}}\) D. \(\frac{3}{{1000}}\ )

__Solution guide:__

The digit 3 is in the tenths. So the value of the digit 3 in the given decimal is 310310.

Choose answer B.

Lesson 2 Textbook page 89

There are 25 fish in the tank, including 20 carp. The ratio of the number of carp to the number of fish in the tank is:

A.5% B.20% C.80% D.100%

__Solution guide:__

The ratio of the number of carp to the number of fish in the tank is:

20 : 25 = 0.8 = 80%

Choose answer C.

Lesson 3 Textbook page 89

2800g equals how many kilograms?

A. 280kg B. 28kg C. 2.8kg D. 0.28kg

__Solution guide:__

We have: 1kg = 1000g or \(1g = \frac{1}{{1000}}kg\)

Therefore: \(2800g = \frac{{2800}}{{1000}}kg = 2.8kg\)

Choose the answer C

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