Area and perimeter : corrected math exercises in 5th grade.

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The answer key to the math exercises on areas and perimeters of figures in 5th grade. Know how to calculate the perimeter and area of usual figures (square, rectangle, rhombus, parallelogram).

Exercise 1:

$P=(((6+3)\times \,2)-4)+(2\times \,\pi\times 1)$

$P=14+2\pi$

Exercise 2:

$P=4,5+6+1,5+3+1,5+1,5+1,5+1,5=21\,cm$

$A=4,5\times \,1,5+3\times \,1,5+3\times \,1,5=15,75\,cm^2$

Exercise 3:

$R=280:2=140\,cm$

$A=\frac{(B+b)\times \,h}{2}$

$A=\frac{(280+150)\times \,80}{2}$

$A=\frac{34400}{2}$

$A=17200\,cm^2$

Calculate the area of the rectangle.

$A=L\times \,l=280\times \,95=26600\,cm^2$

Calculate the area of the half-disk.

$A=\frac{\pi\times \,R^2}{2}$

$A=\frac{\pi\times \,140^2}{2}$

$A\simeq\,30\,787,61\,cm^2$

Calculate the total area.

$A_{totale}\simeq\,17200+26600+30\,787,61\,cm^2$

$A{\color{DarkRed}\,_{totale}\simeq\,74587,61\,cm^2}$

Exercise 4:

$A_{triangle}=\frac{base\times \,hauteur}{2}=\frac{17\times \,90}{2}=765\,mm^2$

$A_{rectangle}=Longueur\times \,largeur=17\times \,16=272\,mm^2$

$A_{totale}=765+272=1037\,mm^2$

Exercise 5:

$A_{losange}=42\times \,26=1092\,cm^2$

$A_{disque}=\pi\times \,R\times \,R=\pi\times \,10\times \,10=100\pi$

$A_{totale}=A_{losange}-A_{disque}=1092-100\pi\simeq\,777,84\,cm^2$

Exercise 6:

First parallelogram, a rectangle, whose area is calculated by the product of two consecutive sides.

Area (ABCD) = 3*4 = 12 area units

To calculate the areas of the other parallelograms let’s use this diagram:

In general hxAB gives us the Area.

Second parallelogram, we need to determine h

We see that from top to bottom, h is 4 and the base is 3. So the area is 12 area units.

Third parallelogram, we need to determine h

From top to bottom, h is always 4 area units while the base is 3 area units

The area is therefore 12 area units.

Fourth and last parallelogram, we have to determine h

From left to right, h is 3 area units, and its base (one of the vertical sides) is 3 area units. The area is still 12 area units!

Exercise 7:

km².

hm²

dam²

m²

dm²

cm².

mm².

1 m² = 100 dm² = 10 000cm² = 1 000 000 mm².

1 m² = 0,01 dam² = 0,000 1 hm² = 0,000 001 km²

ca = 1 m² 1 a = 1 dam² 1 ha = 1 hm²

1. Complete:

360 cm² = 3,6 dm² 1 km² = 1 000 000 m² 10 000 m² = 1 hm²

8 m² = 800 dm ² = 80 000 cm² .

145 cm² = 0,0145 m² = 14 500 mm².

0.1 dam ² = 10 m² = 0.000 01 km²

2. Complete:

15.4 m ² = 1 540 dm ² (1,000 sq ft)	154 km ² = 15,400,000,000 dm ².
0.02 cm ² = 2 mm ².	2,024 mm ² = 0.002,024 m ².
3,5 dam ² = 3 500 000 cm ².	6 325 cm ² = 0.632 5.m ².

Exercise 8:

Calculate the area of the following rectangles:

Area of a rectangle in general is the product of two consecutive sides.
Note that a square is a particular rectangle so what applies above to a rectangle applies to the square. If the lengths are in cm, the area will be expressed in cm².

Area(figure1) = 8×5 = 40 cm².

Area (figure2) = 9×6,5 = 58,5 m².

Area (figure3) = 5×5 = 25 dm²

Exercise 9:

The inscription “90 g/m²” on a ream of paper means

that 1 m² of this paper weighs 90 g.

How much does a ream of 500 sheets of A4 paper weigh, in kg

(21 cm x 29.7 cm rectangle) of this paper?

A leaf has a surface of $21\times \,29,7=623,7\,cm^2$ =0,06237 m²

The ream of 500 sheets represents a surface of 0,06237×500=31,185 m².

The mass of a ream is 31,185×90=2806,65 g or 2,806 kg.

Exercise 10:

A triangle of area 0.1 dam² has a side length of 800 cm.

Calculate the height relative to this side .

0,1 dam²= 10 m² and 800 cm = 8 m .

The area is such that :

$\frac{base\times \,hauteur}{2}=10$

$\frac{8\times \,hauteur}{2}=10$

$8\times \,hauteur=10\times \,2$

$8\times \,hauteur=20$

$hauteur=\frac{20}{8}$

${\color{DarkRed}\,hauteur=2,5\,m}$

Exercise 11:

$A=\frac{(base+BASE)\times \,hauteur}{2}$

$A=\frac{(35+54)\times \,30}{2}$

${\color{DarkRed}\,A=1335\,m^2}$

Exercise 12:

a) 2.6 m²=260 dm²=26000cm².

b) 3 cm² =0.03 dm² = 0.0003 m²

c) 0,574 km² = 57,4 hm² = 574 000 m².

Exercise 14:

There are two times three quarters of a disc:

$2\times \,\frac{3}{4}\pi\times \,1^2\simeq\,4,7\,cm^2$

There are two rectangles:

$2\times \,1\times \,200=400\,cm^2$

There are three quarters of a strip

$\frac{3}{4}\times \,200\times \,2\pi\times \,1\simeq\,942,5\,cm^2$

942,5+400+4,7 $\simeq\,1347,2\,cm^2$

Exercise 15:

The area of the large disk is : $\pi\times \,R^2=\pi\times \,0,8^2\simeq\,2\,cm^2$

$6\pi\times \,R^2=6\pi\times \,0,4^2\simeq\,3\,mm^2$

The area of this bung is :

200-3=197

Exercise 16:

1° Calculate the area of the parallelogram MNOP represented opposite.

$A=base\times \,hauteur=ON\times \,KM=4\times \,3,2=12,8\,cm^2$

2° Calculate PO (round to the nearest 0.1).

$PO\times \,3,5=12,8$

$PO\,=\frac{12,8}{3,5}\simeq\,3,66\,cm$

Exercise 17:

The figure below is a parallelogram

Let us observe the diagram below:

1° Calculate its area.

In figure number 2, [AE] is the height relative to the side [DC]. And the area of ABCD is simply the product of the side length and its relative height.

Here Area = 6×3,5=21cm².

2. Calculate its perimeter.

The perimeter is the sum of the lengths of the sides, i.e. 2xAD+2xAB

Here Perimeter = 2×4+2×6 = 8+12=20 cm.

Exercise 18:

The Ariane 5 rocket is 57 m high.

a) How high is his scale model?

b) The diameter of the model is 5.7 cm. What is the real diameter of the rocket ?.

Exercise 19:

Calculate the perimeter of the following figure:

Let’s take a single parallelogram.

as it has two parallelograms on each side.

The perimeter corresponding to a parallelogram is :

4+3+(4-3)=4+3+1=8 cm

There are 9 parallelograms

The perimeter is therefore 9×8 = 72

Exercise 20:

$A=\frac{base\times \,hauteur}{2}$

$\frac{base\times \,hauteur}{2}=210$

$\frac{21\times \,hauteur}{2}=210$

$hauteur=\frac{2\times \,210}{21}$

$hauteur=\frac{420}{21}$

A – ” The Intruder”:

On the grid below, we have drawn six figures.

Knowing that the unit of area is the square, calculate the area of each of the 6 figures and find the intruder.

$A_1=9\,u.a$

$A_2=9\,u.a$

${\color{DarkRed}\,A_3=8\,u.a}$

$A_4=9\,u.a$

$A_5=9\,u.a$

$A_6=9\,u.a$

Exercise 22:

(AD) is a median of triangle ABC.

Property : the median of a triangle divides, this triangle, in two other triangles of same area.

$A_{ADB}=A_{ACB}$

$A_{ACB}=14$

$A_{ADB}=14-3=11\,unites\,d'\,aire$

Exercise 23:

Exercise 24:

$P=2+3+2+2+3+2+2\times \,\pi\times \,1\simeq\,20,3cm$

$A=3\times \,6-\pi\times \,1^2\simeq\,15cm^2$

Exercise 25:

Calculate the area of the following field in

This figure consists of a right-angled trapezoid from which a semicircle is removed and a triangle is added.

The area of this figure is given by the following numerical expression:

$A=\frac{(80+50)\times \,45}{2}+\frac{15\times \,20}{2}-\frac{\pi\times \,10^2}{2}$

$A=\frac{130\times \,45}{2}+\frac{15\times \,20}{2}-\frac{\pi\times \,10^2}{2}$

$A=\frac{130\times \,45}{2}+\frac{15\times \,20}{2}-\frac{\pi\times \,100}{2}$

$A=2925+150-50\pi$

$A=3075-50\pi$ is the exact value of the field area.

An approximate value is :

$A\simeq\,2\,92$

Exercise 27:

A soccer field represented to scale is a rectangle 23.1 cm long and 13.6 cm wide.

What are the actual dimensions of this soccer field?

It is not possible to answer, you need the value of the scale…

Exercise 28:

Performing conversions:

a. 12m² =1 200 dm² b. 1.32dm² =132 cm².

c. 4.5 cm² =0.00045 m². d. 8,552m²=0.008,552 km²

The answer key to the math exercises on areas and perimeters in 5th grade.

After having consulted the answers to these exercises on the calculation of the perimeter and the area of a figure in 5th grade, you can return to the exercises in 5th grade.

Fifth grade exercises.

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