#1 English Task (Problems of Mathematics from Book)

CHAPTER 1 (ALGEBRA AND ITS APPLICATION)

Page 37 No. 16

The length of two sides in a triangle are (x+a) cm and (3x-2a) cm respectively. The perimeter of the triangle is 2(2x+a) cm. The length of the third side in the triangle is …..

Solution:

The formula of perimeter of the triangle is the first side plus the second side plus the third side.

We can include the length of two sides and the perimeter in a triangle that we know into the formula of perimeter of the triangle.

2(2x + a) cm = (x + a) cm + (3x - 2a) cm + S

We can operate the formula with operation of algebraic expression. For the left side, you can calculate with two multiply two x plus two multiply a equals to four x plus two a cm. Then for the right side, you can group the like terms, x plus 3x plus a subtract two a plus S equals to four x subtract a cm plus S.

(4x + 2a) cm = (4x – a) cm + S

Then to find the value of S, we can operate by subtract the right side from the left side and you also must group the like terms. So, we get four x plus two a cm subtract four x subtract a cm, and the value of S is three a cm.

(4x + 2a) cm – (4x – a) cm = S

S = (4x – 4x + 2a + a) cm

S = 3a cm
So, the length of the third side in the triangle is three a cm.

CHAPTER 3 (THE EQUATION OF A STRAIGHT LINE)

Page 99 No. 1 a

Determine following the equation of lines.

A gradient of 4 and throught point (-7,8)

Solution:

Point (-7,8) has values x1=-7 and x2=8

The gradient of the line is 4. Therefore, m =4

Then to find the equation line, we must include the point and the gradient that we know into the formula y – y1 = m (x – x1)

So, y – 8 = 4 (x – (-7))

We calculate the equation. For the right side we can calculate four multiply x plus four multiply seven equals to four x plus twenty eight.

y – 8 = 4x + 28

and then we can find y with move the value of minus eight to right side. So, we calculate right side again, y equals to four x plus twenty eight plus eight. The result is y equals to four x plus thirty six.

y = 4x + 28 + 8

So, the equation of lines is
 y = 4x + 36

CHAPTER 4 (THE SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES)

Page 115 No. 2

The price of 2 kg of oranges and 3 mangos is Rp 38.000,00. Determine the price of 1 kg of orange if known that 1 kg of mango is Rp 7.000,00.

Solution:

Suppose oranges as x and mangos as y.

Now we have two x plus three y equals to 38.000.

2x + 3y = 38.000

The price of one kg of mango (y) that we know is seven thousand. We can include this price into the equation that we get. So, two x plus three multiply seven thousand equals to thirty eight thousand

2 x + 3 (7.000)  = 38.000

We can calculate and the result is two x plus three multiply twenty thousand equals to thirty eight thousand. Then to find the value of y, we subtract left side from the right side is thirty thousand eight subtract twenty one thousand equals to seventeen thousand. And the last step we divide seventeen thousand with two. The result is eighty thousand five hundred.

2x + 21.000 = 38.000

2x = 38.000 – 21.000

2x = 17.000

x = 8.500
So, the price of 1 kg of orange is Rp 8.500,00

CHAPTER 5 (THE PHYTAGOREAN THEOREM)

Page 144 No. 1 a

Suppose the length of sides of a right triangle is a, b, and c, where c is the hypotenuse. Determine the unknown length of side of the following problem.

a = 12 length units and b = 20 length units

Solution:

The lengths of the other two sides are given, and we must find the hypotenuse (c). We can include the length of sides of a right triangle that we know into the Pythagorean Theorem. The Pythagorean Theorem for the solution of the problem as follows.

 

Then, we must include the length of side of a right triangle that we know into the Pythagorean Theorem. So, c the power of two equals to twelve the power of two plus twenty the power of two. We can calculate this solution, and the result is one hundred forty four plus four hundred equals to five hundred forty four.

So, the result is twenty three point thirty two length units.

CHAPTER 7 (POLYHEDRAL)

Page 273 No. 2

The side length of the base of a square based pyramid T.ABCD is 10 cm. If the height is 9 cm determine the volume of the pyramid.

Solution:

We can complete the problem with the formula of the volume of pyramid is one over three multiply base area multiply height.

V = 1/3 x base area x height

We calculate base area of a square is ten multiply ten equals to one hundred. Then, we find the volume of pyramid with include height and base area into the formula of the volume of pyramid.

Base area = 10 cm x 10 cm

Base area = 100 cm^2

V = 1/3 x 100 cm x 9 cm

V = 300 cm^3

So, we get one over three multiply one hundred multiply nine equals to three hundred cm^3.


These problems are taken from:
Mathematics for Junior High School Year VIII by Dr. Marsigit M.A