EXERCISE AND SOLUTION MATHEMATIC PROBLEM

1. Mathematic for 3 grade junior high school yudhistira, page 12, number 1
Parallelogram ABCD and parallelogram EFGH are congruent. If the perimeter of the parallelogram ABCD is 10 cm, determine the value of x, the length of side EF, FG, GH, and HE.
solution:
Side - the side corresponding to the AD coincide with EF, BC corresponds to the FG, CD corresponds to the GH, DA corresponds to HE. Because the parallelogram ABCD is congruent to parallelogram EFGH,
The length of AB = EF = (3x - 3) cm
The length of BC = FG = x cm
The length of CD = GH = (3x - 3) cm
The length DA = HE = x cm
Thus, the value of
AB = CD = EF = GH = (3x - 3) cm
BC = FG = DA = HE = x cm
ABCD is a parallelogram while the circumference of 10 inches,
then
DE + CD + AB + BC = 10 centimeters
(3x - 3) + x + (- 3x 3) + x = 2 [(3x - 3) + x] = 10
2 [4x - 3] = 10
4x - 3 = 5
4x = 8
x = 2

Thus, the length of the
EF = (3X - 3)
EF = (3.2 - 3)
EF = 3 cm
FG = x
FG = 2 cm
GH = EF = 3 cm
HE = DA = 2 cm

2. Mathematic for 3 grade junior high school yudhistira, page 12, number 2
Pentagonal ABCDE and pentagonal FGHIJ are congruent. If EA = 2 cm, determine the length of side FG, GH, HI, IJ, and the perimeter of pentagon FGHIJ.
solution:
Side - the side corresponding to the AB corresponding to the FG, BC corresponds to the GH, CD corresponds to HI, DE corresponds to the IJ, JF Because EA corresponds to the terms limma pentagon ABCDE is congruent with FGHIJ then,
The length of AB = FG = (2x plus 1) cm
The length of BC = GH = (2x - 1) cm
The length of CD = HI
The length of DE = IJ = x cm
The length of EA = JF = (3x - 1) = 2 cm
then,
EA = JF = (3x - 1) = 2 cm
3x = 3
x = 1
FG = (2x + 1) = 2.1 + 1 = 3 cm
GH = (2x - 1) = 2.1 - 1 = 1 cm
IJ = x = 1 cm

To calculate the length of the HI, then use the formula phitagoras ita, HI as the hypotenuse, and the two other sides of the triangle we get from the reduced elbow reduced FG and FJ JI reduced GH.

then:
HI^2 = (FG - JI)^2 + (FJ - GH)^2
HI^2 = (3-1)^2 + (2-1)^2
HI^2 = (2)^2 + (1)^2
HI^2 = 5
HI = root 5

Thus, the length of the
FG = 3 cm
GH = 1 cm
HI = root 5 cm
IJ = 1 cm
And, traveling up the pentagon FGHIJ FG + GH + HI + IJ + JF = 3 + 1 + root 5 + 1 + 2 = 7 root 5 cm


3. Mathematic for 3 grade junior high school yudhistira, page 12, number 3
Parallelogram ABCD and parallelogram EFGH are congruent. If a large angle DAB = to 45 degrees, decide the size of the HEF, the angle EFG, FGH angle, and the angle GHE.

Solution:
Reduced angle corresponding angle is the angle corresponding to the angle HEF DAB, ABC corresponds to the EFG, FGH corresponding BCD, CDA corresponds to ghe. Because the parallelogram ABCD is congruent to parallelogram EFGH,
then,
HEF = DAB = angle of 45 degrees
Angle ABC = EFG
BCD = angle FGH
Angle CDA = GHE
Due to the nature of the parallelogram is the parallelogram, the two opposite corners of equal magnitude, then the angle DAB = angle HEF = angle BCD = angle FGH = 45 degrees, then angle ABC = angle EFG = angle CDA = angle GHE.

Then the properties of a rectangle that is the four corners of 360 degrees, then:
the Angle HEF + the angle GHE + the angle EFG + the angle FGH = 360
The angle EFG + 45 + 45 + the angle GHE = 360
90 + the angle GHE + angle EFG = 360
the angle EFG + the angle GHE = 270

Because the angle EFG = the angle GHE, the large angle EFG = 270/2 = 135 degrees, as well as the large angle GHE = 270/2 = 135 degrees.
Thus, the large angle HEF = 45 degrees, the angle EFG = 135 degrees, angle FGH = 45 degrees, angle = 135 degrees ghe.

4. Mathematic for 3 grade junior high school yudhistira, page 152, number 1
(6 X 3)^7

Method 1:
(6 X 3)^7= (6)^7 X (3)^7
= 6 X 6 X 6 X 6 X 6 X 6 X 6 X3 X3 X3 X3 X3 X3 X3
= X18 X18 X18 X18 18 X18 X18 = (18)^7
Method 2:
(6 X 3)^7=(18)^7

5. Mathematic for 3 grade junior high school yudhistira, page 152, number 2
(-a^5b^5^)3=
Method 1:
((-a)^5b^5)3=((-a)^5)^3 X ((b)^5)^3=(-a)^5 X (-a)^5 X (-a)^5 X (b)^5 X (b)^5 X (b)^5= (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (-a) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b) X (b)=(-a)^15(b)^15
Method 2:
((-a)^5b^5)^3= ((-a)^5)^3 X (b^5)^3
=(-a)^(5 X 3).(b)^(5X 3)
=(-a)^15.(b)^15