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