MY EXERCISE

NAME : NURAIDA LUTFI HASTUTI

NIM : 11301241031

PRODI : MATHEMATICS EDUCATION 2011

EXERCISE 1 (PAGE 47,MARSIGIT’S BOOK, MATHEMATICS FOR JUNIOR HIGH SCHOOL YEAR IX)

1.        ∆ABC and ∆CDA are congruent. The fulfilled requirements are …

 AB=DC, BC=DA, AC=AC (coincides).

The answer is side, side, side.

2.        Consider the following figure.

Given, <CAD=<DBE, ∆AEC are congruent because they fulfill the requirements angle, side, angle. <CAD=<DBE=45°, AE=EB, <AEC=<DEB because they are opposite to each other.

3.        ∆KLO and ∆KMN are congruent because they fulfill the requirements angle, side, angle.

Because <KOL=<KNM=100°, KO=KN, <OKL=<NKM (coincides)

4.        ∆GHI and ∆XYZ are congruent.

The value of m is 25° because m=<GIH=<XZY. <XZY=180°-95°-60° and the answer is 25°.

5.        –

6.        AB || CD. The Length of AB is 18 cm.

Because ∆ECD and ∆EAB are similar, so

AB/CD = AE/CE

AB/12 = 12/3

       AB = 18

7.        The length of AC is 7 cm.

Because ∆ABC and ∆DEF are similar, so

AC/DF = BC/EF

AC/14 = 8/16

AC = 7

8.        Consider the following figure.

Because both triangle are similar (angle, side, angle), so

KL/EF = KM/FG

KL/4 = 9/6

KL = 6

The answer is 6 cm.

9.        For the following figure, let AB=18 cm, PQ=6cm, PR=5cm, <ABC=<PQR, and <BCA=<QRP. The length of BC is 15 cm.

Because both triangle are similar (angle, side, angle), so

BC/QR = AB/PQ

BC/5 = 18/6

BC = 15

10.    In the following figure, let the length of CD = 9 cm, CE = 6 cm, and BC = 12 cm. the length of AC is 12 cm.

∆ADC similar ∆BEC (similar angle, angle, side), so

AC/BC = DC/EC

AB/12 = 9/6

AC = 12

11.    The proportion between the length of two similar sides is 2:3. If the diagonal length of the small rectangle is 30 cm, so the length of the diagonal of the bigger rectangle is  45 cm. Solution : if x is the length of the diagonal of the bigger rectangle. We can state that 30/x = 2/3 then x = 45

12.    A flagpole with 5 m high is at a distance of 12 m from a tower and collinear with the shadow of the tower. The length of the flagpole’s shadow is 3 m. The height of the tower is 25 m.

Solution : The height of a tower is comparable with the height of the flagpole, then the sum of shadow of tower and flagpole is comparable with the shadow of flagpole. If x is the height of the tower then x/5 = 15/3 then x = 25.

13.    A painting is put on a triplex. The size of the triplex is 30 cm x 50 cm. we found that, on the top, left, and right of the painting there are still the remains of the triplex which is uncovered. If the painting is similar with triplex, so the area of uncovered triplex is 360, 5 cm area.

Solution: The size of painting is 50 cm – 7 cm = 43 cm, 30 cm – 3.5 cm = 26,5 cm. the area of painting is 43 x 26.5 = 1139.5. The area of triplex is 50 x 30 = 1500. Then the area that uncovered is 1500 – 1139.5 = 360.5 cm area.

14.    ∆ABC and ∆DBE are similar. The correct statement is AC/DE = CB/EB.

Because they are similar so, their sides can be compared, then

AC/DE = CB/EB = AB/DB

15.    ∆ABC and ∆ADE are similar. Let, AB = 7 cm, AD = 5 cm, and DE = 6 cm. the length of BC is 8.40 cm

because they are similar so, BC/DE = AB/AD = AC/AE

to find BC then

BC/DE = AB/AD

BC/6 = 7/5

BC = 8.40

16.    ∆ABC and ∆DEF are congruent, the length of the sides of  ∆DEF  are DE = 3 cm, EF = 2.5 cm, and FD = 2 cm.

Solution: because they are congruent so it has 3 pairs side that have same size (length). They are AB=DE, EF=BC, FD=AC.

17.    If ∆ABC and ∆BAD are congruent, so the values of x and y are 45° and 1 cm.

solution: <DAB = <ABC

          (3x – 90)° = 45°

                      3x = 135°

                        x = 45°

then AD = BC

      y + 1 = 2

            y = 1 cm

18.    ∆ABC and ∆PQR are congruent. The length  of PQ is 4 cm.

solution: they are congruent side, side, side

so first we must find x

because PR=AC, QR=BC, PQ=AB, then we can use AC = PR

2x – 2 = 2

x = 2

then QR = AB

QR = 3x – 2

QR = 4

19.    See the figure below. Let AC = 2 cm, BC = 4.5cm. if DE = 1 cm, so the length of BE is 2.25 cm.

solution: ∆ABC and ∆DEB are similar, so

BE/BC=DE/AC=BD/BA

To find BE, it use

BE/BC = DE/AC

BE/4.5 = ½

BE = 2.25

20.    Given the parallelogram PQRS, if the area of parallelogram is 8 cm area, so the length of QR is 3 cm.

solution: the area of parallelogram is PQ multiply the height.

A = PQ . 2

8 = (2x + 2) . 2

x = 1

after we find x, then we can find QR by substitute x to QR

QR = 8x – 5

QR = 8 – 5

QR = 3

MATHEMATICS IN OUR LIFE AND HOW TO DEVELOPE IT

Name : Nuraida Lutfi Hastuti
NIM : 11301241031
Class : Mathematics Education 2011

            Mathematics is around us. We always meet mathematics in our activities. Even though we live together with mathematics, but mathematics is not a living creature. Then what happens with mathematics? Mathematics can also experiencing growth. To facilitate the mathematics of human life was made in such a way that it can develope. Of course there are people who was instrumental in developing the mathematical sciences.

            To develop the science of mathematics is not as easy as turning the palm of the hand. Because mathematics is one of the fields of science then there are the steps in its development. It consists of:
            ANALYSIS STEPS
            In this process of problem solving research following the steps that perk do:
1. Problem Definition
At this step there are three main elements to be identified:
(A) Function Tujuan: state to help the destination for directing efforts to fulfill the objectives to be achieved.
(B) Functions Limits / constraints: constraints that affect the issue of the objectives to be achieved.
(C) the decision variables: variables that influence the decision-making problems.
2. Model Development
Collect data to assess the magnitude of the parameters of the problems faced affected. These estimates are used to construct and evaluate a mathematical model of the problem dart.
3. Solving Model
In formulating this problem typically use analytical models, the mathematical model that produces the equation, so that the optimum solution is achieved.
4. Validity Testing Model
Determine whether the model has been constructed to describe real situation accurately. If not, fix or create a new model.
5. Implementation of the final basil
Study or calculation basil translate into everyday language that is easily understood.

              MODELS IN RESEARCH
              In research recognized some form of a model that describes the characteristics and form a system problem. Various kinds of models in between:
Iconic Model
Is a physical replica as the original form with a much smaller scale. Example: The maker of the building, automotive models, and model airplanes.
Analog Model
A physical model but do not have a shape similar to that modeled. Example: thermometer gauge that indicates the level of the model temperatur.
Symbolic Model
Is a model that uses symbols (letters, numbers, shapes, images, etc.) which presents the characteristics and properties of a system darts. Comb: network (network diagrams), flow charts, flow charts, and others.
Mathematical Model
Includes models that represent the real situation of the system mathematical functions. Example: P = a'. Po state population models of living things.

              From the above, it is no less important is its application in life being so close to the mathematical problems in life. Starting from the initial steps that the role of college students is also important, especially students majoring in mathematics. Students can do simple research on the development of mathematics, or by raising issues that arise in the life of mathematics.

              This is one example of someone who has done research in the field of mathematics that is Wittgenstein. Wittgenstein is a unique philosopher. He criticizes his own opinion. Philosophers divide his views in two periods that are early and later. The early of Wittgentein’s view set by Tractatus Logico Philosophicus and the later set by Philosophical Investigations. Wittgenstein's conceptions of mathematics fall in three periods, that are early, midle and later. The early period Wittgenstein's conception of mathematics set by Tractatus Logico Philosophicus, the midle period set by Philosophical Grammar dan Philosophical Remarks, and the later period set by Remarks on the Foundations of Mathematics. The Wittgenstein’s view on mathematics is not belonging to logicism, formalism, or intuitionism. The later Wittgenstein on mathematics is that “Mathematics as a human invention”. He maintains his rejection for infinitely in mathematics.

              There are many other figures in mathematics research. In its development efforts as well as through a variety of ways. We may follow them, but with a different object and discover something new. With the existence of mathematical research is expected to solve problems in everyday life. In addition to the development of mathematics and produce researchers, especially those from Indonesia.
              Math is easy, math is beautiful, and mathematics is the breath of life.