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