Tuesday, June 30, 2009

Similar Triangle Problems

We present several problems on similar triangles with detailed solutions.



Problem 1:In the triangle ABC shown below, A'C' is parallel to AC. Find the length y of BC' and the length x of A'A












Solution to Problem 1:
BA is a transversal that intersects the two parallel lines A'C' and AC, hence the corresponding angles BA'C' and BAC are congruent. BC is also a transversal to the two parallel lines A'C' and AC and therefore angles BC'A' and BCA are congruent. These two triangles have two congruent angles are therefore similar and the lengths of their sides are proportional. Let us separate the two triangles as shown below.


We now use the proportionality of the lengths of the side to write equations that help in solving for x and y.

(30 + x) / 30 = 22 / 14 = (y + 15) / y
An equation in x may be written as follows.

(30 + x) / 30 = 22 / 14
Solve the above for x.

420 + 14 x = 660 x = 17.1 (rounded to one decimal place).
An equation in y may be written as follows.

22 / 14 = (y + 15) / y
Solve the above for y to obtain.

y = 26.25





Problem 2:

A research team wishes to determine the altitude of a mountain as follows: They use a light source at L, mounted on a structure of height 2 meters, to shine a beam of light through the top of a pole P' through the top of the mountain M'. The height of the pole is 20 meters. The distance between the altitude of the mountain and the pole is 1000 meters. The distance between the pole and the laser is 10 meters. We assume that the light source mount, the pole and the altitude of the mountain are in the same plane. Find the altitude h of the mountain.

Solution to Problem 2:
We first draw a horizontal line LM. PP' and MM' are vertical to the ground and therefore parallel to each other. Since PP' and MM' are parallel, the triangles LPP' and LMM' are similar. Hence the proportionality of the sides gives: 1010 / 10 = (h - 2) / 18
Solve for h to obtain h = 1820 meters.




Problem 3:

The two triangles are similar and the ratio of the lengths of their sides is equal to k: AB / A'B' = BC / B'C' = CA / C'A' = k. Find the ratio BH / B'H' of the lengths of the altitudes of the two triangles.






Solution to Problem 3:
If the two triangles are similar, their corresponding angles are congruent. Hence angle BAH and B'A'H are congruent. We now examine the triangles BAH and B'A'H'. These triangles have two pairs of corresponding congruent angles: BAH and B'A'H' and the right triangles BHA and B'H'A'. The triangles are similar and therefore: AB / A'B' = BH / B'H' = k


Graphs of Functions and Algebra - Interactive Tutorials

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Algebra Formulae

Algebra
Factors and expansions a, b real; n positive integer
(a ± b)² = a² ± 2ab + b²
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³
(a ± b)4 = a4 ± 4a³b + 6a²b² ± 4ab³ + b4
a² - b² = (a - b)(a + b)
a² + b² = (a - bi)(a + bi)
a³ - b³ = (a - b)(a² + ab + b²)
a³ + b³ = (a + b)(a² - ab + b²)
a4 + b4 = (a² + 2½ab + b²)(a² - 2½ab + b²)
an - bn = (a - b)(an-1 + an-2b + ... + bn-1), for n even
an + bn = (a + b)(an-1 - an-2b - ... - bn-1), for n odd

Powers and roots a > 0, b > 0, x, y real
axay = ax+y
ax / ay = ax-y
axbx = (ab)x
ax / bx = (a / b)x
a-x = 1 / ax

Proportion
If a / b = c / d, then (a ± b) / b = (c ± d) / d, and (a - b) / (a + b) = (c - d) / (c + d)

Common Irrational Numbers (includes the first 12 prime numbers)
To find the root of a number which is not prime – do a prime factorization of the number and multiply / exponentiate appropriately.

Monday, June 15, 2009

Program Peningkatan Prestasi PMR 2009


l. Analisis Kesalahan Umum
–Pelajar akan diberikan set soalan berserta dengan jawapan yang salah. Pelajar kemudiannya dikehendaki mengenalpasti kesalahan atau kesilapan yang wujud dalam jawapan tersebut.
–Program ini dapat melatih kepekaan pelajar untuk tidak mengulangi kesilapan yang sama berulang kali.
2. Analisis Item
–Kelemahan pelajar akan dikenalpasti mengikut topik-topik yang tertentu.
–Pelajar akan diberikan pemerhatian yang khusus untuk mendalami topik tersebut.
3. Kelas Tahanan
–Pelajar yang memperoleh markah yang rendah dalam ujian dimestikan untuk mengulangi ujian tersebut sehingga mencapai markah lulus yang ditetapkan. Markah lulus adalah berbeza pada setiapkali ujian tersebut diulang.
–Pelajar akan memberikan tumpuan yang penuh agar tidak mengulangi kesilapan yang sama dalam melakukan ujian tersebut.
4. Mentor Dan Mentee
–Pelajar akan dibahagikan kepada berberapa kumpulan. Setiap kumpulan akan mempunyai seorang mentor sebagai tempat rujukan bagi pelajar yang digelar sebagai mentee. Mentor ini pula akan merujuk kepada duta yang dilantik oleh guru. Duta yang dilantik adalah seorang yang lebih berkemahiran di antara pelajar yang lain.
5. Eye To Eye
-Seringkali pelajar akan berasa segan atau malu untuk bertanya di dalam kelas.Guru akan mengambil inisiatif untuk bertemu dengan pelajar secara individu untuk mengetahui masalah yang berkaitan dengan lebih jelas. Dengan ini, guru dapat mengenalpasti kelemahan yang pelajar hadapi dan berusaha untuk mengatasi masalah tersebut

Ahli-ahli Panitia Mathematics PMR

GKMP Sains & Math
Pn. Hjh. Rafidah Kemat
Ketua Panitia
En Mohd Yussairi Bin Jaafar

Ahli-ahli Panitia

En Mohd Hafiz Bin Mahat

Cik Zaihasra Azwa Bt Zaharim

Cik Farah Falina Bt Ahmad Sanusi


Pn Suziana Bt Mohd Sultan