F4 Maths - -

2009-11-22 10:03 pm

In the figure, triangle ABC is a right-angled triangle with AB = 6 cm and AC = 10 cm. P is a point on AC such that a rectangle PQBR is formed. Let PR = x cm.

Express PQ and the area of the rectangle PQBR in terms of x.

Ans: PQ = ( 6 -3x / 4 ) cm, area = ( - 3x^2 / 4 + 6x ) cm^2

更新1:

b) Find the lengths of PR & PQ if the area if PQBR is a maximum http://img339.imageshack.us/content.php?page=done&l=img339/3251/trianglem.png&via=mupload

回答 (1)

用戶9112

2009-11-22 10:09 pm

✔ 最佳答案

Pythagora's theorem, AC^2 = AB^2 + BC^2
100 = 36 + BC^2 => BC = 8
Let PQ = y
SInce ABC and PQC are similar triangles
PQ/QC = AB/BC
y / (BC - x) = 6/8
y / (8 - x) = 6/8
8y = 6(8 - x) = 48 - 6x
y = (6 - 3x/4) cm
Area = xy = x(6 - 3x/4)
= (- 3x^2/4 + 6x) cm^2

2009-11-22 14:43:11 補充：
A = -(3/4)(x^2 - 8x + 16 - 16)
= (-3/4)(x - -4)^2 + 12
Max is 12 when x= 4
PQ = 4 and PR = 3

2009-11-22 14:43:21 補充：
A = -(3/4)(x^2 - 8x + 16 - 16)
= (-3/4)(x - -4)^2 + 12
Max is 12 when x= 4
PQ = 4 and PR = 3

👍 0 👎 0