F.1 maths question (急!!!20點)

ABC is a triangle with BC (base) = AD (height) = 18cm. If MNPQ is a square such that QM lies on BC, P lies on AB and N lies on AC, is the area of square PQMN half of that of ABC ? Explain your answer.
Solution : As QM lies on BC, △APN～△ABC.Let side of the square PQMN be s. As △APN～△ABC, and BC = AD = 18cm,so PN = AX (height of △APN) = sAX + XD = 18==> s + s = 18==> s = 9So, area of △ABC = 18 x 18 / 2 = 162cm², andarea of the square PQMN = 9 x 9 = 81cm², which is half of that of △ABC.

Given BC = AC = 18,
Than MNPQ is a square,
And QM = BC,
So MQ = BC = PQ = AC = 18,

Than we know :

MNPQ = 18 x 18 = 324

ABC = (18 x 18) / 2 = 162

So, the area of square PQMN is half of that of ABC

其實本題對於中一的程度來說也算太難了吧？？？

2015-07-25 03:21:19 補充：
晉堯，why AC⊥BC?

2015-07-25 21:52:28 補充：
浪費了 Eddie 的苦心～

2015-08-05 17:03:27 補充：
各位網友，知識家將會有變動：

https://hk.knowledge.yahoo.com/plus/notice/notice_2015.html

(⊙＿⊙)

Given BC = AC = 18,
Let s be the side length of square MNPQ (i.e. PQ=QM=MN=PN=s), and h be the height of triangle APN.
∆APN∼∆ABC (AAA),
Height of ∆APN:Height of ∆ABC=PN:BC (ratio of sides and heights of similar triangles are the same)
h/(h+s)=s/18
18h=sh + s^2
h=s^2/(18-s)
Area of ∆APN=sh/2=s[s^2/(18-s)]/2=s^3/2(18-s) ……(i)

∵ PN//MQ//BC (PN and MQ are opposite sides of square MNPQ), ∴BPNC is a trapezium with height PQ.
Area of trapezium BPNC=(PN+BC)xPQ/2
=(s+18)s/2
Area of ∆BPQ and ∆MNC = Area of trapezium BPNC – Area of square MNPQ
=(s+18)s/2 – s^2
=(18s-s^2)/2 ……(ii)

Area of ∆APN, ∆BPQ and ∆MNC = s^3/2(18-s) + (18s-s^2)/2 ……(i)+(ii)
=[s^3+(18s-s^2)(18-s)] /2(18-s)
=s(s^2-18s+162)/(18-s)

When the area of square MNPQ is half of triangle ABC, the sum of area of the 3 small triangles ∆APN, ∆BPQ and ∆MNC equals to the area of the square MNPQ, thus
s(s^2-18s+162)/(18-s) = s^2
s^2-18s+162 = s(18-s)
s^2-18s+81=0
(s-9)^2=0
s=9

A feasible solution does exist for s (i.e. square PQMN has sides of 9 cm), so the area of square PQMN is half of that of ABC.

2015-07-24 10:22:03 補充：
The height h of triangle APN is defined as perpendicular to the base PN.