Mathematics Questions 5 marks

Seashore

2008-02-03 4:19 am

更新1:

2. p, q, r are the lengths of the sides of ΔPQR in which∠PRQ=θ. The length r is 8 less that the product of the lengths p and q. The lengths pa nd q are the roots of the quadratic equation x^2 - 8x +m = 0 where m≠0.

更新2:

(i) Show that cosθ = (14=m)/2 (ii) Hence show that ΔPQR exist for 12

更新3:

(i) Show that cosθ = (14=m)/2 (ii) Hence show that ΔPQR exist for 12

更新4:

Just ignore question 2, thanks

更新5:

點解part (iv) 既answer唔係1?就咁睇既話，max area唔係=個unit square既area咩？

回答 (1)

myisland8132

2008-02-03 4:47 am

✔ 最佳答案

1. ABCD is a unit spuare. Points E and F are chosen on AD and DC respectively such that∠AEG =∠FHC, while G and H are the points at which BE and BF respectively cut the diagonal AC. Let AE=p, FC=q, ∠AEG=α and ∠AGE=β.

(i) Express α in terms of p, and β in terms of q.
(ii) Prove that p+q=1-pq
(iii) Show that the area of the quadrilateral EBFD is given by:
1- p/2 + (p-1)/2(1+p)
(iv) What is the maximum value of the area EBFD?
(i)
tanα =1/p
tanβ =1/q
tan(180-α-β )=tan45=1
tan(α+β)=-1
(tanα+tanβ)/(1-tanαtanβ)=-1
[(p+q)/pq][pq/(pq-1)]=-1
p+q=1-pq
(iii)
Since
p+q=1-pq
q=(1-p)/(p+1)
the area of the quadrilateral EBFD
=1-(p/2)-(q/2)
=1- p/2 + (p-1)/2(1+p)
(iv)
Let A=1- p/2 + (p-1)/2(1+p)
dA/dp
=-1/2+(1/2)[(1+p)-(p-1)]/(1+p)^2
=-1/2+1/(1+p)^2
Let dA/dp=0
1/(1+p)^2=1/2
1+p=√2
so p=√2-1
the maximum value of the area EBFD
=1- (√2-1)/2 + (√2-2)/2(√2)
=1- (√2-1)/2 +√2 (√2-2)/4
=1+ (-2√2+2)/4 +(2-2√2)/4
=1-(2√2-2)/2
=1-(√2-1)
=2-√2

2008-02-03 13:51:15 補充：
ABCD才是邊長為1的正方形﹐EBFD是四邊形

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