math problem

[snowbrood]

from book D4 mathematics o level oxford Q.3c and Q.7e of excercise 3d

 

[whatisthist]

Can you post the question please? We won't be able to help without it, thanks.

 

[shezi1995]

Q.3(c). Ans.
Since PQRS is a parallelogram, SR = PQ= 2(2b)=4b.
SN=1/3SR
=4/3b
MN=MP+PS+SN
=-2b+a+4/3b
=a-2/3b
NM=-(a-2/3b)=2/3b-a
Q.7(e). Ans.
BD=BC+CD(you must have found BC in (a) part)
=v-u+3/2u=v+1/2u

 

[snowbrood]

@shezi1995 hey the answer is 2b-2a/3

 

[shezi1995]

@shezi1995 hey the answer is 2b-2a/3

Well my book gives the same answers i got. You are talking about oxford new syllabus mathematics 6th edition book 4(or D4), aren't you? If yes then my answers are correct or you can post the questions if im answering the wrong ones.

 

[snowbrood]

i am extremely sorry... u are right i owe u an apology. thanks for the help

 

[snowbrood]

hey shezi can u solve Q.10 d(iii) and Q.11b of excercise 3d maths d4 oxford

 

[shezi1995]

i am extremely sorry... u are right i owe u an apology. thanks for the help

No problem.

hey shezi can u solve Q.10 d(iii) and Q.11b of excercise 3d maths d4 oxford

Q.10.d(iii) Ans.
The only thing here is to identify the common height or the height which both these triangles have. So lets call that height "h". See the attachment for the rest of the answer because i couldnt explain it in writing.

Q,11.b Ans.
RQ=RP + PQ
RP=-(PR)=-(3a)=-3a
PQ=4b
So RQ=-3a+4b=4b-3a

 

[snowbrood]

hey i got the same answer i spent a hour on this.. i ran mad and now as i am assured of the answer i can relax now.. can u solve Q.11e(ii) of the same excercise and Q.10 of review questions 3

 

[shezi1995]

hey i got the same answer i spent a hour on this.. i ran mad and now as i am assured of the answer i can relax now.. can u solve Q.11e(ii) of the same excercise and Q.10 of review questions 3

 

[snowbrood]

ah u are a kind person thx buddy, u hv been very helpful.. regards syed abdul rehman shah kazmi(snowbrood)

 

[snowbrood]

may Allah bless u Amin