Algebra question concerning a trapezium with an unknown?
Good morning everyone, My daughter came home and asked her old man regarding homework for her math class. I am actually very good at calculating but when it comes to algebra, I am lost ;) The question is regarding a trapezium (sorry, needed to look up the word) or trapezoid in American. So here the question: We...
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answers:
lenpol7 say: a + b + c + d = 14

b = a - 2
c = a - 2
d = 3(a - 2)

Substituting
a + (a - 2) + (a - 2) + 3(a - 2) = 14
6a - 10 = 14
6a = 24
a = 4
b = 2
c = 2
d = 6
Verification
4 + 2 + 2 + 6 = 14
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Steve A say: Three times longer = four times as long. Is that what is meant?
(By analogy, one time longer = twice as long. 50% longer is 150% as long. Bad English yields bad math.)
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Krishnamurthy say: We have the trapezium with 4 lines
The upper line we call line A.
The left side line is B, the right side line is C and the bottom line is D.
The extent of the trapezium is 14 cm.
The sidelines B and C are 2 cm shorter than the upper line A.
The bottom line is 3 times longer than the sidelines, how long are the lines.
A = 14
A - (B + C) = 2 , D = 3B
Solution:
A = 14, B = 12, C = 12, D = 36
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khalil say: four unknowns and three equations ...
write the fourth
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037 G say: I will solve this problem 2 ways. The first solution is based literally on how you stated the problem,
the second is likely how the problem was stated in the text book.
Check to see if they said B and C or B or C is shorter than A and 3 times D.

Part 1 is if sides B+C is 2 shorter than A and 3 times D
Part 2 is if sides B or C is 2 shorter than A and 3 time D
_______________________________________... 1
14 = A + B + C + D................Extent = 14
(B+C) + 2 = A .......................From 2nd statement add 2 to B+C to equal A
D = 3x(B+C)..........................Last statement D is 3 times (B+C)
Now plug in A & D in the first equation to get
14 = [B+C + 2] + B + C + [3x(B+C)]
also since B=C replace all Bs with a C and solve for C which is the same for B, so
14 = [C+C+2]+C+C+[3(C+C)], or C+C+2+C+C+3x(2xC) , so adding the Cs you
have 10 of them and subtract 2 from both sides to get
12 = 10C, divide both sides by 10 to get C by itself
C = 1.2 cm and so is B
so A = B+C+2 or A = 1.2+1.2+2 = 4.4 cm
and D = 3(1.2+1.2) = 7.2cm
---------------------------------------... 2
Be careful how you read the 2nd statement
if it says both B and C are individually 2 cm shorter than A then the equations
change to
B+2 = A and C + 2= A and B=C
and if the last statement meant D is 3 times either of the side lines then
D = 3B or D = 3C
Plugging these equations into the original, to get all Bs for the other unknowns we get
14 = A+B+C+D
14 = (B+2) + B + B + 3B
12 = 6B so B = 2 cm and so is C
since D = 3B then D = 3(2) = 6 cm
and A = B+2 = 2+2 = 4 cm
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atsuo say: Please confirm them .

"the extent of the trapezium is 14 cm"

... "the extent" is concerned with an area , but "14 cm" means a length .
... I think "the extent" means "the perimeter" , so A + B + C + D = 14 .

"The sidelines B+C are 2 cm (inches) shorter than the upper line A"

... I think "B+C" does not mean the sum of B+C , it means "B and C" .
... So B = A - 2 and C = A - 2 .

"The bottom line is 3 times longer than the sidelines"

... I think B = C , so D = 3B (= 3C) .

Therefore , we find

A + B + C + D = 14 ---(#1)
B = A - 2 ---(#2)
C = A - 2 ---(#3)
D = 3B ---(#4)

Substitute (#2) into (#4) , we find

D = 3(A - 2) = 3A - 6 ---(#5)

Substitute (#2) , (#3) and (#5) into (#1) , we find

A + (A - 2) + (A - 2) + (3A - 6) = 14
6A - 10 = 14
6A = 24
A = 4

Substitute it into (#2) , (#3) and (#5) , we find
B = 4 - 2 = 2
C = 4 - 2 = 2
D = 3*4 - 6 = 6

So we found the lenghths of four sides of the trapezoid .
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dxdzv say: wnqjpujf
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