Area of triangle..calculation to verify

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February 13th 2010, 07:54 AM
nikk

3 Attachment(s)

Area of triangle..calculation to verify

Hai ...attached is question and answer. Detail as below

(a) Question.JPG
(b) Method 1 & 2.JPG - my solution but i can't get the ans. i can't know what wrong with my formula. i think is correct. can u help me on that?
(c) Method 3.JPG -my correct solution

tq for your time
February 13th 2010, 08:17 AM
skeeter

the two expressions have the same value when x = 5.
February 13th 2010, 08:26 AM
nikk

Quote:

Originally Posted by skeeter

the two expressions have the same value when x = 5.

tq skeeter for your quick response....
which expressions did u mean have the same value when x = 5...

do u get the same ans as what i get for method 2 and method 1??
February 13th 2010, 12:33 PM
skeeter

Quote:

Originally Posted by nikk

tq skeeter for your quick response....
which expressions did u mean have the same value when x = 5...

do u get the same ans as what i get for method 2 and method 1??

the expression for y using method 1 and method 2
February 13th 2010, 07:59 PM
nikk

is my metod 1 and method 2 is wrong??
February 14th 2010, 01:05 AM
SuperCalculus

I can't see where you went wrong with your original equation, but the way I'd do it is:
$BD = FE = x-2 cm$ .
As $AB = BC$ ,
$BF = x-3cm.$

Hence, area of shaded region:

$(x-2)(x-3) + 0.5[3(x-2)]$

Expand:

$x^2 - 2x - 3x + 6 + 1.5x - 3$

Simplify;
$x^2 - 3.5x + 3 = y$

Which is the original equation.

EDIT: Both your equation and the one in the question seem to work for the area, so maybe it's just a fault of the question.
February 14th 2010, 01:29 AM
HallsofIvy

Both formulas are correct (although since the problem said "show that the area is $x^2- \frac{7}{2}x+ 3$ " the second method would be the "correct answer").

Skeeter pointed out that they both give the same result when x= 5 because, in fact, x must be 5!

I'm going to use "A" to represent the common value of AB and BC so that I can use "x" as a variable.

Set up a coordinate system with (0, 0) at B and the positive x axis extending from B through C. Then point A is at (0, A) and point C is at (A, 0). It is easy to see that the line AC is given by y= A- x.

Point E has coordinates (A-3, A- 2). Since it lies on line AC, we must have

A- 2= A- (A- 3)= 3 so A= 3+ 2= 5.

The "x" in the problem must be 5 and the two formulas give the same result.
February 14th 2010, 01:38 AM
SuperCalculus

Quote:

Originally Posted by HallsofIvy

Both formulas are correct (although since the problem said "show that the area is $x^2- \frac{7}{2}x+ 3$ " the second method would be the "correct answer").

Skeeter pointed out that they both give the same result when x= 5 because, in fact, x must be 5!

I'm going to use "A" to represent the common value of AB and BC so that I can use "x" as a variable.

Set up a coordinate system with (0, 0) at B and the positive x axis extending from B through C. Then point A is at (0, A) and point C is at (A, 0). It is easy to see that the line AC is given by y= A- x.

Point E has coordinates (A-3, A- 2). Since it lies on line AC, we must have

A- 2= A- (A- 3)= 3 so A= 3+ 2= 5.

The "x" in the problem must be 5 and the two formulas give the same result.

Yes, but surely as the question did not account for this, they were looking for an identity for the area? The formula the user submitted is an equation, not an identity, as it is not identically equal to the formula in the question.
While it is admittedly true in this case, I would argue that it is better to get into the habit of finding an identity rather than a formula, as it can then be applied to mathematically similar triangles rather than just that one. In short, yes, correct, but the identity is better mathematical practise in my opinion.