Let #bar(AB)# be cut into equal and unequal segments at #C and D# Show that the rectangle contained by #bar(AD)xxDB# together with the square on #CD# is equal to the square on #CB#?

Answer 1
In the fig C is mid point of AB . So #AC=BC#
Now rectangle contained by #bar(AD) and bar(DB)# together with the square on#bar(CD)#
#=bar(AD)xxbar(DB)+bar(CD)^2#
#=(bar(AC)+bar(CD))xx(bar(BC)-bar(CD))+bar(CD)^2#
#=(bar(BC)+bar(CD))xx(bar(BC)-bar(CD))+bar(CD)^2#
#=bar(BC)^2-cancel(bar(CD)^2)+cancel(bar(CD)^2)#
#=bar(BC)^2->"Square on CB"# Proved
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Answer 2
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Answer 3

To show that the rectangle contained by (AD) and (DB) together with the square on (CD) is equal to the square on (CB), we can use the Pythagorean theorem and the properties of rectangles.

Given that (AB) is a line segment, we can denote its length as (AB). Let (AC = x) and (CB = y), where (x) and (y) represent the lengths of segments (AC) and (CB) respectively.

The length of (AD) can be expressed as (AB - BC), so (AD = AB - y = x). Similarly, the length of (DB) is (CB - CD), so (DB = y - x).

Now, let's find the areas of the figures involved:

  • The rectangle contained by (AD) and (DB) has an area of (AD \times DB = xy - x^2).
  • The square on (CD) has an area of (CD^2 = y^2 - x^2).
  • The square on (CB) has an area of (CB^2 = y^2).

Adding the area of the rectangle and the square on (CD) gives us (xy - x^2 + y^2 - x^2 = xy + y^2 - 2x^2).

Now, we need to prove that this is equal to the area of the square on (CB), which is (y^2).

So, we need to show that (xy + y^2 - 2x^2 = y^2). Simplifying this expression, we get (xy - x^2 = 0), which can be rewritten as (x(y - x) = 0).

Since (x) cannot be zero (as it represents a length), we must have (y - x = 0), which implies (y = x). This means that (AD) and (DB) are equal in length.

Therefore, the rectangle contained by (AD) and (DB) together with the square on (CD) is equal to the square on (CB).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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