The complete set of solution | x² + x| = x² + x is given by
Answers
Answer:
The area of each triangle is half of the area of any parallelogram on the same base and between the same parallels. Thus, the area of the two triangles is the same. Let us formalize this as a theorem:
Theorem: Two triangles on the same base and between the same parallels are equal in area.
Also, we have already seen how to calculate the area of any triangle. Consider a triangle. Take side BC to be the base of this triangle. Now, measure the height of this triangle, which will be the distance between BC and the parallel to BC through A:
Triangles - Height and Base
In other words, the height of this triangle will be the length of the perpendicular AD (from A to BC). Thus, the area of this triangle can be written as:
A = ½ × BC × AD
Note that there is nothing special about BC – we can take any other side as the base as well. Let us take AC as the base. In that case, the height of the triangle will be the length of the perpendicular from B to AC:
Triangles - Height perpendicular to base
The area of the triangle can now be written as:
A = ½ × AC × BE
Example 1: O is any point in the interior of a parallelogram ABCD. Show that:
area(
Δ
A
O
B
ΔAOB) + area(
Δ
C
O
D
ΔCOD)
= area(
Δ
A
O
D
ΔAOD) + area(
Δ
B
O
C
ΔBOC)
= ½ area(ABCD)
Solution: Consider the following figure, which shows O to be any point inside the parallelogram ABCD. We have a draw a line through O which is parallel to AB:
Two Parallelograms and Triangles
We note that ABFE and CDEF are themselves parallelograms (why?). Now, the area of
Δ
A
O
B
ΔAOB will be half of the area of parallelogram ABFE, because they are on the same base AB and between the same parallels AB and EF:
area(
Δ
A
O
B
ΔAOB) = ½ area(ABFE)
Similarly, the area of
Δ
C
O
D
ΔCOD will be half of the area of parallelogram CDEF, because they are on the same base CD and between the same parallels CD and FE:
area(
Δ
C
O
D
ΔCOD) = ½ area(CDEF)
Thus,
area(
Δ
A
O
B
ΔAOB) + area(
Δ
A
O
B
ΔAOB)
= ½ {area(ABFE) + area(CDEF)}
= ½ area(ABCD)
We have shown that the areas of
Δ
A
O
B
ΔAOB and
Δ
C
O
D
ΔCOD add up to half of the area of the parallelogram ABCD. The remaining area is the sum of the areas of
Δ
A
O
D
ΔAOD and
Δ
B
O
C
ΔBOC. It is now obvious that this sum will also be equal to ½ area(ABCD).
Example 2: In what ratio (of areas) does any median divide a triangle?
Solution: Consider the following figure, in which AD is the median through A, and we have drawn a line through A parallel to BC:
Median dividing a triangle
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