Let E be where the diagonals intersect.
Since ABCD is a rectangle, the diagonals are the same length, and bisect each other.
Thus, EB = EC and the triangle BCE is isosceles, making m<DBC = m<ACB
So,
10x = 4x^2
10 = 4x
x = 5/2 = 2.5
<DBC = 25°
<ACB = 25°
If m<DBC=10x and m<ACb=4x^2, find m<ACB.
The quadrilateral ABCD is a rectangle.
B ------------- C
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A -------------|D
There are diagonal bisectors inside the rectangle but I could not draw them in. The diagonals are DB and CA and the point in the middle is E.
2 answers
the coordinates of the vertices of abc are a (2 5) B (6, -1) and C (-4,-2). FInd the perimeter of abc, to the nearest tenth