First consider triangles ACD and BCD.
They share a commom base, CD, and their height is the ditance between the parallel lines AB and CD.
Consequently, area(ACD) = area(BCD) . . . . . . . . eqn 1
But ACD = AOD + OCD and BCD = BOC + OCD
So area(ACD) = area(AOD) + area(OCD)
and area(BCD) = area(BOC) + area(OCD)
Substituting in eqn 1,
area(AOD) + area(OCD) = area(BOC) + area(OCD)
area(AOD) = area(BOC)
No. An isosceles trapezium (isosceles triangle with its apex removed) would have congruent diagonals but it is not a parallelogram.
None and no diagonals
Yes. An isosceles trapezium is like an isosceles triangle whose peak has been cut off by a line parallel to its base.
yes, because perpendicular lines always intersect. all lines intersect unless they are parallel or on separate planes (skew)
I think you might be thinking of a trapezium. A triangle with the top point sliced off by a line parallel to the base.
No. An isosceles trapezium (isosceles triangle with its apex removed) would have congruent diagonals but it is not a parallelogram.
None and no diagonals
A triangle has by definition three intersecting sides. If two of the sides are parallel, they will never intersect, so no triangle can ever be formed.
An isosceles trapezium has one pair of sides parallel to each other and the other pair that are of equal length (but facing in opposite directions). An isosceles trapezium can be imagined as an isosceles triangle whose top has been chopped off by a line parallel to its base.
Yes. An isosceles trapezium is like an isosceles triangle whose peak has been cut off by a line parallel to its base.
none. A triangle, in standard 2 dimensions cannot have any parallel sides. Let's show without going into formal proof. Let's say that you try to make a triangle with 2 parallel sides. -------------side a------------- |C | | |side b | | |A -------------side c--------------- Side a & b intersect at vertex C. Side b & c intersect at vertex A, and since sides a & c are parallel, they will never intersect, so there is no third vertex, so it's not a polygon, much less a triangle.
Suppose the base and parallel sides of the trapezium are labelled a and b. Suppose, also, that the distance between a and b is h. Draw a diagonal. This will split the trapezium into one triangle whose base is the trapezium's base (a) and another upside-down triangle whose base is the trapezium's top (b). The heights of both these triangles will be the same as the distance between the parallel sides of the trapezium (h). The area of the first triangle is 0.5*a*h The area of the second triangle is 0.5*b*h So the area of the trapezium = 0.5*a*h + 0.5*b*h = 0.5*(a+b)*h
yes, because perpendicular lines always intersect. all lines intersect unless they are parallel or on separate planes (skew)
It's a quadrilateral (4 sides) and it has two sides that are parallel. If you draw a triangle and cut off one corner, you'll have two figures. One is a triangle, and the other, the 4-sided one, will be a trapezium.
I think you might be thinking of a trapezium. A triangle with the top point sliced off by a line parallel to the base.
In a general trapezium, the non-parallel sides are not congruent. However, in an isosceles trapezium, the non-parallel sides are congruent. So the shape is like an isosceles triangle with its apex chopped off by a straight line parallel to its base.
there is 2 diagonals in a triangle