Median of a trapezoid is a line segment found on the midpoint of the legs of a trapezoid. It is also known as mid-line or mid-segment. Its basic formula is AB + CD divided by 2.
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Draw a number line (l) and mark the points O, A, B and C such that OA = AB = BC = 1Draw CD ⊥ l, such that CD = 1 units.Join OCIn right ΔOCD,OD2 = OC2 + CD2Taking O as centre and D as radius, draw an arc which cuts l in FNow, draw EF ⊥ l, such that EF = 1 unitsJoin OE'In right ΔOEF,OE2 = OF2 + EF2Taking O as centre and OE as radius, draw an arc which cuts l in HNow, draw GH ⊥ l, such that GH = 1 unitsJoin OGIn right ΔOGH,Taking O as centre and OG as radius, draw an arc which cuts l in J.Now, draw IJ ⊥ l, such that IJ = 1 unitsJoin OI,In right ΔOIJ,Taking O as centre and OI as radius, draw an arc which cuts l in L.The point L represents on the number line.
a b c
Ab and Ba are the same line because there are no endpoints to a line. Therefore, you can reverse the order of the letters. So, Cd and Dc are not the same ray because the first letter is the endpoint. So on ray Cd, point C is the endpoint and d is a point on the line coming from the endpoint. On ray Dc, D is the endpoint.
It is easiest to draw it using two right angled triangles.Draw a line AB that is 2 units long. From B, draw BC which is perpendicular to AB and 2 units long. Join AC. From C, draw CD which is perpendicular to AC (clockwise if BC is clockwise from AB, or anticlockwise if BC is anticlockwise) and make CD 2 uinits long. Then AD is a line segment which is sqrt(12) units long.
Yes, it doesn't matter which end you start with when naming a line segment.
== == 1) Draw a line segment AB of 5 units 2) Draw the perpendicular bisector CD of AB such that Cd meerts AB at C. 3) Mark off CE = 2 units on CD 4) Draw the straight line segments AE & BE. ABE is your triangle. Its base (AB) = 5 and height (CE) = 2, so its area = [base x ht] / 2 = 5 sq units
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CD and DC
Median of a trapezoid is a line segment found on the midpoint of the legs of a trapezoid. It is also known as mid-line or mid-segment. Its basic formula is AB + CD divided by 2.
They define the same line segment, connecting two points.
Let us say the line segment is AB. Then take a compass and spread it so that the distance between the needle around which the compass rotates and the pencil at the other end is a little over half the length of AB. Place the needle of the compass on A and draw small arcs above and below the line AB. Without altering anything on the compass, place the needle on B and draw small arcs as before above and below the line AB such that these arcs intersect the older arcs. Now join the two intersection points of the arcs and call this line CD. CD is the right bisector of AB A----------------|----------------B
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That is false. CD could be a short line segment only a few degrees off from the two circles' common chord while EF could be a long line segment going all the way across the diagram.
Yes, but then again it depends on what or where they are!
20m