20 points!

lucinda is writing a coordinate proof to show that a diagonal of a parallelogram partitions the parallelogram into two equal areas.

enter your answers in the boxes to complete lucinda's proof.

since klmn is a parallelogram and a parallelogram's opposite sides are parallel and congruent, the coordinates for m are ([answer], [answer]).

in △kmn, the length of the base is [answer] and the height is [answer. so an expression for the area of △kmn is [answer].

in △klm, the length of the base is 3x and the height is 2y. so an expression for the area of △klm is [answer].

comparing the area of the two triangles that are formed by a diagonal of the parallelogram shows that a diagonal of a parallelogram partitions the parallelogram into two equal areas.

answer:

step-by-step explanation:

let the straight lines ab, cd be equal and parallel, and let the straight lines ac, bd join their extremities that are on the same side;

then ac, bd are themselves equal and parallel.

parallel lines

draw ad.

then, since ab is parallel to cd, (hypothesis)

and ad meets them,

the alternate angles bad, adc are equal. (i. 29)

and because ab is equal to cd,

and ad is common to triangles adb, adc,

the two sides ba, ad are equal to the two sides cd, da respectively;

and we showed that angle bad is equal to angle adc;

therefore the remaining side ac is equal to the remaining side bd,

and the remaining angles are equal to the remaining angles, namely those that are opposite the equal sides: (s.a.s.)

angle cad is equal to angle dba.

therefore, since the straight line ad meets the two straight lines ac, bd and makes the alternate angles cad, dba equal,

ac is parallel to bd. (i. 27)

and we showed that they were equal.

therefore, if two straight lines etc.   q.e.d.

end of section

the proposition proves that if two sides of a quadrilateral are equal and parallel, then the figure is a parallelogram. (definition 14.) hence we may construct a parallelogram; for, proposition 31 shows how to construct a straight line parallel to a given straight line.

the next theorem has for its hypothesis that a figure is a parallelogram, that is, the opposite sides are parallel. and it proves what is obvious to the eye:   the opposite sides are equal.

this theorem will be fundamental to the theory of equal areas. in fact, as we have pointed out several times, when we say that two figures are "equal," we mean that they are equal areas. see problems 1 and 2 following proposition 4.

the student should look at the figure below and, given that the figure is a parallelogram, it should be clear why those two triangles are equal.

why?   s.a.s.

proposition 34.   theorem

in a parallelogram the opposite sides and angles are equal, and the diagonal bisects the area.

let acdb be a parallelogram, and bc its diagonal;  

then the opposite sides and angles are equal,

and the diagonal bc divides the parallelogram into two equal areas.

parallel lines

since ab is parallel to cd, and cb meets them,

the alternate angles abc, bcd are equal; (i. 29)

and since ac is parallel to bd, and cb meeets them,

the alternate angles acb, cbd are equal.

in triangles abc, dcb, then, the two angles abc, bca  

are equal to the two angles dcb, cbd respectively,

and the side bc is common;

therefore the remaining sides will equal the remaining sides:

ab will equal cd, and ac will equal bd;

and the remaining angle will equal the remaining angle:

angle cab will equal angle bdc. (a.s.a.)

also, because angle abc is equal to angle bcd,

and angle acb is equal to angle cbd,

the whole angle abd is equal to the whole angle acd. (axiom 2)

therefore the opposite sides and angles of a parallelogram are equal.

parallel lines

next, the diagonal bisects the parallelogram.

for, since ab is equal to cd, and bc is common,

the two sides ab, bc are equal to the two sides dc, cb respectively;

and we have shown that angle abc is equal to angle bcd;

therefore triangles abc, dcb are equal. (s.a.s.)

therefore the diagonal bc divides the parallelogram into two equal areas.

therefore, in a parallelogram etc.   q.e.d.

"turn" the page and do some problems.

or

continue on to the next proposition.

previous proposition

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !