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JordanPhrases that break parallel structure use different parts of speech or different sentence constructions than what came before. For example, in the phrase "I like running, jumping, and to skip," the phrase "to skip" breaks the parallel structure established by the words before it.
How do you construct a parallel line?
You place a ruler on a piece of paper and trace both sides. Put the point of your pair of compasses on the line and draw a circle. Then draw another circle, that is the same size, the same way further along the line. You will then be able to use your rule to draw a line that is a tangent to the two circle, and so parallel to your first line. Once you have done this you will realize that you don't need to draw a full circle just a small arc.
What quadrilateral has one pair of parallel sides and no right angles?
There are actually two answers. parallelogram and trapezoid. No. wait, a square or rectangle is a parallelogram, so I can't rule out that entire class. I guess you are left with trapezoid. Oh wait, a trapezoid can have two right angles and still fit the definition, so I can't rule out that entire class. A non-right triangular section would fit. A non-right trapezoid would be the same thing, I guess. A non-right parallelogram would also work.
A rule that is accepted true without proof?
A rule or a statement that is accepted without proof is a postulate.
How do you prove that the midpoints of a convex quadrilateral form a parallelogram?
There are several different ways and these depend on the level of your knowledge of different branches of mathematics. The simplest method uses vectors, but that is probably the topic that is covered last. So this proof is based on a simpler method.Suppose the quadrilateral is ABCD with lengths AB = 2w, BC = 2x, CD = 2y and DA = 2z (I have chosen multiples of two to avoid fractions later), and suppose that P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively.Consider the triangle DSRthen, by the cosine rule,SR^2 = DS^2 + RD^2 - 2*DS*RD*cos(D) = z^2 + y^2 - 2yz*cos(D)Now join AC and consider the triangle DAC. Again, using the cosine rule,AC^2 = DA^2 + CD^2 - 2*DA*DC*cos(D) = 4z^2 + 4y^2 - 4*2yz*cos(D) = 4*SR^2Taking square roots, AC = 2*SRTherefore the sides of triangles DS are all half the lengths of the corresponding sides in triangle DAC.Therefore the two triangles are similar and so the corresponding angles are equal.It follows that SR is parallel to AC.Similarly it can be shown that triangle ABC is similar to triangle PBQ so thatPQ is half of AC and PQ is parallel to AC.Thus, SR and PQ are both parallel to AC and half its length. They are therefore parallel to one another and the same length.Next, by drawing BD, it can be shown that QR and PS are parallel and equal.Thus, both pairs of opposite sides of PQRS are equal and parallel which proves that the quadrilateral is a parallelogram
How does one differentiate between the sine rule and the cosine rule?
The sine rule is a comparison of ratios: (sin A)/a = (sin B)/b = (sin C)/c. The cosine rule looks similar to the theorem of Pythagoras: c2 = a2 + b2 - 2ab cos C.
