If BC = 10.4
and CD = 14.4
, what is BD?
In triangle BCD, the sides are formed by the rays BD and CD, with point B being one endpoint of ray BD and point C being one endpoint of ray CD. The vertex D is where the two rays meet. Therefore, the sides of triangle BCD consist of the segments BD and CD, along with the segment BC.
Yes, the expression AC + AD + BC + BD can be factored as (A + C)(B + D). This is evident by applying the distributive property, where expanding (A + C)(B + D) yields AB + AD + BC + CD. The terms AC and BD are not present, so the expression can be expressed in a different form, but the original expression itself represents a different factoring structure.
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To find AC, you need to add the lengths of segments BC and CD together. If BC is 7 and CD is 24, then AC = BC + CD = 7 + 24 = 31. Therefore, AC is 31 units long.
The diagonals arenotthe sides. They're lines you draw from one angle of the parallelogram to the angle opposite it. So if you have parallelogram ABCD, your diagonals are AC and BD, because AB, BC, CD, and DA are all sides.
No, four lines can can have 6 total intersections. They would be Ab, Ac, Ad, Bc, Bd, and Cd
Yes. Since they are rational numbers, let's call the first one a/b and the second one c/d where a,b,c, and d are integers. Now we can subtract by finding a common denominator. Let's use bd. So we have ad/bd-cb/bc= (ad-bc)/CD which is rational since we know ad and bc are integers being the product of integers and CD is also an integers. Call ad-bd=P and call CD=Q where P and Q are integers. We now see the difference is of two rationals is rational.
Six. If the sides are labelled a, b, c and d then the congruent pairs are: ab, ac, ad, bc, bd and cd
Yes, the expression AC + AD + BC + BD can be factored as (A + C)(B + D). This is evident by applying the distributive property, where expanding (A + C)(B + D) yields AB + AD + BC + CD. The terms AC and BD are not present, so the expression can be expressed in a different form, but the original expression itself represents a different factoring structure.
You cannot prove it since it is not true for a general quadrilateral.
segments AB, CD, BD.
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shift
To find AC, you need to add the lengths of segments BC and CD together. If BC is 7 and CD is 24, then AC = BC + CD = 7 + 24 = 31. Therefore, AC is 31 units long.
The diagonals arenotthe sides. They're lines you draw from one angle of the parallelogram to the angle opposite it. So if you have parallelogram ABCD, your diagonals are AC and BD, because AB, BC, CD, and DA are all sides.
A Rectangle is a quadrilateral (four sided polygon) with two pairs of equal and parallel sides (opposite sides are parallel and equal, one pair is usually a different length from the other pair but if they are equal it is called a square), and all angles are right angles (90°). It has two diagonals which have the properties:The diagonals are always congruent (of equal length);The diagonals bisect each other (cut each other into two equal parts);The diagonals do not bisect the angles (unless the rectangle is a square when they do);The diagonals are not perpendicular (unless the rectangle is a square when they do).PROOF of the diagonals congruent:Take a rectangle ABCD with diagonals AC and BD.Using Pythagoras on the triangles ACD and BCD:AC² = AD² + CD²BD² = BC² + CD²But as ABCD is a rectangle AD = BC since they are opposite and parallel; thus:AC² = AD² + CD² = BC² + CD² = BD²Thus, as AC and BD are the diagonals, they are equal.Therefore the diagonals of a rectangle are congruent.
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