In each diagram, BD bisects

scherezade29pc
2022-07-28
Answered

In each diagram, BD bisects

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nuramaaji2000fh

Answered 2022-07-29
Author has **18** answers

a)

since the angle is bisected, both smaller angles are equal

3x+17 = 7x-39

56=4x

x=14

$m<abd=3x+17=3(14)+17={59}^{\circ}$

$m<ABC=2\ast 59={118}^{\circ}$

b)

5x+16 = 8x-23

39=3x

x=13

$m<abd=5x+16=5(13)+16={81}^{\circ}$

$m<ABC=81\ast 2={162}^{\circ}$

c)

17x-10 = 11x+5

6x=15

x=2.5

$m<abd=17x-10=17(2.5)-10={42.5}^{\circ}$

$m<ABC=2\ast 42.5={85}^{\circ}$

since the angle is bisected, both smaller angles are equal

3x+17 = 7x-39

56=4x

x=14

$m<abd=3x+17=3(14)+17={59}^{\circ}$

$m<ABC=2\ast 59={118}^{\circ}$

b)

5x+16 = 8x-23

39=3x

x=13

$m<abd=5x+16=5(13)+16={81}^{\circ}$

$m<ABC=81\ast 2={162}^{\circ}$

c)

17x-10 = 11x+5

6x=15

x=2.5

$m<abd=17x-10=17(2.5)-10={42.5}^{\circ}$

$m<ABC=2\ast 42.5={85}^{\circ}$

asked 2022-06-17

I've looked all over and I can't find a good proof of why the diagonals of a rhombus should intersect at right angles. I can intuitively see its true, just by drawing rhombuses, but I'm trying to prove that the slopes of the diagonals are negative reciprocals and its not working out.

I'm defining my rhombus as follows: $[(0,0),(a,0),(b,c),(a+b,c)]$

I've managed to figure out that $c=\sqrt{{a}^{2}-{b}^{2}}$ and that the slopes of the diagonals are $\frac{\sqrt{{a}^{2}-{b}^{2}}}{a+b}$ and $\frac{-\sqrt{{a}^{2}-{b}^{2}}}{a-b}$

What I can't figure out is how they can be negative reciprocals of one another.

I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form.

I'm defining my rhombus as follows: $[(0,0),(a,0),(b,c),(a+b,c)]$

I've managed to figure out that $c=\sqrt{{a}^{2}-{b}^{2}}$ and that the slopes of the diagonals are $\frac{\sqrt{{a}^{2}-{b}^{2}}}{a+b}$ and $\frac{-\sqrt{{a}^{2}-{b}^{2}}}{a-b}$

What I can't figure out is how they can be negative reciprocals of one another.

I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form.

asked 2022-07-02

The property I'm talking about is:

There is some partition of the plane figure P into $n$ congruent figures for any $n$.

Is it true that only discs, sectors of discs, annuli, sectors of annuli and parallelograms have this property?

There is some partition of the plane figure P into $n$ congruent figures for any $n$.

Is it true that only discs, sectors of discs, annuli, sectors of annuli and parallelograms have this property?

asked 2022-07-27

Consider an incidence geometry in which every line has at least three distinct points (rather than at least two as requiblack by the second axiom of Incidence Geometry). What are the least number of points and lines that must exist? Note: You can show the existence of such a "minimum" geometry by constructing a model for it. On the other hand, you have to prove that there cannot be a model with less points or lines

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How does ${d}_{1}$ equal $\frac{1}{2}\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$ ?

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The formula for local midpoint rule in the interval $[-\frac{h}{2},\text{}\frac{h}{2}]$

asked 2022-06-27

The midpoints of the sides of $\mathrm{\u25b3}ABC$ along with any of the vertices as the fourth point make a parallelogram of area equal to what? The answer is, obviously, $\frac{1}{2}\mathrm{area}(\mathrm{\u25b3}ABC)$

In the method, I took $\mathrm{\u25b3}ABC$ with $D$, $E$, and $F$ as midpoints of $AB$, $AC$, and $BC$, respectively; and I joined $DE$ and $EF$ so that I get a parallelogram $\u25fbDEFB$.

I know what the answer is because one can easily make that out. Also, those four triangles (four because the parallelogram can still be divided into two triangles and the rest two triangles add up to four) so it's simple that the area of $\u25fbDEBF$ will be 1/4 of $\mathrm{\u25b3}ABC$, but how?

Can anyone explain me this?

In the method, I took $\mathrm{\u25b3}ABC$ with $D$, $E$, and $F$ as midpoints of $AB$, $AC$, and $BC$, respectively; and I joined $DE$ and $EF$ so that I get a parallelogram $\u25fbDEFB$.

I know what the answer is because one can easily make that out. Also, those four triangles (four because the parallelogram can still be divided into two triangles and the rest two triangles add up to four) so it's simple that the area of $\u25fbDEBF$ will be 1/4 of $\mathrm{\u25b3}ABC$, but how?

Can anyone explain me this?

asked 2022-07-23

Let x and y be two vectors in ${\mathbb{R}}^{2}$. The parallelogram law of vector addition says that the vector corresponding to the diagonal of the parallelogram formed by these vectors is x+y. For the same parallelogram consider the other diagonal directed from x to y. Call this diagonal $\overrightarrow{D}$ .

The vector y−x can be identified with $\overrightarrow{D}$ since we see geometrically that it has the same length and the same slope. This is also intuitively obvious because a rigid motion (translation) can superimpose y−x on $\overrightarrow{D}$ .

However we may identify $\overrightarrow{D}$ with x−y as well since they too have the same slope and length. Moreover a rotation and a translation can superimpose x−y on $\overrightarrow{D}$ as well.

Yet it is geometrically obvious that x−y is the opposite direction of $\overrightarrow{D}$ .

Can someone explain why we should identify $\overrightarrow{D}$ with y−x only? Is it correct to say that we should identify a vector with a directed line segment if the two are related by a translation only? If so, what is the justification behind this and is there a name for the geometry arising from studying properties invariant under translation only?

The vector y−x can be identified with $\overrightarrow{D}$ since we see geometrically that it has the same length and the same slope. This is also intuitively obvious because a rigid motion (translation) can superimpose y−x on $\overrightarrow{D}$ .

However we may identify $\overrightarrow{D}$ with x−y as well since they too have the same slope and length. Moreover a rotation and a translation can superimpose x−y on $\overrightarrow{D}$ as well.

Yet it is geometrically obvious that x−y is the opposite direction of $\overrightarrow{D}$ .

Can someone explain why we should identify $\overrightarrow{D}$ with y−x only? Is it correct to say that we should identify a vector with a directed line segment if the two are related by a translation only? If so, what is the justification behind this and is there a name for the geometry arising from studying properties invariant under translation only?