MHB How to find a length of line in quadrilateral?

[asmmanikanda]

AB and DC not parallel line., AB=55,BC=65,DC=76,DA=48
Ax=28,xD=20,By=37,yC=28

How to find length of xy?

 

[I like Serena]

Hi asmmanikanda, welcome to MHB!

If I draw your trapezium to scale, I get the following diagram, which looks rather different from yours.\begin{tikzpicture}[scale=1.5,font=\Large]
\path[orange] (137:4.8) coordinate[label=A] (A) -- ++(5.5,0) coordinate[label=B] (B) -- ++(-30:6.5) coordinate[label=below:C] (C) -- (0,0) coordinate[label=below: D] (D);
\path[red] (137:2.0) coordinate[label=left:x] (X) (B) ++(-30:3.7) coordinate[label=right:y] (Y);
\draw[black, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[orange, ultra thick] (X) -- (Y);
\path[green!70!black] (A) -- node[above] {55} (B) -- node[above right] {65} (C) -- node[below] {76} (D) -- node[below left] {48} (A);
\path[purple] (D) -- node[above right] {20} (X) -- node[above right] {28} (A) (C) -- node[below left] {28} (Y) -- node[below left] {37} (B);
\end{tikzpicture}

Can you confirm that this is the intended diagram?
Is there perhaps a typo?

 

[skeeter]

If I draw your trapezium to scale, I get the following diagram, which looks rather different from yours.

Not a trapezium ... OP stated that AB is not parallel to DC.

 

[I like Serena]

Not a trapezium ... OP stated that AB is not parallel to DC.

Ah okay... then we have too much freedom I think... and line x-y won't have a unique length.

 

[skeeter]

I made the same mistake (thinking trapezoid) ... sometimes one has to read the "fine print"

 

[asmmanikanda]

Ah okay... then we have too much freedom I think... and line x-y won't have a unique length.

OK sir. If trapezium mean how to find the length of xy?

 

[asmmanikanda]

Hi asmmanikanda, welcome to MHB!

If I draw your trapezium to scale, I get the following diagram, which looks rather different from yours.

\begin{tikzpicture}[scale=1.5,font=\Large]
\path[orange] (137:4.8) coordinate[label=A] (A) -- ++(5.5,0) coordinate[label=B] (B) -- ++(-30:6.5) coordinate[label=below:C] (C) -- (0,0) coordinate[label=below: D] (D);
\path[red] (137:2.0) coordinate[label=left:x] (X) (B) ++(-30:3.7) coordinate[label=right:y] (Y);
\draw[black, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[orange, ultra thick] (X) -- (Y);
\path[green!70!black] (A) -- node[above] {55} (B) -- node[above right] {65} (C) -- node[below] {76} (D) -- node[below left] {48} (A);
\path[purple] (D) -- node[above right] {20} (X) -- node[above right] {28} (A) (C) -- node[below left] {28} (Y) -- node[below left] {37} (B);
\end{tikzpicture}

Can you confirm that this is the intended diagram?
Is there perhaps a typo?

Yes sir. I posted rough diagram only sir. Your diagram may be your correct. My questions 1. How to find a length of xy if not trapezium?
2.how to find a length of xy if trapezium?

 

[anemone]

Can you post the original question here? If it has a diagram in the question, you can take a picture of it and upload it here, so we can get a better idea of what is being asked. :)

 

[asmmanikanda]

Can you post the original question here? If it has a diagram in the question, you can take a picture of it and upload it here, so we can get a better idea of what is being asked. :)

Image is in not to scale. 1. How to find length of xy if AB And CD is parallel ?
2.how to find length of xy if AB and CD is not parallel?

 

[I like Serena]

We can divide the quadrilateral into triangles like this.

\begin{tikzpicture}[scale=0.15,font=\Large,
declare function={
sideFromCos(\cosalpha,\b,\c) = {sqrt((\b)^2 + (\c)^2 - 2*(\b)*(\c)*\cosalpha)};
cosFromSides(\a,\b,\c) = {((\a)^2 + (\b)^2 - (\c)^2) / (2 * (\a) * (\b))};
}
]
\usetikzlibrary{angles,quotes}

\def\angleDelta{80}
\def\AB{55}
\def\BC{65}
\def\CD{76}
\def\AD{48}
\def\AX{28}
\def\DX{20}
\def\BY{37}
\def\CY{28}
\def\AC{sideFromCos(cos(\angleDelta), \AD, \CD)}
\def\CX{sideFromCos(cos(\angleDelta), \DX, \CD)}
\def\angleBeta{acos(cosFromSides(\AB, \BC, \AC))}
\def\angleZeta{acos(cosFromSides(\CD/10, \AC/10, \AD/10))}
\def\angleEta{acos(cosFromSides(\AC/10, \BC/10, \AB/10))}
\def\angleGamma{(\angleZeta+ \angleEta)}
\def\angleAlpha{360-\angleBeta-\angleGamma-\angleDelta}
\def\angleEpsilon{acos(cosFromSides(\CD/10, \CX/10, \DX/10))}

\path[orange] (\angleDelta:\AD) coordinate[label=A] (A) -- ++({\angleAlpha+\angleDelta-180}:\AB) coordinate[label=B] (B) -- (\CD,0) coordinate[label=below:C] (C) -- (0,0) coordinate[label=below: D] (D);
\path[red] (\angleDelta:\DX) coordinate[label=left:X] (X) (C) ++({180-\angleGamma}:\CY) coordinate[label=right:Y] (Y);

\draw[help lines] (C) -- +({180-\angleZeta}:{\AC});
\draw[help lines] (C) -- +({180-\angleEpsilon}:{\CX});

\draw[black, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[orange, ultra thick] (X) -- (Y);
\path[green!70!black] (A) -- node[above] {\AB} (B) -- node[above right] {\BC} (C) -- node[below] {\CD} (D) -- node[below left] {\AD} (A);
\path[purple] (D) -- node[above right] {\DX} (X) -- node[above right] {\AX} (A) (C) -- node[ left ] {\CY} (Y) -- node[ left ] {\BY} (B);

\pic [draw, "$\delta$", angle radius=1cm, angle eccentricity=0.6] {angle = C--D--A};
\pic [draw, "$\beta$", angle radius=1cm, angle eccentricity=0.6] {angle = A--B--C};
\pic [draw, "$\gamma$", angle radius=1cm, angle eccentricity=0.7] {angle = B--C--D};
\pic [draw, "$\zeta$", angle radius=1.5cm, angle eccentricity=1.15] {angle = A--C--D};
\pic [draw, "$\eta$", angle radius=1.3cm, angle eccentricity=1.15] {angle = B--C--A};
\pic [draw, "$\epsilon$", angle radius=2.2cm, angle eccentricity=1.15] {angle = X--C--D};

\end{tikzpicture}

Now we can repeatedly apply the cosine rule to find the various sides and angles.
The general cosine rule is:
$$c^2=a^2+b^2-2ab\cos\alpha \tag 1$$
And we can invert it to find the angle:
$$\alpha=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab} \right) \tag 2$$

The length XY depends on the angle $\delta$, and its formula form is rather long, so I'm choosing not to write it out at this time.

If we have the constraint that we have a trapezium, then we have $\beta + \gamma = 180^\circ$ due to so called Z-angles.

\begin{tikzpicture}[scale=0.15,font=\Large,
declare function={
sideFromCos(\cosalpha,\b,\c) = {sqrt((\b)^2 + (\c)^2 - 2*(\b)*(\c)*\cosalpha)};
cosFromSides(\a,\b,\c) = {((\a)^2 + (\b)^2 - (\c)^2) / (2 * (\a) * (\b))};
}
]
\usetikzlibrary{angles,quotes}

\def\angleDelta{137}
\def\AB{55}
\def\BC{65}
\def\CD{76}
\def\AD{48}
\def\AX{28}
\def\DX{20}
\def\BY{37}
\def\CY{28}
\def\AC{sideFromCos(cos(\angleDelta), \AD, \CD)}
\def\CX{sideFromCos(cos(\angleDelta), \DX, \CD)}
\def\angleBeta{acos(cosFromSides(\AB, \BC, \AC))}
\def\angleZeta{acos(cosFromSides(\CD/10, \AC/10, \AD/10))}
\def\angleEta{acos(cosFromSides(\AC/10, \BC/10, \AB/10))}
\def\angleGamma{(\angleZeta+ \angleEta)}
\def\angleAlpha{360-\angleBeta-\angleGamma-\angleDelta}
\def\angleEpsilon{acos(cosFromSides(\CD/10, \CX/10, \DX/10))}

\path[orange] (\angleDelta:\AD) coordinate[label=A] (A) -- ++({\angleAlpha+\angleDelta-180}:\AB) coordinate[label=B] (B) -- (\CD,0) coordinate[label=below:C] (C) -- (0,0) coordinate[label=below: D] (D);
\path[red] (\angleDelta:\DX) coordinate[label=left:X] (X) (C) ++({180-\angleGamma}:\CY) coordinate[label=right:Y] (Y);

\draw[help lines] (C) -- +({180-\angleZeta}:{\AC});
\draw[help lines] (C) -- +({180-\angleEpsilon}:{\CX});

\draw[black, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[orange, ultra thick] (X) -- (Y);
\path[green!70!black] (A) -- node[above] {\AB} (B) -- node[above right] {\BC} (C) -- node[below] {\CD} (D) -- node[below left] {\AD} (A);
\path[purple] (D) -- node[above right] {\DX} (X) -- node[above right] {\AX} (A) (C) -- node[ left ] {\CY} (Y) -- node[ left ] {\BY} (B);

\pic [draw, "$\delta$", angle radius=1cm, angle eccentricity=0.6] {angle = C--D--A};
\pic [draw, "$\beta$", angle radius=1cm, angle eccentricity=0.6] {angle = A--B--C};
\pic [draw, "$\gamma$", angle radius=1cm, angle eccentricity=0.7] {angle = B--C--D};
\pic [draw, "$\zeta$", angle radius=1.5cm, angle eccentricity=1.15] {angle = A--C--D};
\pic [draw, "$\eta$", angle radius=1.3cm, angle eccentricity=1.15] {angle = B--C--A};
\pic [draw, "$\epsilon$", angle radius=2.2cm, angle eccentricity=1.15] {angle = X--C--D};

\end{tikzpicture}
Working through the cosine rules gives us then that $\delta \approx 137^\circ$, $\gamma \approx 30^\circ$, and $XY\approx 66.4$.

 

[asmmanikanda]

We can divide the quadrilateral into triangles like this.

\begin{tikzpicture}[scale=0.15,font=\Large,
declare function={
sideFromCos(\cosalpha,\b,\c) = {sqrt((\b)^2 + (\c)^2 - 2*(\b)*(\c)*\cosalpha)};
cosFromSides(\a,\b,\c) = {((\a)^2 + (\b)^2 - (\c)^2) / (2 * (\a) * (\b))};
}
]
\usetikzlibrary{angles,quotes}

\def\angleDelta{80}
\def\AB{55}
\def\BC{65}
\def\CD{76}
\def\AD{48}
\def\AX{28}
\def\DX{20}
\def\BY{37}
\def\CY{28}
\def\AC{sideFromCos(cos(\angleDelta), \AD, \CD)}
\def\CX{sideFromCos(cos(\angleDelta), \DX, \CD)}
\def\angleBeta{acos(cosFromSides(\AB, \BC, \AC))}
\def\angleZeta{acos(cosFromSides(\CD/10, \AC/10, \AD/10))}
\def\angleEta{acos(cosFromSides(\AC/10, \BC/10, \AB/10))}
\def\angleGamma{(\angleZeta+ \angleEta)}
\def\angleAlpha{360-\angleBeta-\angleGamma-\angleDelta}
\def\angleEpsilon{acos(cosFromSides(\CD/10, \CX/10, \DX/10))}

\path[orange] (\angleDelta:\AD) coordinate[label=A] (A) -- ++({\angleAlpha+\angleDelta-180}:\AB) coordinate[label=B] (B) -- (\CD,0) coordinate[label=below:C] (C) -- (0,0) coordinate[label=below: D] (D);
\path[red] (\angleDelta:\DX) coordinate[label=left:X] (X) (C) ++({180-\angleGamma}:\CY) coordinate[label=right:Y] (Y);

\draw[help lines] (C) -- +({180-\angleZeta}:{\AC});
\draw[help lines] (C) -- +({180-\angleEpsilon}:{\CX});

\draw[black, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[orange, ultra thick] (X) -- (Y);
\path[green!70!black] (A) -- node[above] {\AB} (B) -- node[above right] {\BC} (C) -- node[below] {\CD} (D) -- node[below left] {\AD} (A);
\path[purple] (D) -- node[above right] {\DX} (X) -- node[above right] {\AX} (A) (C) -- node[ left ] {\CY} (Y) -- node[ left ] {\BY} (B);

\pic [draw, "$\delta$", angle radius=1cm, angle eccentricity=0.6] {angle = C--D--A};
\pic [draw, "$\beta$", angle radius=1cm, angle eccentricity=0.6] {angle = A--B--C};
\pic [draw, "$\gamma$", angle radius=1cm, angle eccentricity=0.7] {angle = B--C--D};
\pic [draw, "$\zeta$", angle radius=1.5cm, angle eccentricity=1.15] {angle = A--C--D};
\pic [draw, "$\eta$", angle radius=1.3cm, angle eccentricity=1.15] {angle = B--C--A};
\pic [draw, "$\epsilon$", angle radius=2.2cm, angle eccentricity=1.15] {angle = X--C--D};

\end{tikzpicture}

Now we can repeatedly apply the cosine rule to find the various sides and angles.
The general cosine rule is:
$$c^2=a^2+b^2-2ab\cos\alpha \tag 1$$
And we can invert it to find the angle:
$$\alpha=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab} \right) \tag 2$$

The length XY depends on the angle $\delta$, and its formula form is rather long, so I'm choosing not to write it out at this time.

If we have the constraint that we have a trapezium, then we have $\beta + \gamma = 180^\circ$ due to so called Z-angles.

\begin{tikzpicture}[scale=0.15,font=\Large,
declare function={
sideFromCos(\cosalpha,\b,\c) = {sqrt((\b)^2 + (\c)^2 - 2*(\b)*(\c)*\cosalpha)};
cosFromSides(\a,\b,\c) = {((\a)^2 + (\b)^2 - (\c)^2) / (2 * (\a) * (\b))};
}
]
\usetikzlibrary{angles,quotes}

\def\angleDelta{137}
\def\AB{55}
\def\BC{65}
\def\CD{76}
\def\AD{48}
\def\AX{28}
\def\DX{20}
\def\BY{37}
\def\CY{28}
\def\AC{sideFromCos(cos(\angleDelta), \AD, \CD)}
\def\CX{sideFromCos(cos(\angleDelta), \DX, \CD)}
\def\angleBeta{acos(cosFromSides(\AB, \BC, \AC))}
\def\angleZeta{acos(cosFromSides(\CD/10, \AC/10, \AD/10))}
\def\angleEta{acos(cosFromSides(\AC/10, \BC/10, \AB/10))}
\def\angleGamma{(\angleZeta+ \angleEta)}
\def\angleAlpha{360-\angleBeta-\angleGamma-\angleDelta}
\def\angleEpsilon{acos(cosFromSides(\CD/10, \CX/10, \DX/10))}

\path[orange] (\angleDelta:\AD) coordinate[label=A] (A) -- ++({\angleAlpha+\angleDelta-180}:\AB) coordinate[label=B] (B) -- (\CD,0) coordinate[label=below:C] (C) -- (0,0) coordinate[label=below: D] (D);
\path[red] (\angleDelta:\DX) coordinate[label=left:X] (X) (C) ++({180-\angleGamma}:\CY) coordinate[label=right:Y] (Y);

\draw[help lines] (C) -- +({180-\angleZeta}:{\AC});
\draw[help lines] (C) -- +({180-\angleEpsilon}:{\CX});

\draw[black, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[orange, ultra thick] (X) -- (Y);
\path[green!70!black] (A) -- node[above] {\AB} (B) -- node[above right] {\BC} (C) -- node[below] {\CD} (D) -- node[below left] {\AD} (A);
\path[purple] (D) -- node[above right] {\DX} (X) -- node[above right] {\AX} (A) (C) -- node[ left ] {\CY} (Y) -- node[ left ] {\BY} (B);

\pic [draw, "$\delta$", angle radius=1cm, angle eccentricity=0.6] {angle = C--D--A};
\pic [draw, "$\beta$", angle radius=1cm, angle eccentricity=0.6] {angle = A--B--C};
\pic [draw, "$\gamma$", angle radius=1cm, angle eccentricity=0.7] {angle = B--C--D};
\pic [draw, "$\zeta$", angle radius=1.5cm, angle eccentricity=1.15] {angle = A--C--D};
\pic [draw, "$\eta$", angle radius=1.3cm, angle eccentricity=1.15] {angle = B--C--A};
\pic [draw, "$\epsilon$", angle radius=2.2cm, angle eccentricity=1.15] {angle = X--C--D};

\end{tikzpicture}
Working through the cosine rules gives us then that $\delta \approx 137^\circ$, $\gamma \approx 30^\circ$, and $XY\approx 66.4$.

Can you explain step by step till answer pls