How to find a length of line in quadrilateral?

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  • Thread starter asmmanikanda
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In summary, to find the length of xy, we can divide the quadrilateral into triangles and apply the cosine rule repeatedly. If the quadrilateral is a trapezium, we can use an additional constraint to simplify the calculation.
  • #1
asmmanikanda
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IMG_20200808_221717.jpg

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?
 
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  • #2
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?
 
Last edited:
  • #3
Klaas van Aarsen said:
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.
 
  • #4
skeeter said:
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.
 
  • #5
I made the same mistake (thinking trapezoid) ... sometimes one has to read the "fine print"
 
  • #6
Klaas van Aarsen said:
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?
 
  • #7
Klaas van Aarsen said:
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?
 
Last edited by a moderator:
  • #8
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. :)
 
  • #9
anemone said:
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?
 

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  • #10
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 $XY\approx 66.4$.
 
Last edited:
  • #11
Klaas van Aarsen said:
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
 

FAQ: How to find a length of line in quadrilateral?

1. How do you find the length of a line in a quadrilateral?

To find the length of a line in a quadrilateral, you can use the Pythagorean theorem or the distance formula. Both of these methods involve using the coordinates of the quadrilateral's vertices to calculate the length of the line.

2. Can you find the length of a line in a quadrilateral without knowing the coordinates of its vertices?

No, in order to find the length of a line in a quadrilateral, you need to know the coordinates of at least two of its vertices. Without this information, it is not possible to accurately calculate the length of the line.

3. What is the Pythagorean theorem and how is it used to find the length of a line in a quadrilateral?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In a quadrilateral, you can use this theorem to find the length of a line by treating it as the hypotenuse of a right triangle and using the coordinates of the vertices to calculate the lengths of the other two sides.

4. How do you use the distance formula to find the length of a line in a quadrilateral?

The distance formula is a mathematical formula used to calculate the distance between two points in a coordinate plane. To find the length of a line in a quadrilateral using this formula, you would first identify the coordinates of the two endpoints of the line. Then, you would plug these coordinates into the distance formula to calculate the length of the line.

5. Can the length of a line in a quadrilateral be negative?

No, the length of a line in a quadrilateral cannot be negative. Length is a measure of distance, which is always a positive value. If you use the distance formula or the Pythagorean theorem to calculate the length of a line in a quadrilateral, the result will always be a positive value.

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