Answer

Verified

383.7k+ views

**Hint:**First of all, draw a quadrilateral ABCD. Then we will draw a diagonal in the quadrilateral. After drawing the diagonal, the quadrilateral is divided into two triangles. Then we will find the area of each of the triangles by using the formula of the area of triangle having three points i.e. $\dfrac{1}{2}\left| \begin{matrix}

{{x}_{1}} & {{y}_{1}} & 1 \\

{{x}_{2}} & {{y}_{2}} & 1 \\

{{x}_{3}} & {{y}_{3}} & 1 \\

\end{matrix} \right|$ and then we will add the area of two triangles to get the area of the quadrilateral ABCD.

**Complete step-by-step solution:**

Let us draw a quadrilateral ABCD with coordinates of A, B, C and D as follows:

Now, joining A and C to get the diagonal AC and joining B and D to get the diagonal BD in the above diagram we get,

Now, we are going to find the area of the triangles ACD and ABC.

We know that formula for area of triangle having three points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is as follows:

$\dfrac{1}{2}\left| \begin{matrix}

{{x}_{1}} & {{y}_{1}} & 1 \\

{{x}_{2}} & {{y}_{2}} & 1 \\

{{x}_{3}} & {{y}_{3}} & 1 \\

\end{matrix} \right|$

Now, we are going to find the area of triangle ACD by taking A (-5, 7), C (-1, -6), D (4, 5) as $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ in the above formula and we get,

$\Rightarrow \dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-1 & -6 & 1 \\

4 & 5 & 1 \\

\end{matrix} \right|$

Applying the following operations in the above determinant we get,

$\begin{align}

& {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\

& {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\

\end{align}$

$\begin{align}

& \Rightarrow \dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-1-\left( -5 \right) & -6-7 & 1-1 \\

4-\left( -5 \right) & 5-7 & 1-1 \\

\end{matrix} \right| \\

& =\dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-1+5 & -13 & 0 \\

4+5 & -2 & 0 \\

\end{matrix} \right| \\

& =\dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

4 & -13 & 0 \\

9 & -2 & 0 \\

\end{matrix} \right| \\

\end{align}$

Expanding the above determinant along third column we get,

$\begin{align}

& =\dfrac{1}{2}\left[ 4\left( -2 \right)-9\left( -13 \right) \right] \\

& =\dfrac{1}{2}\left[ -8+117 \right] \\

& =\dfrac{1}{2}\left[ 109 \right] \\

\end{align}$

From the above, we have calculated the area of triangle ACD as $\dfrac{109}{2}$sq. units.

Now, we are going to find the area of triangle ABC by taking A (-5, 7), B (-4, -5), C (-1, -6), as $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ in the above formula and we get,

$\Rightarrow \dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-4 & -5 & 1 \\

-1 & -6 & 1 \\

\end{matrix} \right|$

Applying the following operations in the above determinant we get,

$\begin{align}

& {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\

& {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\

\end{align}$

$\begin{align}

& \Rightarrow \dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-4-\left( -5 \right) & -5-7 & 1-1 \\

-1-\left( -5 \right) & -6-7 & 1-1 \\

\end{matrix} \right| \\

& =\dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-4+5 & -12 & 0 \\

-1+5 & -13 & 0 \\

\end{matrix} \right| \\

& =\dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

1 & -12 & 0 \\

4 & -13 & 0 \\

\end{matrix} \right| \\

\end{align}$

Expanding the above determinant along third column we get,

$\begin{align}

& =\dfrac{1}{2}\left[ 1\left( -13 \right)-4\left( -12 \right) \right] \\

& =\dfrac{1}{2}\left[ -13+48 \right] \\

& =\dfrac{1}{2}\left[ 35 \right] \\

\end{align}$

From the above, we have calculated the area of triangle ABC as $\dfrac{35}{2}$sq. units.

Now, adding the areas of triangles ABC and ACD we get,

$\begin{align}

& \Rightarrow \Delta ABC+\Delta ACD \\

& =\dfrac{35}{2}+\dfrac{109}{2} \\

& =\dfrac{144}{2} \\

& =72 \\

\end{align}$

**Hence, we have calculated the area of quadrilateral ABCD is 72 sq. units.**

**Note:**The possible mistake that could be possible in the above problem is the calculation mistake as you can see that rigorous calculation is involved in finding the area of the two triangles which is made by drawing the diagonals. Also, in evaluating the determinant in the above problem we have used the row operations to minimize the calculations.

$\Rightarrow \dfrac{1}{2}\left| \begin{matrix}

-5 & 7 & 1 \\

-4 & -5 & 1 \\

-1 & -6 & 1 \\

\end{matrix} \right|$

Applying the following operations in the above determinant we get,

$\begin{align}

& {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\

& {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\

\end{align}$

The other way to evaluate the above determinant is not to apply the row operations and do the expansion of the determinant along the first row but then chances of making calculations will be higher than in the ones which we have done above.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Establish a relation between electric current and drift class 12 physics CBSE

Guru Purnima speech in English in 100 words class 7 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Copper is not used as potentiometer wire because class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE