# Find the area of the quadrilateral whose vertices taken in order, are \[\left( { - 4, - 2} \right),\left( { - 3, - 5} \right),\left( { - 3, - 2} \right){\text{ and }}\left( {2,3} \right)\].

Last updated date: 14th Mar 2023

•

Total views: 303k

•

Views today: 4.84k

Answer

Verified

303k+ views

Hint- To solve this question we will use the formula given by $\Delta = \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$ , we will divide the quadrilateral into two triangles and then apply the formula.

Complete step-by-step solution -

Given that vertices of quadrilateral are \[\left( { - 4, - 2} \right),\left( { - 3, - 5} \right),\left( { - 3, - 2} \right){\text{ and }}\left( {2,3} \right)\]

Let \[A\left( { - 4, - 2} \right),B\left( { - 3, - 5} \right),C\left( { - 3, - 2} \right){\text{ and D}}\left( {2,3} \right)\] be the vertices of the quadrilateral.

As we know that the area of the quadrilateral ABCD= area of $\Delta ABC$ + area of $\Delta ACD$

Area of $\Delta ABC$ is given by

$ = \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$

In $\Delta ABC$

${x_1} = - 4,{x_2} = - 3,{x_3} = 3,{y_1} = - 2,{y_2} = - 5,{y_3} = - 2$

Substituting above values in the formula, we get

$

= \dfrac{1}{2}\left| { - 2( - 3 - 3) - 5(3 + 4) - 2( - 4 + 3)} \right| \\

= \dfrac{1}{2}\left| { - 2( - 6) - 5(7) - 2( - 1)} \right| \\

= \dfrac{1}{2}\left| {12 - 35 + 2} \right| \\

= \dfrac{1}{2}\left| { - 21} \right| \\

= 10.5{\text{ sq}}{\text{.units}} \\

$

Area of $\Delta ACD$ is given by

$ = \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$

In $\Delta ACD$

${x_1} = - 4,{x_2} = 3,{x_3} = 2,{y_1} = - 2,{y_2} = - 2,{y_3} = 3$

Substituting above values in the formula, we get

$

= \dfrac{1}{2}\left| { - 2( - 3 - 2) - 2(2 + 4) - 3( - 4 - 3)} \right| \\

= \dfrac{1}{2}\left| { - 2 - 12 - 21} \right| \\

= \dfrac{1}{2}\left| { - 35} \right| \\

= 17.5{\text{ sq}}{\text{.units}} \\

$

Therefore, the area of the quadrilateral is = $10.5 + 17.5 = 28{\text{ sq}}{\text{. units}}$

Hence, the area of the quadrilateral is $28.0{\text{ sq}}{\text{. units}}$

Note- The given problem is the coordinate geometry problem and the vertices of the quadrilateral are given. In order to solve such questions, try to break the quadrilateral in two triangles and apply the formula to calculate the area of the triangle.

Complete step-by-step solution -

Given that vertices of quadrilateral are \[\left( { - 4, - 2} \right),\left( { - 3, - 5} \right),\left( { - 3, - 2} \right){\text{ and }}\left( {2,3} \right)\]

Let \[A\left( { - 4, - 2} \right),B\left( { - 3, - 5} \right),C\left( { - 3, - 2} \right){\text{ and D}}\left( {2,3} \right)\] be the vertices of the quadrilateral.

As we know that the area of the quadrilateral ABCD= area of $\Delta ABC$ + area of $\Delta ACD$

Area of $\Delta ABC$ is given by

$ = \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$

In $\Delta ABC$

${x_1} = - 4,{x_2} = - 3,{x_3} = 3,{y_1} = - 2,{y_2} = - 5,{y_3} = - 2$

Substituting above values in the formula, we get

$

= \dfrac{1}{2}\left| { - 2( - 3 - 3) - 5(3 + 4) - 2( - 4 + 3)} \right| \\

= \dfrac{1}{2}\left| { - 2( - 6) - 5(7) - 2( - 1)} \right| \\

= \dfrac{1}{2}\left| {12 - 35 + 2} \right| \\

= \dfrac{1}{2}\left| { - 21} \right| \\

= 10.5{\text{ sq}}{\text{.units}} \\

$

Area of $\Delta ACD$ is given by

$ = \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$

In $\Delta ACD$

${x_1} = - 4,{x_2} = 3,{x_3} = 2,{y_1} = - 2,{y_2} = - 2,{y_3} = 3$

Substituting above values in the formula, we get

$

= \dfrac{1}{2}\left| { - 2( - 3 - 2) - 2(2 + 4) - 3( - 4 - 3)} \right| \\

= \dfrac{1}{2}\left| { - 2 - 12 - 21} \right| \\

= \dfrac{1}{2}\left| { - 35} \right| \\

= 17.5{\text{ sq}}{\text{.units}} \\

$

Therefore, the area of the quadrilateral is = $10.5 + 17.5 = 28{\text{ sq}}{\text{. units}}$

Hence, the area of the quadrilateral is $28.0{\text{ sq}}{\text{. units}}$

Note- The given problem is the coordinate geometry problem and the vertices of the quadrilateral are given. In order to solve such questions, try to break the quadrilateral in two triangles and apply the formula to calculate the area of the triangle.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

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

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India