Question

if p(-5,-3),q(-4,-6),r(2,-3),s(1,2) are the vertices of a quadrilateral pqrs . find its area

Answer 1

By triangle formula you can find the area of triangle pqs and qsr and add the are of both triangle resulting value is the are of quadrilateral

Answer 2

Answer:

• Area of quadrilateral PQRS = 28 sq. units.

Step-by-step explanation:

Given:

• PQRS is a quadrilateral.
• Vertices of P = (-5, -3)
• Vertices of Q = (-4, -6)
• Vertices of R = (2, -3)
• Vertices of S = (1, 2)

To Find:

• Area of quadrilateral.

Construction:

• Join PR and then two triangles form named as ΔPQR and ΔRSP.

Now, Area of quadrilateral = Area of ΔPQR + Area of ΔRSP.

Now, first we will find the area of ΔPQR.

$\longrightarrow \sf Area\;\triangle PQR = \Bigg|\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle PQR = \Bigg|\dfrac{1}{2}[-5(-6+3)-4(-3+3)+2(-3+6)]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle PQR = \Bigg|\dfrac{1}{2}[-5(-3)-4(0)+2(3)]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle PQR = \Bigg|\dfrac{1}{2}[15-0+6]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle PQR = \Bigg|\dfrac{21}{2}\Bigg|$

$\longrightarrow \sf Area\;of\;\triangle PQR = 10.5\;sq.\;units$

$\longrightarrow \sf Area\;\triangle RSP = \Bigg|\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle RSP = \Bigg|\dfrac{1}{2}[-5(2+3)+1(-3+3)+2(-3-2)]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle RSP = \Bigg|\dfrac{1}{2}[-5(5)+1(0)+2(-5)]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle RSP = \Bigg|\dfrac{1}{2}[-25-0-10]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;\triangle RSP = \Bigg|\dfrac{-35}{2}\Bigg|$

$\longrightarrow \sf Area\;of\;\triangle RSP = 17.5\;sq.\;units$

Now, Area of quadrilateral = Area of ΔPQR + Area of ΔRSP.

⇒ Area of quadrilateral = 10.5 + 17.5

⇒ Area of quadrilateral = 28 sq. units

Hence, area of quadrilateral PQRS = 28 sq. units.

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