Find the area of the quadrilateral whose vertices, taken in order
are (-4, - 2), (-3, - 5), (3, - 2) and (2, 3)
Answers
Let A( -4, -2), B( -3, -5), C( 3, -2), D( 2, 3). Divide the quadrilateral in two parts by joining the diagonals A and D. { Two Δ's are formed ABD and BCD}
Area of Quadrilateral ABCD= Area Δ ABD +Area Δ BCD … (1)
Using formula to find area of triangle:
Area of ABD= [−4 (−5 − 3) – 3 {3 − (−2)} + 2 {−2 − (−5)}]
= (32 – 15 + 6)
= (23) = 11.5 sq units … (2)
Again using formula to find area of triangle:
Area of △BCD = [−3 (−2 − 3) + 3 {3 − (−5)} + 2 {−5 − (−2)}]
= (15 + 24 − 6)
= (33) = 16.5 sq units … (3)
Putting (2) and (3) in (1), we get
Area of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units
Answer:
28 Sq.units.
Step by step solution:
Draw the quadrilateral, mark the co-ordinates and join AC.
Now, two triangles ADC and ABC are formed, we find the area for each, and add them up.
Area of ΔADC.
ar(ADC) = ¹/₂ [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
ar(ADC) = ¹/₂ [-4(3 + 2) + 2(-2 + 2) + 3(-2 - 3)]
ar(ADC) = ¹/₂ [-4(5) + 2(0) + 3(-5)]
ar(ADC) = ¹/₂ [-20 - 15]
ar(ADC) = ¹/₂ [-35]
ar(ADC) = -17.5
But area cannot be negative.
∴ ar(ADC) = 17.5 sq.units.
Area of ΔABC.
ar(ABC) = ¹/₂ [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
ar(ABC) = ¹/₂ [-4(-5 + 2) + -3(-2 + 2) + 3(-2 + 5)]
ar(ABC) = ¹/₂ [-4(-3) + -3(0) + 3(3)]
ar(ABC) = ¹/₂ [12 + 9]
ar(ABC) = ¹/₂ [21]
ar(ABC) = 10.5 sq.units.
Area of Quadrilateral = ar(ABC) + ar(ADC)
Area of Quadrilateral = 17.5 + 10.5
Area of Quadrilateral = 28 sq.units.
∴ The area of the quadrilateral is 28 square units.
