Advertisements
Advertisements
Question
Find the area of the quadrilateral whose vertices are A(–3, 1), B(–2, –2), C(4, 1), D(2, 3).
Sum
Advertisements
Solution
A(–3, 1), B(–2, –2), C(4, 1), D(2, 3)
A(`square`ABCD) = A(ΔABC) + A(ΔACD)
Area of triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
A(ΔABC) = `1/2|(-3, 1, 1),(-2, -2, 1),(4, 1, 1)|`
= `1/2[-3(-2 - 1) - 1(-2 - 4) + 1(-2 + 8)]`
= `1/2(9 + 6 + 6)`
∴ A(ΔABC) = `21/2` sq. units'.
∴ A(ΔACD) = `1/2|(-3, 1, 1),(4, 1, 1),(2, 3, 1)|`
= `1/2[-3(1 - 3) - 1(4 - 2) + 1(12 - 2)]`
= `1/2(6 - 2 + 10)`
∴ A(ΔACD) = 7 sq. units
∴ A(`square`ABCD) = A(ΔABC) + A(ΔACD)
= `21/2 + 7`
= `35/2` sq. units.
shaalaa.com
Is there an error in this question or solution?
