find the area of the quadrilateral abcd, co-ordinates of whose vertices are a (5,-2) b (-3,-1) c (2,1) d (6,0)
Answers
1) Find the area of \(\Delta ABC\) whose vertices are:
(i) A(1, 2), B(-2, 3) and C(-3, -4)
(ii) A(-5, 7), B(-4, -5) and C(4, 5)
(iii) A(3, 8), B(-4, 2) and C(5, -1)
(iv) A(10, -6), B(2, 5) and C(-1, 3)
i) A (1, 2), B (-2, 3) and C (-3, -4) are the vertices of \(\Delta ABC\). Then
(x1 = 1, y1 = 2), (x2 = -2, y2 = 3), (x3 = -3, y3 = -4)
Area of triangle ABC
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ 1(3-(-4))+(-2)(-4-2)+(-3)(2-3) \right \}\)
= \(\frac{1}{2}*\left\{1(3+4)-2(-6)-3(-1)\right\}\)
= \(\frac{1}{2}\left \{7+12+3\right \}\)
= \(\frac{1}{2}\left \{22\right \}\)
= 11sq.units
ii) A(-5, 7), B(-4, -5) and C(4, 5)
A (-5, 7), B (-4, -5) and C (4, 5) are the vertices of \(\Delta ABC\). Then
(x1 = -5, y1 = 7), (x2 = -4, y2 = -5), (x3 = 4, y3 = 5)
Area of triangle ABC
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ -5 (-5–5) + (-4)(5-7) + 4 (7-(-5)) \right \}\)
= \(\frac{1}{2}*\left\{-5 (-10) -4 (-2) + 4 (12)\right\}\)
= \(\frac{1}{2}\left \{50 + 8 + 48\right \}\)
= \(\frac{1}{2}\left \{106\right \}\)
= 53 sq.units
iii) A(3, 8), B(-4, 2) and C(5, -1)
A (3, 8), B (-4, 2) and C (5, -1) are the vertices of \(\Delta ABC\). Then
(x1 = 3, y1 = 8), (x2 = -4, y2 = 2), (x3 = 5, y3 = -1)
Area of triangle ABC
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ 3 (2 – (-1)) + (-4)(-1 – 8) + 5 (8 – 2)) \right \}\)
= \(\frac{1}{2}*\left\{3 (2 + 1) – 4 (-9) + 5 (6)\right\}\)
= \(\frac{1}{2}\left \{9 + 36 + 30\right \}\)
= \(\frac{1}{2}\left \{75\right \}\)
= 37.5 sq.units
iv) A(10, -6), B(2, 5) and C(-1, 3)
A (10, -6), B (2, 5) and C (-1, -3) are the vertices of \(\Delta ABC\). Then
(x1 = 10, y1 = -6), (x2 = 2, y2 = 5), (x3 = -1, y3 = 3)
Area of triangle ABC
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ 10 (5 -3) + (2)(3 – (-6)) + (-1) (-6 – 5) \right \}\)
= \(\frac{1}{2}*\left\{10 (2 ) + 2 (9) – 1 (11)\right\}\)
= \(\frac{1}{2}\left \{20 + 18 + 11\right \}\)
= \(\frac{1}{2}\left \{49\right \}\)
= 24.5 sq.units
2) Find the area of quadrilateral ABCD whose vertices are A(3, -1), B(9, -5), C(14, 0) and D(9, 19).
By joining A and C, we get two triangles ABC and ACD
Let A (3, -1), B (9, -5) and C (14, 0) and D (9, 19)
(x1 = 3, y1 = -1), (x2 = 9, y2 = -5), (x3 = 14, y3 = 0), (x4 = 9, y4 = 19)
Then
Area of triangle ABC
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ 3 (-5 -0) + (9)(0 + 1) + (14) (-1 + 5) \right \}\)
= \(\frac{1}{2}*\left\{3 (-5 ) + 9 (1) + 14 (4)\right\}\)
= \(\frac{1}{2}\left \{-15 + 9 + 56\right \}\)
= \(\frac{1}{2}\left \{50\right \}\)
= 25 sq.units
Area of triangle ACD
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ 3 (0 -19) + (14)(19 + (1)) + (9) (-1 – 0)) \right \}\)
= \(\frac{1}{2}*\left\{3 (-19 ) + 14 (20) + 9 (-1)\right\}\)
= \(\frac{1}{2}\left \{-57 + 280 – 9\right \}\)
= \(\frac{1}{2}\left \{214\right \}\)
= 107 sq.units
So, the area of the quadrilateral is 25 + 107 = 132 sq. units
3) Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q (- 4, – 6) and R (2, -3) and S (1, 2)
By joining P and R, we get two triangles PQR and PRS
Let P (-5, -3), Q (- 4, – 6) and R (2, -3) and S (1, 2)
(x1 = -5, y1 = -3), (x2 = – 4, y2 = – 6), (x3 = 2, y3 = -3), (x4 = 1, y4 = 2)
Then
Area of triangle PQR
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ -5 (-6 + 3) -4(-3 + 3) + 2 (-3 + 6) \right \}\)
= \(\frac{1}{2}*\left\{-5 (-3 ) – 4 (0) + 2 (3)\right\}\)
= \(\frac{1}{2}\left \{15 – 0 + 6\right \}\)
= \(\frac{1}{2}\left \{21\right \}\)
= 10.5 sq.units
Area of triangle PRS
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ -5 (-3 -2) + 2(2 + 3) + 1 (-3 + 3) \right \}\)
= \(\frac{1}{2}*\left\{-5 (-5 ) + 2 (5) + 1 (0)\right\}\)
= \(\frac{1}{2}\left \{25 + 10 + 0\right \}\)
= \(\frac{1}{2}\left \{35\right \}\)
= 17.5 sq.units
So, the area of the quadrilateral is 10.5 + 17.5 = 28 sq. units
4) Find the area of quadrilateral ABCD whose vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3, 4).
By joining A and C, we get two triangles ABC and ACD
Let A (-3, -1), B (-2, -4) and C (4, -1) and D (3, 4)
(x1 = -3, y1 = -1), (x2 = -2, y2 = -4), (x3 = 4, y3 = -1), (x4 = 3, y4 = 4)
Then
Area of triangle ABC
= \(\frac{1}{2}\left\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right\}\)
= \(\frac{1}{2}*\left \{ -3 (-4 + 1) – 2(-1 + 1) + 4 (-1 + 4) \right \}\)
= \(\frac{1}{2}*\left\{-3 (-3 ) -2 (0) + 4 (3)\right\}\)
= \(\frac{1}{2}\left \{9 + 0 + 12\right \}\)
= \(\frac{1}{2}\left \{21\right \}\)
= 10.5 sq.units