Let A(x_{1},y_{1}), B(x_{2}.y_{2}) and C (x_{3.}y_{3}) be the vertices of a triangle. Draw \(\perp\)AL, BM and CN from the vertices A, B, and C respectively to the X-axis.
Then,
OL = x_{1} OM = x_{2} ON =x_{3}AL = y_{1} BM = y_{2} CN = y_{3}Here, LN = ON - OL = x_{3}-x_{1}NM = OM - ON = x_{2}-x_{3 }and
LM = OM -OL = x_{2}-x_{1}Now, Area ofΔABC is equal to
Area of trapezium ALNC + Area of trapezium CNMB - Area of trapezium ALMB.
= ½LN(AL+CN) + ½NM(CN+BM) - ½ LM(AL+BM)
= ½(x_{3-}x_{1})(y_{1}+y_{3})+½(x_{2}-x_{3})(y_{3}+y_{2})-½(x_{2}-x_{1})(y_{1}+y_{2})
= ½(x_{3}y_{1}+x_{3}y_{3}-x_{1}y_{1}-x_{1}y_{3}+x_{2}y_{3}+x_{2}y_{2}-x_{3}y_{3}-x_{3}y_{2}-x_{2}y_{1}-x_{2}y_{2}+x_{1}y_{1}+x_{1}y_{2})
= ½(x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3})
Let A (x_{1},y_{1}), B(x_{2},y_{2}), C(x_{3,}y_{3}) and D (x_{4},y_{4})be the vertices of quadrilateral ABCD. Join AC. Then the diagonal AC divides the quadrilateral into two triangles ABC and ACD.
Now,
Area of quadrilateral ABCD = Area of \(\triangle\)ABC + Area of \(\triangle\)ACD.
\(\square\)ABCD = ½\(\begin{vmatrix}x_1&x_2&x_3&x_1\\y_1&y_2&y_3&y_1 \end{vmatrix}\) + ½ \(\begin{vmatrix}x_1&x_3&x_4&x_1\\y_1&y_3&y_4&y_1 \end{vmatrix}\)
= ½ (x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3}) + ½ (x_{1}y_{3}-x_{3}y_{1}+x_{3}y_{4}-x_{4}y_{3}+x_{4}y_{1}-x_{1}y_{4})
= ½ (x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{4}-x_{4}y_{3}+x_{4}y_{1}-x_{1}y_{4})
Note: To find the area of a quadrilateral by using this formula, the vertices of quadrilateral should be taken in order. So it is better to plot the points roughty before applying the formula. Otherwise the result may be wrong.
∴ \(\square\)ABCD = ½\(\begin{vmatrix}x_1&x_2&x_3&x_1\\y_1&y_2&y_3&y_1 \end{vmatrix}\) + ½ \(\begin{vmatrix}x_1&x_3&x_4&x_1\\y_1&y_3&y_4&y_1 \end{vmatrix}\)
=½ (x_{1} y_{2}-x_{2} y_{1}+x_{2} y_{3}-x_{3} y_{2}+x_{3} y_{1}-x_{1} y_{3}) +½ (x_{1} y_{3}-x_{3} y_{1}+x_{3} y_{4}-x_{4} y_{3}+x_{4} y_{1}-x_{1} y_{4})
=½ (x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{4}-x_{4}y_{3}+x_{4}y_{1}-x_{1}y_{4})
Note : To find the area of a quadrilateral by using this formula, the vertices of quadrilateral should be taken in order. So it is better to plot the points roughty before applying the formula. Otherwise the result may be wrong.
Note: To find the area of a quadrilateral, we divide it into two triangles by joining a diagonal. Then the sum of the areas of the two triangles will be equal to the area of the quadrilateral.
.
Given vertices are of triangle are:
(x_{1}, y_{1}) = (3, -4)
(x_{2},y_{2}) = (-2, 3)
(x_{3}, y_{3}) = (4, 5)
We have,
\begin{align*} \text {Area of triangle} &= \frac 12 |[(x_1y_2 - x_2y_1) + (x_2y_3 - x_3y_2) + (x_3y_1 - x_1y_3)]|\\ &= \frac 12 |[{3 × 3 - (-2) × (-4)} + {(-2) × 5 - 4 × 3} + {4 × (-4) - 3 × 5}]|\\ &= \frac 12 |[(9 - 8) + (-10 - 12) + (-16 - 15)]|\\ &= \frac 12 |[1 - 22 - 3]|\\ &= \frac 12 |1 - 53|\\ &= \frac 12 ×|-52|\\ &= \frac 12 × 52\\ &= 26 \;\text{square units}_{Ans}\\ \end{align*}
If the area of a triangle having the verticles (a,1), (3,a) and (a,-1) be 2 Square units, calculate the value of a.
Points p(3,-5),Q(-2,a) and R(18,1) are collinear.Calculate the value of a.
-6
6
3
-7
Find the area of triangles having the following vertices:
(2,0), (0,2) and (-2,-1)
1sq.units
8 sq.units
6 sq.units
5 sq.units
Find the area of triangles having the following vertices:
(2,3), (4,-1) and (-2,1)
13 sq.units
18 sq.units
10 sq.units
15 sq.units
Find the area of triangles having the following vertices:
(-3,0), (0,0) and (0,3)
6 sq. units
4 sq. units
8 sq. units
4.5 sq. units
Find the area of triangles having the following vertices:
A(0,3), B(6,2) and C(4,7)
9 sq. units
17 sq. units
14 sq. units
11 sq. units
Find the value of k if the following points lie on the same straight line:
(k,-5),(2,3) and (-6,5)
77
34
56
21
Find the value of k if the following points lie on the same straight line:
A(1,3),B(k,1)and C(6,-2)
8
1
5
3
Find the value of k if the following points lie on the same straight line:
(4,2),(5,k) and (8,7)
21/7
13/7
11/2
12/4
Find the value of k if the following points lie on the same straight line:
(-4,1),(k,5) and (k+5,9)
4
12
8
1
For what value of k the area quadrilateral, with vertices(k,-3),(6,4),(5,6) and (-3, 5),is 41 sq. units?
k=8
k=4
k=10
k=1
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Formula
Mar 23, 2017
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Subas Tripathi
How the formula of area of quadrilateral derived?
Mar 20, 2017
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