## Note on Area of Triangle and Quadrilateral

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#### Area of Triangle

Let A(x1,y1), B(x2.y2) and C (x3.y3) 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 = x1 OM = x2 ON =x3
AL = y1 BM = y2 CN = y3
Here, LN = ON - OL = x3-x1
NM = OM - ON = x2-x3 and
LM = OM -OL = x2-x1
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(x3-x1)(y1+y3)+½(x2-x3)(y3+y2)-½(x2-x1)(y1+y2)
= ½(x3y1+x3y3-x1y1-x1y3+x2y3+x2y2-x3y3-x3y2-x2y1-x2y2+x1y1+x1y2)
= ½(x1y2-x2y1+x2y3-x3y2+x3y1-x1y3)

Let A (x1,y1), B(x2,y2), C(x3,y3) and D (x4,y4)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 (x1y2-x2y1+x2y3-x3y2+x3y1-x1y3) + ½ (x1y3-x3y1+x3y4-x4y3+x4y1-x1y4)
= ½ (x1y2-x2y1+x2y3-x3y2+x3y4-x4y3+x4y1-x1y4)

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 (x1 y2-x2 y1+x2 y3-x3 y2+x3 y1-x1 y3) +½ (x1 y3-x3 y1+x3 y4-x4 y3+x4 y1-x1 y4)

=½ (x1y2-x2y1+x2y3-x3y2+x3y4-x4y3+x4y1-x1y4)

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.

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### Very Short Questions

Given vertices are of triangle are:

(x1, y1) = (3, -4)

(x2,y2) = (-2, 3)

(x3, y3) = (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*}

0%

6
3
5
7

-7

3

6

-6

6 sq.units

5 sq.units

8 sq.units

1sq.units

13 sq.units

10 sq.units

18 sq.units

15 sq.units

• ### Find the area of triangles having the following vertices:(-3,0), (0,0) and (0,3)

8 sq. units

6 sq. units

4.5 sq. units

4 sq. units

• ### Find the area of triangles having the following vertices:A(0,3), B(6,2) and C(4,7)

17  sq. units

11 sq. units

14 sq. units

9  sq. units

34

77

56

21

3

5

8

1

21/7

12/4

13/7

11/2

8

4

1

12

k=10

k=1

k=8

k=4

• ## You scored /11

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