Class 9 Opt Maths: Area Of Triangle

Area of triangle

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)

Area of a Quadrilateral

Area of quadrilateral

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 △$\mathrm{△}$ABC + Area of △$\mathrm{△}$ACD.
□$◻$ABCD = ½∣∣∣x1y1x2y2x3y3x1y1∣∣
∣

$\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|$ + ½ ∣∣∣x1y1x3y3x4y4x1y1∣∣∣$\left|\begin{array}{cccc}{x}_1& x_3& x_4& x_1\\ y_1& y_3& y_4& y_1\end{array}\right|$
= 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.

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