# CBSE Class 10 Maths Paper | 2018 Q15B

###### CBSE Solved Question Paper 3 Mark | Coordinate Geometry

Question 15B: If A(- 5, 7), B(- 4, - 5), C(- 1, - 6) and D(4, 5) are the vertices of a quadrilateral. Find the area of the quadrilateral ABCD.

## NCERT Solution to Class 10 Maths

### Explanatory Answer | Class 10 Maths 2018 Q15 B

A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are vertices of a quadrilateral.
Join AC. We get 2 triangles ABC and ADC.

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC.

#### Step 1: Compute Area of Triangle ABC

Area of ∆ABC = $$frac{1}{2}$ {x1$y2 – y3) + x2 (y3 - y1) + x3(y1 – y2)}
A(-5, 7), B(-4, -5) and C(-1, -6)
Area of ∆ABC = $$frac{1}{2}$ {$-5) (-5 – (-6) + (-4) (-6 – 7) + (-1) (7 – (-5))}
Area of ∆ABC = $$frac{1}{2}$ {$-5) (1) + -4(-13) + (-1)(12)}
Area of ∆ABC = $$frac{1}{2}$ {-5 + 52 – 12} = $\frac{1}{2}$ × 35 = 17.5 sq units #### Step 2: Compute Area of Triangle ADC Area of triangle ADC = $\frac{1}{2}$ {x1$y2 – y3) + x2 (y3 - y1) + x3(y1 – y2)}
A(-5, 7), C(-1, -6), and D(4, 5)
Area of ∆ADC = $$frac{1}{2}$ {$-5)(-6 - 5) + (-1)(5 – 7) + 4(7 – (-6))}
Area of ∆ADC = $$frac{1}{2}$ {$-5)(-11) + (-1) (-2) + 4(13)}
Area of ∆ADC = $\frac{1}{2}$ {55 + 2 + 52} = $\frac{109}{2}$ = 54.5 sq units

#### Step 3: Compute Area of Quadrilateral ABCD

Area of quadrilateral ABCD = 17.5 + 54.5 = 72 sq units

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