Question 1 of 70intermediate🔧 ApplyNumerical4 marks

Draw a quadrilateral in the Cartesian plane, whose vertices are (−4,5),(0,7),(5,−5)(-4, 5), (0, 7), (5, -5) and (−4,−2)(-4, -2). Also, find its area.

Correct Answer

60.5 square units

Exercise: EXERCISE 9.1 | Q: 1 | (Page 8)
For More Understanding

Explanation

The problem requires finding the area of a quadrilateral given its vertices. The provided context mentions a 'recall of coordinate geometry' which includes important formulae studied in earlier classes. This refers to the formula for the area of a triangle given its vertices. To find the area of the quadrilateral, we divide it into two triangles by drawing a diagonal. We then calculate the area of each triangle using the formula and sum them up.

Solution Steps

  1. Step 1: Plotting the Vertices The vertices of the quadrilateral are given as A(−4,5)A(-4, 5), B(0,7)B(0, 7), C(5,−5)C(5, -5), and D(−4,−2)D(-4, -2). These points are plotted on the Cartesian plane and joined in order to form quadrilateral ABCD.

  2. Step 2: Dividing the Quadrilateral To find the area of quadrilateral ABCD, we draw a diagonal, say BD. This divides the quadrilateral into two triangles: ΔABD\Delta ABD and ΔBCD\Delta BCD.

  3. Step 3: Area of Triangle ABD The vertices of ΔABD\Delta ABD are (−4,5)(-4, 5), (0,7)(0, 7), and (−4,−2)(-4, -2). Using the area formula: Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

    Area(ΔABD)=12∣−4(7−(−2))+0(−2−5)+(−4)(5−7)∣\text{Area}(\Delta ABD) = \frac{1}{2} | -4(7 - (-2)) + 0(-2 - 5) + (-4)(5 - 7) | =12∣−4(9)+0+(−4)(−2)∣= \frac{1}{2} | -4(9) + 0 + (-4)(-2) |

    =12∣−36+8∣=12∣−28∣=14= \frac{1}{2} | -36 + 8 | = \frac{1}{2} |-28| = 14 sq. units.

  4. Step 4: Area of Triangle BCD The vertices of ΔBCD\Delta BCD are (0,7)(0, 7), (5,−5)(5, -5), and (−4,−2)(-4, -2).

    Area(ΔBCD)=12∣0(−5−(−2))+5(−2−7)+(−4)(7−(−5))∣\text{Area}(\Delta BCD) = \frac{1}{2} | 0(-5 - (-2)) + 5(-2 - 7) + (-4)(7 - (-5)) | =12∣0+5(−9)+(−4)(12)∣= \frac{1}{2} | 0 + 5(-9) + (-4)(12) |

    =12∣−45−48∣=12∣−93∣=46.5= \frac{1}{2} | -45 - 48 | = \frac{1}{2} |-93| = 46.5 sq. units.

  5. Step 5: Total Area The area of quadrilateral ABCD is the sum of the areas of ΔABD\Delta ABD and ΔBCD\Delta BCD. Area(ABCD)=14+46.5=60.5\text{Area}(ABCD) = 14 + 46.5 = 60.5 sq. units.

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