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Exercise 9.1 · Q1

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

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The area of a quadrilateral can be found by splitting it into two triangles and summing their areas. Using the shoelace formula on the given vertices, the area is 60.5\boxed{60.5} square units.

The key idea here is that a quadrilateral is just two triangles glued together along a diagonal. If you can find the area of each triangle separately and add them, you get the area of the whole shape. The neat part is that you don't even need to draw the figure perfectly — the coordinates alone give you everything.

For any triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), (x3,y3)(x_3, y_3), the area is half the absolute value of the determinant:

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

This formula works because it's essentially the cross product of two side vectors — it measures the parallelogram area, then halves it.

For a quadrilateral, you pick a diagonal that splits it cleanly. The vertices given are (−4,5)(-4, 5), (0,7)(0, 7), (5,−5)(5, -5), (−4,−2)(-4, -2). Let's label them in order: A (−4,5)(-4, 5), B (0,7)(0, 7), C (5,−5)(5, -5), D (−4,−2)(-4, -2). The diagonal AC or BD will work. Let's use AC.

  1. Split the quadrilateral into two triangles.

    Triangle 1: A (−4,5)(-4, 5), B (0,7)(0, 7), C (5,−5)(5, -5)

    Triangle 2: A (−4,5)(-4, 5), C (5,−5)(5, -5), D (−4,−2)(-4, -2)

  2. Find area of triangle ABC.

    Plug into the formula:

AreaABC=12∣(−4)(7−(−5))+0((−5)−5)+5(5−7)∣\text{Area}_{ABC} = \frac{1}{2} \left| (-4)(7 - (-5)) + 0((-5) - 5) + 5(5 - 7) \right|

Simplify inside:

=12∣(−4)(12)+0+5(−2)∣= \frac{1}{2} \left| (-4)(12) + 0 + 5(-2) \right|

=12∣−48−10∣=12×58=29= \frac{1}{2} \left| -48 - 10 \right| = \frac{1}{2} \times 58 = 29

  1. Find area of triangle ACD.

AreaACD=12∣(−4)(−5−(−2))+5((−2)−5)+(−4)(5−(−5))∣\text{Area}_{ACD} = \frac{1}{2} \left| (-4)(-5 - (-2)) + 5((-2) - 5) + (-4)(5 - (-5)) \right|

Simplify:

=12∣(−4)(−3)+5(−7)+(−4)(10)∣= \frac{1}{2} \left| (-4)(-3) + 5(-7) + (-4)(10) \right|

=12∣12−35−40∣=12∣−63∣=31.5= \frac{1}{2} \left| 12 - 35 - 40 \right| = \frac{1}{2} \left| -63 \right| = 31.5

  1. Add the two areas.

Area of quadrilateral=29+31.5=60.5\text{Area of quadrilateral} = 29 + 31.5 = 60.5

Watch out

A common mistake is to forget the absolute value or to misorder the vertices. If you list them in a different order, the diagonal might cross, giving a wrong area. Always check that the diagonal you choose lies inside the quadrilateral.

Tip

You can also use the shoelace formula directly on all four vertices in order. List them as (−4,5)(-4,5), (0,7)(0,7), (5,−5)(5,-5), (−4,−2)(-4,-2), then repeat the first. Compute sum of products down-right minus down-left, take half the absolute value. It's faster once you're comfortable.

✓Final answer

The area of the quadrilateral is 60.5\boxed{60.5} square units.

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