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    CAT Application of Heron’s Formula in Finding Areas of Quadrilaterals - Practice Questions & MCQ

    Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

    Quick Facts

    • 3 Questions around this concept.

    Solve by difficulty

    QuestionMEDIUM

    The area of the given trapezium is: (in cm2)

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    The area of a quadrilateral ABCD in which AC =17 m, BC = 15 m, CD = 12 m, DA = 9m and $\angle$B = 90º is: (in m2)

    View Solution

    The adjacent sides of a parallelogram ABCD are AB = 34 m, BC = 20 m and diagonal AC = 42 m. Then the area of the parallelogram is: (in m2)

    View Solution

    Concepts Covered - 1

    Application of Heron’s Formula in Finding Areas of Quadrilaterals

    Many a time, the figures are in the shape of quadrilaterals. To find the area of quadrilateral, we need to divide the quadrilateral in triangular parts and then use the formula for area of the triangle. Let us look at this problem:

    The area of the quadrilateral ABCD in which, AB = 12 cm, BC = 5 cm, CD = 14 cm, DA = 19 cm and ∠ B = 90°. 

    Let ABCD be the given quadrilateral

    In ∆ ABC, by Pythagoras’ theorem, we have

                AC2 = AB2 + BC2

                AC2 = 122 + 52 = 144 + 25  = 169 

                 AC = 13

    \begin{aligned} \text { Area of } \Delta A B C &=\frac{1}{2} \times A B \times B C \\ &=\left(\frac{1}{2} \times 12 \times 9\right) \mathrm{cm}^{2}=54 \mathrm{cm}^{2} \end{aligned}

    In ∆ ABD, Let a = AC = 13 cm, b = CD = 14 cm and  c = AD = 19 cm.

    \therefore \;\text{s}=\frac{1}{2}\left ( 13+14+19 \right )\text{cm}=23\text{ cm} 

    (s - a) = (23 - 13) = 10 cm

    (s - b) = (23 - 14) = 9 cm

    (s - c) = (23 - 19) = 4 cm

    \\\therefore \quad \text { area }(\Delta A C D)=\sqrt{s(s-a)(s-b)(s-c)}\\\text{ }\quad\quad\quad\quad\quad\quad\quad\;\;\;\;=\sqrt{23 \times 10 \times 9 \times 4} \mathrm{cm}^{2} \\\text{ }\quad\quad\quad\quad\quad\quad\quad\;\;\;\;=\sqrt{8280} \mathrm{cm}^{2}\\\text{ }\quad\quad\quad\quad\quad\quad\quad\;\;\;\;=6\sqrt{230}\mathrm{cm}^{2}

    \\\therefore \quad \text { area of quad. } A B C D=\operatorname{area}(\triangle A B C)+\operatorname{area}(\triangle A C D)\\\text{}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;\;=(54+6\sqrt{230})\text{ cm}^2

     

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