Question

Find the area of a trapezium ABCD in which AB║DC, AB = 32cm, BC = 16cm, CD = 20cm and DA = 10cm.

Answer 1

       $\underline{\bold{Given:}}$

       $ABCD \text{ is a trapezium}$

       $AB \parallel CD$

       $AB = 32 \text{ cm}$

       $BC = 16 \text{ cm}$

       $CD = 20 \text{ cm}$

       $AD = 10 \text{ cm}$

       $\underline{\bold{To -Find:}}$

       $\text{The area of a trapezium }ABCD$

       $\underline{\bold{Previous-knowledge:}}$

       $\text{Area of trapezium} = \dfrac{1}{2}(\text{Sum of parallel sides})(\text{Height})$

       $\text{Area of triangle}=\dfrac{1}{2}(\text{base)(\text{height)}}$

       $\text{Heron's formula} \implies \sqrt{s(s-a)(s-b)(s-c)} = \text{Area of triangle}$

       $\underline{\bold{Construction:}}$

       $\text{Draw a line }DE \text{ such that }DE \text{ is perpendicular to }AB \text{ and }E \text{ lies on } AB$

       $\text{Draw a line }DF \text{ such that }DF = BC \text{ and }DF \parallel BC, F \text{ lies on }AB$

       $\underline{\bold{Solution:}}$

       $DF = BC \text{ and }DF \parallel BC \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\text{Construction})$

$\implies FBCD \text{ is a parallelogram as it has a pair of equal and parallel sides}$

$\implies BF = DC = 20\text{ cm}$      

$\implies AF = AB - BF = 12\text{ cm}$

       $\text{In }\triangle AFD$

       $\text{Applying Heron's formula}$

$\implies \sqrt{s(s-a)(s-b)(s-c)} = \text{Area of }\triangle$

       $\text{In our case:}$

       $a = AD = 10 \text{ cm}$

       $b = AF = 12 \text{ cm}$

       $c = DF = 16 \text{ cm}$

       $s = \dfrac{a+b+c}{2} = \dfrac{10 + 16 + 12}{2} = \dfrac{38}{2} = 19 \text{ cm}$

$\implies \text{Ar}\triangle AFD = \sqrt{19(19-10)(19-12)(19-16)}$

$\implies \text{Ar}\triangle AFD = \sqrt{19(9)(7)(3)}$

$\implies \text{Ar}\triangle AFD = \sqrt{3591}$

$\implies \text{Ar}\triangle AFD = 59.92 \text{ cm}^{2} = 60 \text{ cm}^{2} (\text{approx.)}$

       $\text{Applying area of triangle formula}$

$\implies \text{Ar}\triangle AFD = \dfrac12(AF)(DE)$

$\implies 60 = \dfrac12(12)(DE)$

$\implies 10 \text{ cm}= DE$

       $\text{Applying area of trapezium formula}$

$\implies\text{Ar}ABCD = \dfrac{1}{2}(AB + BC)(DE)$

$\implies\text{Ar}ABCD = \dfrac{1}{2}(32 + 20)(10)$

$\implies\text{Ar}ABCD = (52)(5)$

$\implies\boxed{\text{Ar}ABCD = 260 \text{ cm}^{2} \text{ }(\text{approx.})}$