Find the area of a trapezium whose parallel sides are 47 cm and 60cm and non parallel sides are 25cm and 26fm
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Step-by-step explanation:
\\Given:\\ Sides\ of\ a\ trapezium\ AB=47cm, CD=60cm,\ and\ BC=26cm, AD=25cm,\\ Draw\ the\ a\ parallel\ line\ such\ that\ BE\parallel AD\\\\ Now,\ ABED\ is\ a \ parallelogram\\\\ So, AD=BE= 25cm\ and\ AB=DE=47cm\\\\ \Rightarrow EC= DC-DE=60-47=13cm\\\\ By\ heron's\ formula:\\\\ Area\ of\ triangle\ (A)=\sqrt{s(s-a)(s-b)(s-c)}\\\\ Here,\ a,\ b,\ c\ are\ sides\ of\ triangle\ and\ s=\frac{a+b+c}2\\\\\\ \Rightarrow s=\frac{25+26+13}2=\frac{64}2=32cm\\\\\\ Area\ of\ \Delta BCE= \sqrt{32(32-25)(32-26)(32-13)}\\\\ = \sqrt{32\times7\times6\times19}=8\sqrt{399}=159.8cm^2\\\\\\ Area\ of\ a \ parallelogram\ ABED=AB\times AD=47\times25=1175cm^2\\\\ Now,\\ Area\ of\ a \ parallelogram\ ABCD=Area\ of\ a \ parallelogram\ ABED+Area\ of\ \Delta BCE\\\\ =1175+159.8=1334.8cm^2
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Kishore Kumar
Let ABCD be the trapezium with AB = 60 cm, BC = 26 cm, CD = 47 cm and AD = 25 cm Complete parallelogram AECD Hence CE = 25 cm and AE = 47 cm Draw CF⊥ BC From the figure BE = AB – AE ⇒ BE = 60 – 47 = 13 cm First we will find the area of ΔCEB using Heron’s formula Hence area of ΔCEB = 8√399 sq cm Recall that area of triangle = 8√399 = (1/2) × 13 × CF Area of parallelogram AECD = AE × CF Hence area of trapezium ABCD = area of parallelogram AECD + area of DCEB
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