The area of the adjoining trapezium is 276 [tex]cm^{2}[/tex].
Given the sides of trapezium be 16 cm, 15 cm, 30 cm, 13 cm.
We are required to find the area of adjoining trapezium.
Draw two perpendiculars on AD and the points will be E and F.
From triangles ABE and CFD.
let the length of AE=x, FD=30-16-x=14-x.
BE=[tex]\sqrt{169-x^{2} }[/tex], CF=[tex]\sqrt{225-(14-x)^{2} }[/tex]
BE=CF
[tex]\sqrt{169-x^{2} }[/tex]=[tex]\sqrt{225-(14-x)^{2} }[/tex]
Squaring both sides.
169-[tex]x^{2}[/tex]=225-196-[tex]x^{2}[/tex]+28x
140=28x
x=5 cm.
Put in BE=[tex]\sqrt{169-x^{2} }[/tex]
BE=[tex]\sqrt{169-25}[/tex]
=[tex]\sqrt{144}[/tex]
=12 cm.
Area of trapezium=1/2 (sum of parallel sides)*height
=1/2 (30+16)*12
=23*12
=276 [tex]cm^{2}[/tex]
Hence if the sides of trapezium are 16 cm, 15 cm, 30 cm, 13 cm then the area of the adjoining trapezium is 276 [tex]cm^{2}[/tex].
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