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Short Note
Find the area of the shaded region in the given figure.
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Solution
We are given the following figure with dimensions.
Area of shaded region = Area of ΔABC – Area of ΔADB
Now in ΔADB
`⇒ AB62 = AD^2 + BD^2` --(i)
⇒ Given that AD = 12 cm BD = 16 cm
Substituting the values of AD and BD in the equation (i), we get
`AB^2=12^2+16^2`
`AB^2=144+256`
`AB=sqrt400`
`AB=20cm`
∴ Area of triangle = `1/2xxADxxBD`
`=1/2xx12xx16`
`=96cm^2`
Now
In ΔABC, S =`1/2(AB+BC+CA)`
`=1/2xx(52+48+20)`
`1/2(120)`
`60cm`
By using heron’s formula
We know that, Area of Δle ABC `=sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(60(60-20)(60-48)(60-52))`
`=sqrt(60(40)(12)(8))`
`=480cm^2`
`Area of shaded region = Area of ΔABC – Area of ΔADB`
`=(480-96)cm^2`
`384 cm^2`
∴ Area of shaded region = 384 `cm^2`
Concept: Area of a Triangle by Heron's Formula
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