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Question
Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD =9 cm, CD = l2cm, ∠ACB = 90° and AC=l5cm.
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Solution
The sides of a quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm
Here, By using Pythagoras theorem
`BC=sqrt(17^2-15^2) = sqrt(289-225)=sqrt64=8cm`
Now, area of ΔABC=`1/2xx8xx15=60cm^2`
For area of Δle ACD,
Let a = 15 cm, b = 12 cm and c = 9 cm
Therefore, S =`(15+12+9)/2=36/2=18cm`
Area of ACD =`sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(18(18-15)(18-12)(18-9))`
`=sqrt(18xx18xx3xx3)`
`=sqrt((18xx3)^2)`
`=54cm^2`
∴ Thus, the area of quadrilateral ABCD = `60 + 54 = 114 cm^2`
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