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Question
Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB=90° and AC = 15 cm.
Solution
In the right angled : ΔACB
`AB^2=BC^2+AC^2`
⇒ `17^2=BC^2+15^2`
⇒ `17^2-15^2=BC ^2`
⇒ `64=BC^2`
⇒`BC=8 cm`
Perimeter=`AB+BC+CD+AD`
=`17+8+12+9`
= `46 cm^`
Area of `ΔABC=1/2(bxxh`)
=`1/2(8xx15)`
=`60 cm^2`
In : ΔADC
`AC^2=AD^2+CD^2`
So, ΔADC is a right- angled triangle at D
Area of `ΔADC=1/2xxbxxh`
=`1/2xx9xx12`
=`54 cm^2`
∴ Area of the quadrilateral =` Area of ΔABC+Area of ΔADC`
=`60+54`
=`114 cm^2`
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