Question

Find the perimeter and area of a quadrilateral ABCD in which BC=12cm ,CD=9cm,BD=15cm,DA= 17cm and angle ABD=90°.

Answer 1

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Answer 2

Area of ABCD = $\bold {114\ cm^2}$  

Step-by-step explanation:

In quadrilateral ABCD,

∠ABD = 90°,

Area of quadr ABCD = Area of ΔABD + Area of ΔBCD

In ΔABD,

By using pythagoras theorem,

$(AD)^2 = (AB)^2+(BD)^2$

$(17)^2 =(AB)^2+(15)^2$

$289 = (AB)^2+ 225$

$289-225 = (AB)^2$

$64 =(AB)^2$

AB = 8cm

So,Area of ΔABD = $\frac{1}{2}\times Base\times Height$

                            = $\frac{1}{2}\times 15\times 8$

                            = $60\ cm^2$

Now,in ΔBCD,

$s = \frac{12+9+15}{2}$

  s= 18

Now,Area of ΔBCD = $\sqrt{s(s-a)(s-b)(s-c)}$

                                = $\sqrt{18(18-12)(18-9)(18-15)}$

                                 = $\sqrt{18(6)(9)(3)}$                                

                               = $54\ cm^2$

Area of quadrilateral ABCD = Area of ΔABD+Area of ΔBCD

                                              =$60\ cm^2 + 54\ cm^2$

                                              =$114\ cm^2$

Hence,Area of ABCD = $\bold {114\ cm^2}$