Question

Find the quadrilatral whose sides are ab=8cm,bc=15cm,cd=12cm,ad=25cm and angle b=90degree

Answer 1

In right ΔABC,

$AB^2 + BC^2 = AD^2$

$AC^2 = 15^2 + 8^2$

$AC = \sqrt {15^2 + 8^2}$

$= \sqrt {225 + 64}$

$= \sqrt 289$

= 17 cm

Length of AC = 17 cm

$\small{Area \: of \: ΔABC = \frac {1}{2} (8 \times 15) = 60 cm^2}$

Now, In ΔACD, AD = 25 cm, AC = 17 cm & CD = 12 cm

Using Heron's Formula, semiperimeter,

$s = \frac {a + b + c}{2}$

$= \frac {17+12+25}{2}$

$\implies \frac {54}{2} = 27 cm$

Length of AD = 27 cm

$\small{Area \: of \: ΔACD = \sqrt {s(s-a)(s-b)(s-c)} }$

$\small \bold{=\sqrt {27(27-17)(27-12)(27-25)} }$

$= \sqrt {27 \times 10 \times 15 \times 2}$

$\small{= \sqrt {3 \times 3 \times 3 \times 2 \times 5 \times 3 \times 5 \times 2} }$

$= 3 \times 3 \times 2 \times 5$

$= 90 cm^2$

Hence, Area of Quadrilateral ABCD = Area of ΔABC + Area of ΔACD

= 60 + 90

$= 150 cm^2$