Question

A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Answer 1

hey there,

C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
BD is joined. 



In ΔBCD,
By applying Pythagoras theorem,
BD2 = BC2 + CD2  
⇒ BD2 = 122 + 52 
⇒ BD2 = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m2
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD  = √s (s-a) (s-b) (s-c)
                                       = √15(15 - 13) (15 - 9) (15 - 8) m2
                                       = √15 × 2 × 6 × 7 m2
                                       = 6√35 m2 = 35.5 m2(approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m2 + 35.5m2 = 65.5m2   

Hope this helps!

Answer 2

HELLO DEAR,


RIGHT TRIANGLE (BCD), WE HAVE;
${bd}^{2} = {bc}^{2} + {cd}^{2} \\ = > bd = \sqrt{ {12}^{2} + {5}^{2} } = 13m \\ and \: area \: of \: triangle(bcd) \\ = > \frac{1}{2} \times bc \times cd = \frac{1}{2 } \times 12 \times 5 = 30 {m}^{2} \\ = >$

NOW PERIMETER OF TRIANGLE ABD = AB + BD + AD = 9 m + 13 m + 8 m = 30 m
so, SEMI PERIMETER OF TRIANGLE ABD;

$s = \frac{30}{2} = 15m \\ = >$

USING HERON'S FORMULA WE HAVE;
AREA OF TRIANGLE (ABD) = 
$\sqrt{s(s - a)(s - b)(s - c)} \\ = > \sqrt{15(15 - 9)(15 - 13)(15 - 8)} \\ = > \sqrt{15 \times 6 \times 2 \times 7} \\ = > \sqrt{1260} = 35.496 {m}^{2}$
AREA OF QUADRILATERAL ABCD = AREA OF TRIANGLE ABD + AREA OF TRIANGLE BCD

$= 30 {m}^{2} + 35.496 {m}^{2} \\ = > 65.496 {m}^{2}$
I HOPE ITS HELP YOU DEAR,
THANKS