Question

Plzz do the first part of my respective question!plzz do it fast as u can plzz !

Answer 1

HEY MATE
hear is your answer
FOR TRIANGLE DBC WE WILL WRITE PYTHAGORAS THEOREM
${bc}^{2} \: + {bd}^{2} = {dc}^{2} \\ = > {bd}^{2} + {8}^{2} = {17}^{2} \\ = > bd = \sqrt{289 - 64} \\ = > bd = \sqrt{225} \\ = 15$
Now as bd =15 cm = height of ∆DBC= hypotenuse of ∆DBA
Now side for side AB
${ad}^{2} + {ab}^{2} = {db}^{2} \\ = > {9}^{2} + {ab}^{2} = {15}^{2} \\ = > ab = \sqrt{225 - 81} \\ = > ab = \sqrt{144} \\ = 12$

Therefore AB = 12
now are of quadrilateral ABCD
= area of ∆ABD+area of triangle DBC
= 1/2× 12×9 + 1/2 × 15 ×8
= 6×9 + 15×4
=54+ 60
= 114 cm sq.

Answer 2

by pgt in ∆ bcd

dc^2=bd^2 + bc^2
17*17=bd^2 + 8*8
289 - 64 = bd^2
225 = bd^2
bd= √225 =√15x15= 15

by pgt in ∆bad
bd^2 =ba^2 + ad^2
225= 81 + ba^2
ba^2 =225-81 =144
ba = 12

now area of ∆ abd =1/2*12*9
=54cm^2.........eq1
now area of ∆dbc = 1/2*8*15
= 60cm^2........eq2
adding eq 1 and 2
we get
area of quad = 54 + 60
and now you can solve it...... you can get your answer
hope it help you