Pqrs and abrs are parallelogram and x is any point on br show that 1) area(pqrs)=area(abrs) 2) area(axs)=1/2 area(pqrs)

Step-by-step explanation:

Given PQRS and ABRS are parallelogram and X is any point on BR. we have to prove that  

1)

2)

In ∆ASP and ΔBRQ  

∠SPA = ∠RQB [Corresponding angles]

∠PAS = ∠QBR [Corresponding angles]

PS = QR [Opposite sides of the parallelogram PQRS]

By AAS rule, ∆ASP≅ΔBRQ  

∴ ar(ASP) = ar(BRQ)      (Congruent triangles have equal area)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ]  

                           = ar(QRB) + ar(ASRQ]  = ar(ABRS)  

So, ar(PQRS)=ar(ABRS)

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

⇒   (∵ar(PQRS)=ar(ABRS))  

Hence Proved.

29 is the correct answer

step-by-step explanation:

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!