Question

Can u solve the question 1(b) and 2(b) pls exam tomorrow

Answer 1

(A) x=1 and x=2
putting value of x=1 in the equation
(1)³-a(1)+b=0
1-a+b=0
-a+b= -1
b= -1+a. ......................................(I)
Now put value of x=2
(2)³-a(2)+b=0
8-2a+b=0
-2a+b=-8
a+b= -8/-2
a+b=4
from equation (I)
a-1+a=4
2a-1=4
2a=4+1
a=5/2
putting in equation (I)
b= -1+5/2
b= -2+5/2
b=3/2
(B) (i) In ΔBEF and Δ ABD
.......
.......
.......
.......
hence the ΔBEF similar to ΔABD
By BPT
AE:EB=AD:EF(given that AE:EB=1:2)
since AD:EF=1:2
so EF:AD=2:1
(ii) area ΔBEF: area ΔABD
1/2×EF×EB/1/2×AB×AD (so value of AD=AE+EB=1+2=3)
(1/2 get cancelled)
putting the values
we get 4:3
I HOPE THIS WILL HELP YOU MARK ME BRAINLIEST

Answer 2

Aloha!


>>---------------☆-------------------<<

1(b)

here is the answer:

given: EF is parallel to AD, ratio of AE to EB is 1:2 and ABCD is a parallelogram.

to find: EF:AD

solution:

we are going to apply tables theorem to the triangle ABD

as EF is parallel to AD

we get

BE/AE = BF/FD

now doing reciprocal and adding 1 to both sides we get

AE/BE + 1= FD/BF + 1
AE/BE + BE/BE = FD/BF + BF/BF
AB/BE = BD/BF

again doing reciprocal we get:

BE/AB = BF/BD-----------------------☆1☆

now, in △ ABD and △ EBF
We see that,


BE/AB = BF/BD from ☆1☆
also,

angle B is common

so, △ ABD and △ EBF are similar. ( by SAS ) ( You can also use AA criteria where corresponding angles would be equal )^^ it would be more easy.

so, we can say that sides would be in proportion

EB /AB = BF/BD = EF/AD-------------☆2☆

and the ratio that we are given is

AE/EB = 1/2

adding 1 to both sides and we will get:

(AE + EB)/ EB = (1 + 2)/2

AB/EB = 3/2


doing reciprocal,

EB/AB = 2/3

now by ☆2☆ we get,


EB /AB = EF/AD
2/3 = EF/AD


hence found :)


now we have to find the area

we know
the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

area△BEF/area△ABD = EF^2/AD^2
area△BEF/area△ABD = 2(2)/ 3(3)
area△BEF/area△ABD = 4/9


>>---------------☆-------------------<<

2(b)


here is the answer:

in △ BCG

50 + 90 + angle BCG = 180. ( angle sum property )
angle BCG = 180 - 140
angle BCG = 40˚


now,


angle DCB is 90˚

so,

angle BCG + angle ACB = angle DCB
40 + angle ACB = 90
angle ACB = 90 - 40
angle ACB = 50˚


>>-------------☆---------------<<


hope it helps :)