Question

please solve this question

Answer 1

19 -yes both the graphs are correct
to find the point where the both line intersect just mix both equation
first equation - y=1
second equation - 2x +3y=6
put value of "y" from equation one to equation two
So
2x +3y=6
2x +3(1)=6
2x +3=6
2x=6-3
2x=3
x=3/2
x=1.5
so the point where lines meet is ( 1.5 , 1 )
area covered
we can see that it is right angle triangle
with height 1.5units and base 1units
so area = 1/2(1.5 x 1)
= 0.75 sq. units

20- equation is 5y + 10x = 50

22 - to solve this question first make all four triangle (AOB , BOC , COD , DOA ) congurent
this can be done all four side of square is equal
and its diagonal are equal and bisect each other (AO=BO=CO=DO)
SO by SSS rule you can proof that all this triangle are congurent
after it
AOB + BOC + COD + DOA = ABCD
all triangles are equal as they are congurent
AOB + AOB + AOB + AOB = ABCD
4AOB=ABCD
AOB = ABCD/4
AOB:ABCD
=ABCD/4:ABCD
ABCD will get cancelled
1/4:1
1:4 is the answer

23-Given: Arc AB. Point C on the circle is outside AB.
To prove: ∠AOB = 2 × ∠ACB
Construction: Draw a line CO extended till point D.
Proof: In ΔOAC in each of these figures,
∠AOD = ∠OAC + ∠OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)
OA = OC (Radii of same circle)
Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal)
∠AOD = ∠OAC + ∠OCA
⇒∠AOD = 2 × ∠OCA
Similarly, in ΔOBC, ∠BOD = 2 × ∠OCB
∠AOD = 2 × ∠OCA and ∠BOD = 2 × ∠OCB
⇒ ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB
∠AOD + ∠BOD = 2 × (∠OCA + ∠OCB)
or ∠AOB = 2 × ∠ACB
Hence, the theorem is proved.
24-
population=4000
water per day per head =150
tank = 20*15*6
total water for one day = 4000*150
600000liter
volume of tank = 20*15*6= 1800 cb m= 1800000liter
total days = 1800000/600000=3days
values shown - helpfulness, unity etc

25
cone hemisphere and cylinder
by volume
1/3 π r^2 h : 2/3 π r^3 : π r^2 h
1/3 π r^2 h : 2/3 π r^2 r: π r^2 h
as in hemisphere hight is equal to radius so
1/3 π r^2 h : 2/3 π r^2 h : π r^2 h
cancel π r^2 h from all
1/3:2/3:1
multiple 3
1:2:3
by curved surface area
π r l : 2 π r^2 : 2 π r h
l =√h^2+r^2
h = r
l =√r^2+r^2 = √2r^2 = r√2
so
π r^2 √2 : 2 π r^2 : 2 π r^2
cancelled π r^2
√2 : 2 :2
2:4:4

26
imagine it with paper
h= 4
b=30-4=26
l=40-4=36
volume = hbl
= 4*26*36
=3744 cubic centimetres
lateral surface area = 2h(l+b)
=2*4(26+36)
8*62
496 sq cm
cost is RS 8/ sq cm
total cost = 496*8
=₹3968