the point which lies on the straight line y = 2 +2
Answers
Answer:
A cube is cut with 3 straight cuts . Either horizontally or vertically . What is the percent increased in the total surface area of the cube
To Find:-
Find the percent increased in the total surface area of the cube.
Solution:-
First ,
We have to find the total surface area of a cube:-
\sf\dashrightarrow \: Area = 6 \times { Edge }^{ 2 } ⇢Area=6×Edge
2
\sf\dashrightarrow \: Area = 6 \times { 9a }^{ 2 } ⇢Area=6×9a
2
\sf\dashrightarrow \: Area = 54 { a }^{ 2 } ⇢Area=54a
2
Now ,
We have to find T.S.A of three cuboid:-
\sf\dashrightarrow \: T.S.A = 3 \times 2( 3a \times a + a \times 3a + 3a \times 3a ) ⇢T.S.A=3×2(3a×a+a×3a+3a×3a)
\sf\dashrightarrow \: T.S.A = 6( 15 { a }^{ 2 } ) ⇢T.S.A=6(15a
2
)
\sf\dashrightarrow \: T.S.A = 90 { a }^{ 2 } ⇢T.S.A=90a
2
Now ,
We have to find the percent increased in the total surface area of the cube:-
\sf\dashrightarrow \: Percent = \dfrac { Increased \: in \: area \times 100 } { Initial \: percent } ⇢Percent=
Initialpercent
Increasedinarea×100
\sf\dashrightarrow \: Percent = \dfrac { 3600 { a }^{ 2 } } { 54 { a }^{ 2 } } ⇢Percent=
54a
2
3600a
2
\sf\dashrightarrow \: Percent = 66.6 ⇢Percent=66.6
Hence ,
Percentage increased is 66.6 %
Given line is 3
x+y=1
Slope of this line is m
2 =− 3
Let the slope of line L be m 1 .
Now, angle between the lines is 60
0
tan60
0
=∣
1−
3
m
1
m
1
+
3
∣
⇒±
3
=
1−
3
m
1
m
1
+
3
⇒m
1
=0,m
1
=
3
Equation of line L passing through (3,−2) is
y+2=
3
(x−3)
⇒y−
3
x+2+3
3
=0[/tex]