Question

solve the linear equation 3x - 2/10- x+3/7 4x - 7/3=x-1​

Answer 1

Answer:

Correct Question -

The circumference of two circle are in the ratio 2 : 3. Find the ratio of their areas.

Given -

Ratio of their circumference = 2:3

To find -

Ratio of their areas.

Formula used -

Circumference of circle

Area of circle.

Solution -

In the question, we are provided, with the ratios of the circumference of 2 circles, and we need to find the ratio of area of those circle, for that first we will use the formula of circumference of a circle, then we will use the formula of area of circles. We will be writing 1 equation in it too.

So -

Let the circumference of 2 circles be c1 and c2

According to question -

c1 : c2

Circumference of circle = 2πr

where -

π = $\tt\dfrac{22}{7}$

r = radius

On substituting the values -

c1 : c2 = 2 : 3

2πr1 : 2πr2 = 2 : 3

$\tt\dfrac{2\pi\:r\:1}{2\pi\:r\:2}$ = $\tt\dfrac{2}{3}$

$\tt\dfrac{r1}{r2}$ = $\tt\dfrac{2}{3}$$\longrightarrow$ [Equation 1]

Now -

Let the areas of both the circles be A1 and A2

Area of circle = πr²

So -

Area of both circles = πr1² : πr2²

On substituting the values -

A1 : A2 = πr1² : πr2²

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(\pi\:r1)}{(\pi\:r2)}^{2}$

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(r1)}{(r2)}^{2}$

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{(2)}{(3)}^{2}$ [From equation 1]

So -

$\tt\dfrac{A1}{A2}$ = $\tt\dfrac{4}{9}$

$\therefore$ The ratio of their areas is 4 : 9

______________________________________________________

Answer 2

height(H)=120cm

\sf\dashrightarrow \blue{diameter of roller= 84cm}⇢diameterofroller=84cm

\sf\therefore \blue{radius= \dfrac{diameter}{2}}∴radius=

2

diameter

\sf\dashrightarrow \blue{ \dfrac{84}{2}}⇢

2

84

\sf\dashrightarrow \blue{\cancel \dfrac{84}{2}}⇢

2

84

\sf\dashrightarrow \blue{radius= 42cm}⇢radius=42cm

\large\underline\mathfrak{\purple{TO\:FIND,}}

TOFIND,

\sf\dashrightarrow \red{AREA\: OF\:PLAYGROUND }⇢AREAOFPLAYGROUND

FORMULA

\rm{\boxed{\sf{ \circ\:\: C.S.A\: OF\: CYLINDER= 2 \pi rh \:\: \circ}}}

∘C.S.AOFCYLINDER=2πrh∘

\large\underline\mathtt{\purple{SOLUTION,}}

SOLUTION,

© ATQ,

\purple{\text{AREA COVERED BY ROLLER IN 1 REVOLUTION = PERIMETER OF ROLLER}}AREA COVERED BY ROLLER IN 1 REVOLUTION = PERIMETER OF ROLLER

\sf\therefore \pink{AREA \:COVERED \:IN\: ONE\: REVOLUTION= 2 \pi r h}∴AREACOVEREDINONEREVOLUTION=2πrh

\sf\implies \red{ 2 \times \dfrac{22}{7} \times 42 \times 120}⟹2×

7

22

×42×120