Question

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Answer 1

Answer࿐

Given,

3x+2y = -4 ----------(1)

2x+5y = 1 -----------(2)

¶ Find Point of Intersection of these 2 lines

Do 2×(1) - 3×(2)

6x+4y = -8 ----------(3)

6x+15y = 3 ---------(4)

-----------------------

-11y = -11

=> y = 1

Substitute in (1)

3x+2(1) = -4

=> 3x = -4-2

=> 3x = -6

=> x = -2

•°• Point of Intersection of the lines 3x+2y+4= 0 & 2x+5y-1= 0 is (-2,1)

¶ By using the Slope-Intercept form, find the form of equation of line passing through the point (-2,1)

y = mx + c

substitute x = -2 & y = 1

=> 1 = -2m + c

=> c = 1 + 2m

•°• The Required Equations of straight line is of form :

y = mx + 1 + 2m -----------(5)

¶ The perpendicular distance (or simply distance) 'd' of a point P(x1,y1) from Ax+By+C = 0 is given by

Given,

(x1,y1) = (-2,1) & d = 2

=> (4m+2)² = 2(m² + 1)

=> 16m² + 16m + 4 = 2m² + 2

=> 14m² + 16m + 2 = 0

=> 7m² + 8m + 1 = 0

Factorise the equation

=> 7m² + 7m + m + 1 = 0

=> 7m(m+1) + 1(m+1) = 0

=> (m+1)(7m+1) = 0

=> m = -1 and m = -1/7

Now Substitute m = -1 in (5)

=> y = (-1)x+1+2(-1)

=> y = -x + 1 - 2

=> y = -1 - x

(or)

=> -x - y -1 = 0

=> x + y + 1 = 0 ------------(6)

Substitute m = -1/7 in (5)

=> y = (-1/7)x + 1 + 2(-1/7)

=> y = -x/7 + (7-2)/7

=> y = (-x+5)/7

=> 7y = -x+5

or

=> -x - 7y + 5 = 0

=> x + 7y - 5 = 0 ----------(7)

•°• The Required equation of straight lines are :

x + y + 1 = 0 & x + 7y - 5 = 0

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Answer 2

${\orange{\bigstar}} \ {\underline{\green{\textsf{\textbf{Given :-}}}}}$

Diameter of the circle = 42 m

Cost of cleaning per m² = ₹2.35

${\blue{\bigstar}} \ {\underline{\pink{\textsf{\textbf{To Find :-}}}}}$

Cost of cleaning the whole field

${\red{\bigstar}} \ {\underline{\purple{\textsf{\textbf{Formula Used :-}}}}}$

${\boxed{\green{\textsf{\textbf{Area of a circle = }}} {\blue{\sf{\pi r^2}}}}}$

where,

r = Radius

${\sf{\pi = \dfrac{22}{7}}}$

${\orange{\bigstar}} \ {\underline{\blue{\textsf{\textbf{Solution :-}}}}}$

Radius = ${\sf{\dfrac{Diameter}{2}}}$

$\longmapsto {\sf{\dfrac{42}{2}}}$

$\longmapsto {\sf{21 \ m}}$

${\pink{\textsf{\textbf{Radius = 21 m}}}}$

According to the question by using the formula of Area of a Circle, we get,

$\dashrightarrow \ {\green{\sf{Area \ of \ circular \ field = (\pi \times 21^2) \ m^2}}}$

⇢ Solving the above equation,

: $\ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 21^2 \bigg ) \ m^2}}$

: $\ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 21 \times 21 \bigg ) \ m^2}}$

:$\ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 441 \bigg ) \ m^2}}$ ⟹ (

: $\ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{{\cancel{7}}^{ \ 1}} \times {\cancel{441}}^{ \ 63} \bigg ) \ m^2}}$

: $\ \Longrightarrow \ {\sf{(22 \times 63) \ m^2}}$

: $\ \Longrightarrow \ {\sf{1,386 \ m^2}}$

${\blue{\textsf{\textbf{Area of the circular field = 1,386 sq. m.}}}}$

Cost of cleaning the field = ₹ (1,386 × 2.35)

:$\ \Longrightarrow \ {\sf{\purple{Rs. \ 3257.1}}}$

$\fbox \orange{ Cost of cleaning the field is Rs. 3257.1 }$