Question

if the area of arectangular feild is (21a square -7x) and one of its side is7x ,what is its other side​

Answer 1

Answer:

${ \huge{ \underline{ \red{ \bf{Given}}}}}$

${ \to { \sf{Area \: of \: rectangle \: = 21 {a}^{2} - 7x}}}$

${ \to{ \sf{one \: side \: of \: rectangle \: field = 7x}}}$

Find :- Other side

Solution:-

${ \to{ \sf{area \: of \: rectangle \: = l \times b}}}$

${ \to{ \sf{21 {a}^{2} - 7x = 7x \times b}}}$

${ \to{ \sf{ \frac{21 {a }^{2} - 7x}{7} = xb }}}$

${ \to{ \sf{ \frac{7(3 {a}^{2} - x) }{7} = xb}}}$

${ \to{ \sf{ \frac{{ \cancel{7}}(3 {a}^{2} - x) }{{ \cancel{7}}} = xb}}}$

${ \to{ \sf{ \frac{3 {a}^{2} - x}{x} = b }}}$

Therefore breadth of rectangle field is 3a²-x/x

Step-by-step explanation:

Verification:-

${ \to{ \sf{area = l \times b}}}$

${ \to{ \sf{21 {a}^{2} - 7x = 7x \: \times \frac{3 {a}^{2} - x}{x} }}}$

${ \to{ \sf{21 {a}^{2} - 7x = 7{ \cancel{x}} \: \times \frac{3 {a}^{2} - x}{{ \cancel{x}}} }}}$

${ \to{ \sf{21 {a}^{2} - 7x = 7 \times( 3 {a}^{2} - x)}}}$

${ \to{ \sf{21 {a}^{2} - 7x = 21 {a}^{2} - 7x }}}$

L. H. S = R. H. S

Hence proved ✔