Question

The length and width of a rectangle ABCD are 24 cm and 18 cm respectively. Given that rectangle ABCD is similar to rectangle PQRS. Find (i) the width of rectangle PQRS if its length PQ is 36cm.(ii) the length of rectangle PQRS if its width QR is 36cm.

Answer 1

$\sf\large\underline{Given:}$

$\tt{\implies Dimensions\:_{(rectangle\:ABCD)}=24cm\:and\:18cm}$

$\tt{\implies Dimensions\:_{(rectangle\:PQRS)}=36cm\:and\:xcm}$

$\sf\large\underline{To\:Find:}$

$\tt{\implies Width\:_{(rectangle\:PQRS)}=?}$

$\sf\large\underline{Solution:}$

• At first calculate the area of rectangle ABCD than calculate the length or width of rectangle PQRS]

$\tt{\implies Area\:_{(rectangle\:ABCD)}=l*b}$

$\tt{\implies Area\:_{(rectangle\:ABCD)}=24*18}$

$\tt{\implies Area\:_{(rectangle\:ABCD)}=432cm^2}$

• Now calculate the length or width of rectangle PQRS before calculating it's breadth or length w have to notice in the question question. In the given Question rectangle ABCD is similar to rectangle PQRS

To find the wide of rectangle PQRS if it's length is 36cm

We have,

• In rectangle PQRS (for calculate wide)
• Length=36cm
• Area=432

$\tt{\implies wide\:_{(rectangle\:PQRS)}=\dfrac{Area}{length}}$

$\tt{\implies wide\:_{(rectangle\:PQRS)}=\dfrac{432}{36}}$

$\tt{\implies wide\:_{(rectangle\:PQRS)}=12cm}$

To find the length of rectangle PQRS if it's wide is 36cm

We have,

• In rectangle PQRS (for calculate wide)
• wide=36cm
• Area=432

$\tt{\implies Length\:_{(rectangle\:PQRS)}=\dfrac{Area}{wide}}$

$\tt{\implies Length\:_{(rectangle\:PQRS)}=\dfrac{432}{36}}$

$\tt{\implies Length\:_{(rectangle\:PQRS)}=12cm}$

Answer 2

${ \underline{ \red{ \tt{ \huge{ \underline{ QUESTION }}}}}}$




▪ The length and width of rectangle ABCD are 24 cm and 18 cm respectively. Given that rectangle ABCD is similar to rectangle PQRS . Find-
(i) the width of rectangle PQRS , if its length PQ is 36 cm
(ii) the length of rectangle PQRS , if its width QR is 36 cm.





${ \underline{ \red{ \tt{ \underline{ \huge{SOLUTION }}}}}}$




${ \dagger{ \blue{ \bold{ \: \: GIVEN }}}}$




${ \pink{ \sf{ \underline{rectangle \: ABCD }}}}$




${ \star{ \blue{ \sf{ \: \: length = 24 \: cm}}}} \\ \\ { \star{ \blue{ \sf{ \: \: width = 18 \: cm}}}}$



${ \rm{ \underline{\underline{For \: rectangle}}}}$



${ \boxed { \boxed{ \red{ \sf{Area = length \times width}}}}}$



then,



${ \blue{ \sf{area \: of \: rect. \: ABCD = 24cm \times 18cm}}}$



${ \implies{ \underline{ \blue{ \sf{Area \: of \: rect. \: ABCD = 432 \: {cm}^{2} }}}}}$




since, it's given that the rectangle ABCD is similar to rectangle PQRS....


then,



${ \sf{ \pink{area \: of \: rectangle \: PQRS }}} \\ \\ { \sf{ \pink{ = area \: of \: rectangle \: ABCD }}} \\ \\ { \sf{ \red{ = 432 \: {cm}^{2} }}}$



${ \dagger{ \bold{ \blue{ \: \: \: TO \: FIND }}}}$




⛦ The width of rectangle PQRS, if its length PQ is 36 cm



${ \orange{ \sf{area \: of \: rectangle \: PQRS = 432 \: {cm}^{2} }}} \\ \\ { \sf{ \orange{length = 36 \: cm}}}$



${ \red{ \sf{width = \frac{area}{length}}}} \\ \\ { \sf{ \red{width = \frac{432 \: {cm}^{2} }{36 \: cm}}}} \\ \\ { \implies{ \underline{ \sf{ \boxed{ \red{width = 12 \: cm}}}}}}$



⛦ Length of rectangle PQRS, if its width QR is 36 cm



${ \orange{ \sf{area = 432 \: {cm}^{2} }}} \\ \\ { \orange{ \sf{width = 36 \: cm}}}$



${ \sf{ \red{length = \frac{area}{width}}}} \\ \\ { \sf{ \red{length = \frac{432 \: {cm}^{2} }{36 \: cm}}}} \\ \\ { \implies{ \underline{ \sf{\boxed{ \red{length = 12 \: cm}}}}}}$