How do you find the perimeter and area of a rectangle with width of #2sqrt7-2sqrt5# and length of #3sqrt7+3sqrt5#?

4 Answers
May 18, 2017

Answer:

Perimeter #P = 10 sqrt(7) + 2 sqrt(5)#

Area #A = 12#

Explanation:

The perimeter #P# of a rectangle is the sum of all side lengths.

The formula is #P = 2(l + w)#, where #l# is the length and #w# is the width:

#Rightarrow P = 2((3 sqrt(7) + 3 sqrt(5)) + (2 sqrt(7) - 2 sqrt(5)))#

#Rightarrow P = 2(3 sqrt(7) + 2 sqrt(7) + 3 sqrt(5) - 2 sqrt(5))#

#Rightarrow P = 2(5 sqrt(7) + sqrt(5))#

#therefore P = 10 sqrt(7) + 2 sqrt(5)#

The area #A# of a rectangle is the product of the length and the width.

The formula is #A = l w#:

#Rightarrow A = (3 sqrt(7) + 3 sqrt(5)) times (2 sqrt(7) - 2 sqrt(5))#

#Rightarrow A = 3 sqrt (7) times 2 sqrt(7) + 3 sqrt(7) times (- 2 sqrt(5)) + 3 sqrt(5) times 2 sqrt(7) + 3 sqrt(5) times (- 2 sqrt(5))#

#Rightarrow A = 6 times 7 - 6 sqrt(35) + 6 sqrt(35) - 6 times 5#

#Rightarrow A = 42 - 30#

#therefore A = 12#

Therefore, the perimeter is #10 sqrt(7) + 2 sqrt(5)# and the area is #12#.

May 18, 2017

Answer:

Perimeter of rectangle is #2(5sqrt 7+sqrt5)# unit.

Area of rectangle is #12 # sq.unit.

Explanation:

Length of rectangle is #l= 3 sqrt7 +3 sqrt 5#
Width of rectangle is #w= 2 sqrt7 -2 sqrt 5#

Perimeter of rectangle is #P=2l+ 2w = 2(3 sqrt7 +3 sqrt 5) +2(2 sqrt7 -2 sqrt 5)#

#P= sqrt7(6+4) + sqrt5 (6-4) = 10sqrt 7 +2sqrt5 =2(5sqrt 7+sqrt5)#unit

Area of rectangle is # A= l*w= (3 sqrt7 +3 sqrt 5)*(2 sqrt7 -2 sqrt 5) #

#A= 6 * 7 - cancel(6 * sqrt 35) + cancel(6 sqrt35) -6*5 = 42-30 =12 # sq.unit.

Perimeter of rectangle is #2(5sqrt 7+sqrt5)# unit
Area of rectangle is #12 # sq.unit. [Ans]

May 18, 2017

Answer:

12

Explanation:

we know area of rectangle = length*width sq unit.

So, #color(green)(Area)# = #[3sqrt7 +3sqrt5]xx[2sqrt7-2sqrt5]#

#rArr 3[sqrt7+sqrt5]xx2[sqrt7-sqrt5]#

#rArr 3xx2xx[(sqrt7)^2-(sqrt5)^2]#

#rArr 3xx2xx[7-5] = 3xx2xx2 =12# sq unit

#color(red)(perimeter)#= 2*[length+width]unit

#rArr 2[{3sqrt7+3sqrt5}+{2sqrt7-2sqrt5}]#

#rArr2[3sqrt7+3sqrt5+2sqrt7-2sqrt5]#

#rArr2[5sqrt7+sqrt5]#

May 18, 2017

Answer:

Area#color(blue)(=12 color(white)(a)#square units,#color(blue)(P=2(5sqrt7+sqrt5)#

Explanation:

#W=2sqrt7-2sqrt5#

#L=3sqrt7+3sqrt5#

perimeter of a rectangle#=P=L+W+L+W#

#P=2L+2W=2(L+W)#

#P=2[(3sqrt7+3sqrt5)+(2sqrt7-2sqrt5)]#

#P=2[3sqrt7+3sqrt5+2sqrt7-2sqrt5]#

#P=2[5sqrt7+sqrt5]#

Area#=A=W xx L#

#A=(2sqrt7-2sqrt5) xx (3sqrt7+3sqrt5)#

#color(white)(aaaaaaaaaaaaa)##2sqrt7-2sqrt5#
#color(white)(aaaaaaaaaa)## xx underline(3sqrt7+3sqrt5)#
#color(white)(aaaaaaaaaaaaa)##6 xx 7-6sqrt5sqrt7#
#color(white)(aaaaaaaaaaaaaaaaaaaa)##6sqrt5sqrt7-6 xx 5#
#color(white)(aaaaaaaaaaaaaa)##overline(42-30color(blue)(=12 #square units