**A solid rectangle block has a square base of side y cm and a height of x cm. The total surface are of rectangle block is 120cm square and the total length of the 12 edges is 54cm. Show that y^2-9y+20=0 and find the possible values of x and y. Please show clear working and answers, thks!**

**Answer:**

here surface area of upper and lower square= y^2

total surface area of two such squares = 2*y^2

remaining sides = 4

area of each such side= xy

total area of each such sides= 4xy

total surface area of solid cube= 2y^2+4xy= 120 cmsqr

there are 8 edges with side = y cm

remaining 4 edges= x cm

so 8y + 4x = 54

= 4y+ 2x= 27

2x= 27-4y

put it in initial equation 2y^2 + 2y(27-4y)=120

= -6y^2 +54y = 120

= y^2- 9y + 20 = 0

for such type of qs