The length of a rectangle is five more than the width. The perimeter is 170. Find the length and width. Write equation and solve.

Answers

Answer 1

Answer: 170=2l+2w
l=w+5
w=l-5
170=2l+2(l-5)
170=2l+2l-10
180=4l
l=45
w=40

Answer 2

Answer:

Final answer:

To find the length and width of a rectangle given the perimeter and the relationship between the length and width, we can set up an equation. By solving the equation, we can determine the dimensions of the rectangle.

Explanation:

Let's assume the width of the rectangle is x units. Given that the length is 5 more than the width, the length would be x + 5 units. The perimeter of a rectangle is given by 2(length + width), so we can write the equation as 2(x + 5 + x) = 170. Simplifying the equation, we get 4x + 10 = 170. Solving for x, we have x = 40. Therefore, the length is x + 5 = 45 units and the width is x = 40 units.

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Answers

the complete answers in the attached figure

Part 1) we have

$r=4cm\n R=8 cm\n L=6cm$

Find the height h

$h^(2)=L^(2) -(R -r)^(2)\n h^(2)=6^(2) -(8-4)^(2)\n h^(2)=36-16\n h=√(20) cm$

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[8^(2) +4^(2) +8*4]√(20)\n \n V=(1)/(3)\pi[112]√(20)\n \n V=524.52 cm^(3)$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(8+4)*6\n LA=226.19 cm^(2)$

the answer Part 1) is

a) the volume is equal to $524.52 cm^(3)$

b) The Lateral area is equal to $226.19 cm^(2)$

Part 2) we have

$r=4ft\n R=5 ft\n h=100 ft$

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=100^(2) +(5-4)^(2)\n L^(2)=10000+1\n L=√(10001) ft$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(5+4)*√(10001)\n LA=2,827.57 ft^(2)$

the answer part 2) is

a) The Lateral area is equal to $2,827.57 ft^(2)$

Part 3) we have

$V=52\pi ft^(3) \n h=3ft\n R=3r$

Step 1

Find the values of R and r

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h$

substitute $R=3r$ in the formula above

$V=(1)/(3)\pi[(3r)^(2) +r^(2) +(3r)*r]*3$

$V=(1)/(3)\pi[7r)^(2)]*3$

$V=[tex] 52\pi$

$52\pi =\pi [7r^(2) ]\n r^(2) =(52)/(7) \n \n r=2.73 ft$

$R=3*2.73\n R=8.19 ft$

Step 2

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=3^(2) +(8.19-2.73)^(2)\n L^(2)=38.81\n L=6.23 ft$

Step 3

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(8.19+2.73)*6.23 LA=213.73 ft^(2)$

the answer Part 3) is

a) The lateral area is equal to $213.73 ft^(2)$

Part 4) we have

$r=15 in\n R=33 in\n h=24 in$

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=24^(2) +(33-15)^(2)\n L^(2)=576+324\n L=30 in$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(33+15)*30\n LA=4,523.89 in^(2)$

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[33^(2) +15^(2) +33*15]24\n \n V=(1)/(3)\pi[112]24\n \n V=142.83 in^(3)$

the answer is

a) The lateral area is equal to $4,523.89 in^(2)$

b) the volume is equal to $142.83 in^(3)$

Part 5) we have

$r=5 cm\n h=8√3 cm$

Step 1

Find the value of (R-r)

$tan 60=√(3)$

$tan 60=((R-r))/(8√(3)) \n\n R-r= √(3) *8√(3) \n R-r=24 cm\n R=24+r\n R=24+5\n R=29 cm$

Step 2

Find the value of slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=(8√(3))^(2)+(24-5)^(2)\n L^(2)=192+361\n L=23.52 cm$

Step 3

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(24+5)*23.52\n LA=2,142.82 cm^(2)$

Step 4

Find the total area

total area=lateral area+area of the top+area of the bottom

Area of the top

$r=5 cm\n A=\pi *r^(2) \n A=\pi *25\n A=78.54 cm^(2)$

Area of the bottom

$r=24 cm\n A=\pi *r^(2) \n A=\pi *576\n A=1,809.56 cm^(2)$

Total surface area

$SA=2,142.82+78.54+1,809.56\n SA=4,030.92 cm^(2)$

the answer is

a) The total surface area is $4,030.92 cm^(2)$

Part 6)

Part a) Find the volume of the water tank

we have

$r=4 ft\n R=6 ft\n h=8 ft$

Step 1

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[6^(2) +4^(2) +6*4]8\n \n V=(1)/(3)\pi[76]8\n \n V=636.70 ft^(3)$

the answer Part a) is $636.70 ft^(3)$

Part b) Find the volume of the wetted part of the tank if the depth of the water is 5 ft

by proportion find the radius R of the upper side for h=5 ft

$((R1-r))/(8) =((R2-r))/(5) \n\n ((6-4))/(8) =((R2-4))/(5)\n \n(R2-4)= 1.25\n R2=4+1.25\n R2=5.25 ft$

Find the volume for $R2=5.25 ft$

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[5.25^(2) +4^(2) +5.25*4]5\n \n V=(1)/(3)\pi[64.56]5\n \n V=338.05 ft^(3)$

the answer Part b) is $338.05 ft^(3)$

Part 7) we have

$SA=435\pi cm^(2) \n A1=144\pi cm^(2)\n A2=81\pi cm^(2)$

Step 1

Find the value of R and the value of r

$A1=\pi *R^(2) \n 144\pi =\pi *R^(2)\n R=12 cm$

$A2=\pi *r^(2) \n 81\pi =\pi *r^(2)\n r=9 cm$

Step 2

Find the value of lateral area

$LA=SA-A1-A2\n LA=435\pi -144\pi -81\pi \n LA=210\pi cm^(2)$

Step 3

Find the slant height

$LA=\pi (R+r)L\n\n L=(LA)/(\pi(R+r)) \n \n L=(210\pi)/(\pi(12+9)) \n \n L=10 cm$

Find the altitude of the frustum

$h^(2) =L^(2) -(R-r)^(2) \n h^(2) =10^(2) -(12-9)^(2)\n h^(2)=91\n h=9.54 cm$

the answer Part a) is

the slant height is $10 cm$

the answer Part b) is

the altitude of the frustum is $9.54 cm$

Find the volume and the lateral area of a frustum of a right circular cone whose radii are 4 and 8 cm, and slant height is 6 cm.
$h= √(s^2-(R_1-R_2)^2) \n = √(6^2-(4-8)^2) \n = √(36-16) \n = √(20)$
$Volume= (1)/(3) \pi h(R_1^2+R_1R_2+R_2^2) \n = (1)/(3) \pi * √(20) (4^2+4 * 8+8^2) \n = (1)/(3) \pi √(20) (16+32+64) \n = (1)/(3) \pi √(20) (112) \n =524.5cm^3$
Lateral area = Total surface area - area of base - area of top
$Lateral \ area= \pi (R_1+R_2)s \n = \pi (4+8) * 6 \n =12 \pi * 6 \n =72 \pi \n =226.2cm^2$

The length of a swimming pool is 8m longer than its width and the area is 105m2

Answers

$width : w \n length : \ l = w + 8 \n A = 105 \ m^2 \n \nA= w \cdot l \n \nw(w+l)=105 \n \nw^2+w = 105 \n \nw^2+w-105 =0 \n \n(w+15)(w-7)=0 \n \nw+15 =0 \ \ or \ \ w-7 = 0 \n \nw = -15 \ \ or \ \ w=7 \n \n width \ cant \ be \ negative, \ so \n \n w = 7 \ and \ l = w+8 = 7 +8 =15 \n \nchek : \n \n A=w\cdot l \n \nA=7\cdot 15=105 \ m^2$

$l\ \rightarrow\ the\ length\nw\ \rightarrow\ the\ width \n\nl=w+8\ \ \ and\ \ \ l\cdot w=105\ \ \ and\ \ \ l>0,\ w>0\n\n(w+8)\cdot w=105\ \ \ \Rightarrow\ \ \ w^2+8w=105\n\n w^2+2\cdot4w+4^2=105+4^2\n\n (w+4)^2=105+16\ \ \ \Rightarrow\ \ \ (w+4)^2=121\n\nw+4=11\ \ \ or\ \ \ w+4=-11\n\nw=7\ \ \ \ \ \ \ \ \ \ or\ \ \ w=-15<0\n\nw=7\ \ \ \Rightarrow\ \ \ l=7+8=15\n\nAns.\ the\ length\ is\ 15\ m\ \ and\ \ the\ width\ is\ 7\ m.$