Areas and Volumes



Question: The base of a triangle is 3 more than twice the  height. If the area of triangle is 10 square centimeters, find the dimension of the triangle?

 Solution: 

 Let x be the height of triangle.
then the base of triangle is 2x+3.
Given Area of triangle =10 sq cm
Therefore Area of triangle

A=1/2 base*height
10=1/2(2x+3)*x
20=(2x+3)*x
20=2x^2+3x
2x^2+3x-20=0
2x^2+8x-5x-20=0
2x(x+4)-5(x+4)=0
(2x-5)(x+4)=0
Ignore x+4=0
x=-4
Consider 2x-5=0
x=5/2  cm height of triangle
base= 2x+3=2*5/2+3=8 cm base of the triangle.

Verification: A=(1/2)*(5/2)*8=10

------------------------------------------------------------------------------------------------------------


Question: A rectangular box with height 3 cm has a volume  of 24 cubic cms. The length of box is 2 more than the width, find the dimensions and volume of the box?

 Solution:

Let x be the width of rectangular box.
then the length of rectangular box is x+2.
Given height =3 cm
and Volume =24 cubic cms
Therefore Volume of Rectangular box

V=Length*Width*height
24=x*(x+2)*3
8=x^2+2x
x^2+2x-8=0
x^2+4x-2x-8=0
x(x+4)-2(x+4)=0
(x-2)(x+4)=0
Ignore x+4=0
x=-4
Consider x-2=0
x=2  cm width of Rectangular box
length= x+2=2+2=4 cm length of the rectangular box.

Verification: V=2*4*3=24.



------------------------------------------------------------------------------------------------------------


Question: A rectangular garden  is 10 m by 20 m. the owner, who has enough pine bark  to cover 136 sq m, wants a pine bark path  around the outside  of the garden. Find the width  of the path and the outside dimensions.?

 Solution: 

Let x =width of the path
2x+10=total outside width
2x+20=total outside length
10*20=200 m^2= Inner area
136 m^2= Border Area (pine bark)

Equation:

Total Area= Inner area + Border Area
(2x+10)(2x+20)=200+136
4x^2+40x+20x+200=200+136
4x^2+60x=136
x^2+15x=34
x^2+15x-34=0
x^2+17x-2x-34=0
x(x+17)-2(x+17)=0
(x-2)(x+17)=0
Ignore (x+17)=0
x=-17
consider x-2=0
x=2 m width of the path
2x+10=2(2)+10=14 m total outside width
2x+20=2(2)+20=24 m Total outside length
------------------------------------------------------------------------------------------------------------


Question: The volume of cone is 36π cubic centimeters. the radius of the base is twice the height. Find the dimensions of the cone?

 Solution:
 Let x be the height of the cone
radius r= 2x
Given volume of cone is 36π

Therefore Volume of cone V= 1/3*π*r^2*h
36π=1/3*π*(2x)^2*x
9=1/3*x^3
x^3=27
x=27 cm height of the cone.
radius r=2*27=54 cm radius of the cone.

------------------------------------------------------------------------------------------------------------


 Question: The base of the triangle is 3 less than the twice of the height. If the area of the triangle is 10 sqcm, find the dimensions of the triangle/

 Solution: 
Let x be the height of the triangle
base of the triangle is 2x-3
Given area of the triangle is 10 sqcm
we know the Area of triangle

A= 1/2*base*height
10=1/2*(2x-3)*x
20=2x^2-3x
2x^2-3x-20=0
2x^2-8x+5x-20=0
2x(x-4)+5(x-4)=0
(2x+5)(x-4)=0
x=4 cm height of triangle
2x-3 =5 cm base of triangle.

------------------------------------------------------------------------------------------------------------


Question: The height of parallelogram is 3 less than the base. if the area of the parallelogram is 40 sq m, find the dimensions of the parallelogram?

 Solution:

Let x be the base of parallelogram
then x-3 is the height of the parallelogram.
given Area of parallelogram is 40 sqm

Area of parallelogram
A=base*height
40=x*(x-3)
40=x^2-3x
x^2-3x-40=0
x^2-8x+5x-40=0
x(x-8)+5(x-8)=0
(x+5)(x-8)=0
x-8=0
x=8 m base of parallelogram
x-3=8-3=5 m height of parallelogram.

------------------------------------------------------------------------------------------------------------

Question: A rectangular box with height 2 cm has a volume of 50 cubic cm. The length of the box is 5 less than twice the width . Find the dimensions of the box?

 Solution:

 Let x be the width of rectangular box
then 2x-5 is the length of the rectangular box
given Volume of the rectangular box=50 cubic cm
height = 2 cm
Volume of rectangular box
V= length*width* height
50=(2x-5)*x*2
25=2x^2-5x
2x^2-5x-25=0
2x^2-10x+5x-25=0
2x(x-5)+5(x-5)=0
(2x+5)(x-5)=0
x-5=0
x=5 cm width of the rectangular box
2x-5=10-5=5 cm length of the rectangular box


------------------------------------------------------------------------------------------------------------

Question: A rectangular box with height 4 meters  has a volume of 576 cubic meters and a square base. Find the dimensions of the box.

 Solution: 

Let x be the length of the box.
given that box has got a square base it means length=base
base=x
given volume as 576 cubic meters and height as 4 meters.

Therefore volume of rectangular box

V=length*base*height
576=x*x*4
x^2=144
x=12 m length of the box
and 12 m base of the box.

------------------------------------------------------------------------------------------------------------


Question: A rectangular garden  is 10 m by 20 m. the owner, who has enough pine bark  to cover 216 sq m, wants a pine bark path  around the outside  of the garden. Find the width  of the path and the outside dimensions.?

Solution:



Let x =width of the path
2x+10=total outside width
2x+20=total outside length
10*20=200 m^2= Inner area
216 m^2= Border Area (pine bark)

Equation:

Total Area= Inner area + Border Area
(2x+10)(2x+20)=200+216
4x^2+40x+20x+200=200+216
4x^2+60x=216
x^2+15x=54
x^2+15x-54=0
x^2+18x-3x-54=0
x(x+18)-3(x+18)=0
(x-3)(x+18)=0
Ignore (x+18)=0
x=-18
consider x-3=0
x=3 m width of the path
2x+10=2(3)+10=16 m total outside width
2x+20=2(3)+20=26 m Total outside length


------------------------------------------------------------------------------------------------------------

Question: The combined area of an 8" by 10" picture and its frame (border around the outside) is 288 sq inches . find the width and dimension of the picture frame?

 Solution:










------------------------------------------------------------------------------------------------------------

Question: The  8" by 10" picture is surrounded by a matte (border) whose area is 40 sq inches. (That is , the area of the matte is 40 sq inches).Find the width of the matte and the outside  dimensions of the matte?

 Solution:

 Let x be the width of the matte.



2x+8=total outside width of the frame
2x+10=total outside length of the frame.
Inner area=8*10=80 sq inches
Border Area=40 sq inches.

Total Area=Inner Area+ Border Area\
(2x+8)*(2x+10)=80+40
4x^2+16x+20x+80=80+40
4x^2+36x-40=0
x^2+9x-10=0
x^2+10x-x-10=0
x(x+10)-1(x+10)=0
(x-1)(x+10)=0
Therefore x=1 inch width of the frame.
The dimensions
2x+8=10 inches

.------------------------------------------------------------------------------------------------------------


Question: A Cone has a base radius that is twice the height. If the volume of the cone is 36π cubic meters, find the dimensions of the cone.

 Solution: 

Let x be the height of the cone.
then radius r=2x
Given volume of the cone = 36π

The volume of cone V=1/3πr^2 h
36π=1/3π(2x)^2*x
36π=1/3*π*4x^2*x
x^3=27
x=3 meters of height
2*x=6 meters of radius.

------------------------------------------------------------------------------------------------------------

Question: A cylinder has a diameter that is three times the height.If the volume of the cylinder is 144π cubic inches, find the dimensions of the cylinder.

 Solution:

 Let x be the height of the cylinder.
Given that diameter is 3*x.
We know that radius =diameter/2
r=3/2*x
Given volume =144π cubic inches

Volume of the cylinder V=πr^2h
144π=π(3/2x)^2*x
144=9/4*x^3
64=x^3
x^3=64
x=4 inches height of the cylinder.
r=3/2*4=6 inches height of the cylinder.

------------------------------------------------------------------------------------------------------------


Question:  A cone has a base radius that is three times the height. If the volume of the cone is 375π cubic centimeters, find the dimension of the cone.

 Solution: 

Let x be the height of the cone.
then radius r=3*x
Given volume of the cone V=375π
The volume of the cone V=1/3πr^2h
375π=1/3*π*(3x)^2*x
375π=1/3*π*9x^3
375=3x^3
x^3=375/3
x^3=125*3/3
x^3=125
x=5 centimeters of height
r=3*x=15 centimeters of the radius.

------------------------------------------------------------------------------------------------------------

Question: The height of the cone is twice the radius of the base. If the volume of the cone is 144π cubic feet, find the dimension of the cone.

 Solution: 

Let x be the radius of the cone.
then height h=2*x
Given volume V=144π
Then Volume of the cone V=1/3πr^2h
144π=1/3πx^2*2x
144π=2/3π*x^3
x^3=216
x=6 feet of radius
h=2*x=12 feet of height.

------------------------------------------------------------------------------------------------------------

 Question: Find the radius of the sphere whose volume is 36π cubic centimeters?

 Solution: 

  Let x be the radius of the sphere.
then let us assume height of the sphere is equal to the radius say =x

Given Volume of the sphere is V= 36π

Then the volume of the sphere V=4/3πr^2*h

36π=4/3π*x^2*x
x^3=27
x=3 centimeters of the radius.

------------------------------------------------------------------------------------------------------------

Question: Find the radius of the sphere whose volume is 4.5π cubic centimeters?

 Solution: 

 Let x be the radius of the sphere.
then let us assume height of the sphere is equal to the radius say h =x

Given Volume of the sphere is V= 4.5π

Then the volume of the sphere V=4/3πr^2*h

Let x be the radius of the sphere.
then let us assume height of the sphere is equal to the radius say =x

Given Volume of the sphere is V= 36π

Then the volume of the sphere V=4/3πr^2*h

4.5π=4/3π*x^2*x
4.5=4/3*x^3
multiply by 10 on both sides
45=40/3*x^3
9=8/3*x^3
x^3=27/8
x=3/2 or 1.5 centimeters of radius 

------------------------------------------------------------------------------------------------------------
Question: A man want to build a rectangular speaker box of volume 3 cubic feet  for his 15 inch (diameter) of speakers. if the square base is 18 inches on each side , how deep are square boxes?

Solution:

Comments

Post a Comment

Popular posts from this blog

Reasoning-Number Series

Complete the following series 225,224,____,222,221 a)220         b)223         c)219           d)226 Answer: b)      2. 17,19,23,29,31,37,______     a)41               b)43         c)40           d)42    Answer: If you look at the above series they are all prime numbers, next series will be 41        3.225,196,169,____,121,100,81     a)156              b)144         c)136           d)125 Answer: above series can be written as 15^2,14^2,13^2,____,11^2,10^2,9^2. The missing series is 12^2=144.     4.12,14,15,16,18,20     a)22              b)25         c)21           d)23 Answer: consider above series into two alternate parts. Like 12,15,18,____ & 14,16,20. Then the difference of two alternate series as follows: 3,3,3___ & 2,4. Next series after 18 need to be added with 3= 18+3=21.     5.64,125,216,343,__     a)64              b)424         c)317           d)512 Answer: The above series can be written, 4^3,5^3,6^3,7^3, then the next
Read more

Reasoning-Letter Series

Question & Answers 46. abc-a--d-bcdabcd (1)d b b a        (2)d c a a        (3)a b c d     (4)d b c a 47.b-da-cda-c-abc (1)a a b b        (2)d b c a        (3)a d b c     (4)c b b d 48. a a a b -b c - - a a - (1)a b b c        (2)a c b c        (3)b a c a     (4)b c c a 49. -b a b - b c b c - c a - a b (1)a b c c         (2)a b c b        (3)a c a c     (4)a a c b 50.a b c - - c b a a - c d d c - a (1)d b d b        (2)d d b b         (3)b c a d     (4)a d b c 51.E, A _, O, I (1)E        (2)U        (3)X     (4)V 52.D , F ,G, H , J ,K , L, M , N __ (1)O        (2)Q        (3)P     (4)R 53.E, J, O , T, Y, ____ (1)D        (2)Z        (3)C     (4)E 54.G, J, N , Q, U, ____ (1)W        (2)X        (3)Y     (4)Z 55.R, K, F, C, ____ (1)Z        (2)Y        (3)W     (4)A 56. BY,CX,EV, GT, KP,____ (1)JQ        (2)MN        (3)PK     (4)LO 57.AI,EO,IU,__,UE (1)OU        (2)OI        (3)UA     (4)OA 58.AP, BQ, CR, DS, ____ (1)ET        (2)EU        (3
Read more

Multiply Negative numbers in java

For this challenge, you will be given an array and you are asked to count how many numbers are negative and multiply them and print the output to the STDOUT.  Input Format  -  you will be taking a number as an input from STDIN which tells about the length of the array. On another line, array elements should be there with single space between them.  Output Format  -  print the multiplication to the STDOUT.  Sample Test Case:   Sample Input: - 7 -8 -7 6 0 7 -1 6 Sample Output  -  -56  Explanation:   There are three negative numbers present in the array and their multiplication comes out to be -56.
Read more