Test: Perimeters, Areas and Volumes - 1 - UCAT MCQ

r + h = 37 . . .(1) 
Total surface area of cylinder = 2πr(r + h) = 1628 cm² . . . (2) (given)

Substituting equation (1) in (2), we get:

⇒ 2πr × 37 = 1628
⇒ 74 ×22 × r ÷ 7= 1628

⇒ 74 × 22 × r = 11396

⇒ r = 11396 / (74 × 22)

⇒ r = 7

Using equation (1), we get:

h = 37 - 7
⇒ h = 30 cm

Therefore, the volume of the cylinder = πr²h

 ⇒ Volume= π × 7² × 30 

⇒ Volume= 22 × 7 × 30 = 4620 cm³

Hence, the volume of the cylinder is 4620 cm3.

Let the radius of the pond be R metres.

Then, πR² = 616 → R² = 616 × 7 / 22 = 196 → R = 14 m.

Radius of the stage = (14 / 2) m = 7 m.

Area where water is present

= π (14² - 7²) = (22/7 × 21×7) m² 
= 462 m².

Area of the floor of the room

⇒ 5.5 × 3.75 = 20.625 m²

Now, the cost of paving the floor

⇒ 20.625 × 800 = Rs. 16500

∴ The cost of paving the floor is Rs. 16,500.

Let, each side of the field = x

Then, each side with the path = x + 4.5 + 4.5 = x + 9

So, (x + 9)² - x² = 105.75

⇒ x2 + 18x + 81 - x2 = 105.75

⇒ 18x + 81 = 105.75

⇒ 18x = 105.75 - 81 = 24.75

⇒ x = 24.75/18 = 11/8

∴ Each side of the square field = 11/8 m

The perimterer = 4 × (11/8) = 11/2 m

So, the total cost of fencing = (11/2) × 100 = Rs. 550

∴ The cost of fencing of the field is Rs. 550

The volume of cube = The volume of the rectangular container with a length of 16 cm, breadth of 8 cm, and height of the water level rise

Let, the height of the water level will rise = x cm

So, 8³ = 16 × 8 × x

⇒ 512 = 128 × x

⇒ x = 512/128 = 4

∴ The rise of water level (in cm) is 4 cm

⇒ Total water consumption of Kazipet of 1 day = 4000 × 9 = 36000 litres

⇒ Volume of cuboidal tank = 720 m³ = 720 × 1000 litres = 720000 litres

∴ The number of days for which water is available = 720000/36000 = 20 days

Here, let radius of cylinder and radius of cone be r₁ and r₂ respectively

Also, height of the cylinder and cone be h₁ and h2  respectively

As, base of a cylinder = base of a cone

⇒ π r₁² = π r₂²  ⇔ r₁² = r₂²

⇒ r₁ = r₂

Also, it is given height is same. So, h₁ = h_2.

⇒  Volume of the cylinder = π r₁² h₁ 

_{⇒ Volume of the cone = 1/3} π r₂² h₂ 

⇒ Ratio of volume of the cylinder and volume of the cone = π r₁² h₁ : 1/3 π r₁² h₁ = 1 :1/3 = 3 :1.

Hence, the ratio of volume of the cylinder and  volume of the cone = 3 :1.

Radius of circle =  21m

Now, the circumference of the circle = 2πr

2πr = 2 × (22/7) × 21 = 132m    

Here, is given that the wire is bent in the form of a circle to a square .so, circumference of the circle is equal to the perimeter of the square.

Side of square = 4a

4a = 132

a = 132/4

a  = 33

Diagonal of square = a√2 = 33√2   

Hence, the required diagonal of square = 33√2.

Let us assume the radius of the circle be r

⇒ The perimeter of the square = 4 × 22 = 88 cm

⇒ The circumference of the circle = 2 × π ×  r

⇒ 88 = 2 × (22/7) × r

⇒ r = 88 × 7 / 22 × 2

⇒ r = 14 cm

∴ The required result will be 14 cm.

Perimeter = 4 × side

⇒ 148 = 4 × side

⇒ side = 37 cm

In right angled triangle  ΔAOB,

⇒ AB² = AO² + OB²

⇒ (37)² = (12)² + OB²

⇒ 1369 = 144 + OB²

⇒ OB² = (1369 – 144)

⇒ OB² = 1225 cm²

⇒ OB = 35 cm

BD = 2 × OB

⇒ 2 × 35 cm

⇒ 70 cm

Area of Rhombus = (1/2 × 24 × 70) cm²

⇒ 840 cm²

∴ Area of Rhombus is 840 cm²