Test: Perimeters, Areas and Volumes - 2 - UCAT MCQ

As AB and CD are two chords that intersect at O, AP x PB = CP x PD
6 x 4 = CP x 3
CP = 8
From center O draw OM ⊥r AB and ON ⊥r CD.
From the center a line ⊥r to a chord bisects the chord.
So, we have AM = MB = 5 cm
MP = 1 cm, ON = 1 cm, CD = 11 cm, CN = 5.5 cm
ON² + CN² = OC²
1² + 5.5² + r²
1 + 30.25 = r²
Area = πr²
π x 31.25
31.25π = 125π/4 sq cms

The question is "What is the area of the circle?"

Hence, the answer is 125π/4 sq cms.

Choice A is the correct answer.

Each cylindrical can has a diameter of 14 cm and while they are kept erect in the box will occupy height of 30 cm
Number of such cans that can be placed in a row = l /Diameter  = 76 / 14 = 5 (Remaining space will be vacant)
Number of such rows that can be placed = Width / Diameter  = 46 / 14  = 3
Thus 5 x 3 = 15 cans can be placed in an erect position.
However, height of box = 45cm and only 30 cm has been utilized so far
Remaining height = 15 cm > 14 cm (Diameter of the can)
So, some cans can be placed horizontally on the base.
Number of cans in horizontal row = Length of box / Height of can = 76/30 = 2
Number of such rows = Width of box / Diameter of can  = 46/14  = 3
∴ 2 x 3 = 6 cans can be placed horizontally
∴ Maximum number of cans = 15 + 6 = 21

Total surface area of the box = 2(4 x 6 + 1 x 6 + 1 * 4)
= 2(24 + 6 + 4)
= 68 cm²
As the problem says, paper can’t be torn/cut a portion of paper will need to be fold, so, the area of paper required would be greater than 68 cm².
Only option b) gives the area greater 68 cm²

The question is " The dimension of the required paper could be"

Hence, the answer is 12 cm by 6 cm

Given,
Surface area of oil = 36 π = πr²
⇒ r = 6 cm
Now,
∆ABC ~ ∆AED

h = 6 cm
∴ Volume of oil in the cone = (1/3)πr²h
= (1/3)π6² x 6
= 72π
⇒ Time taken = 72π/1
= 72π hours.

∠ABC = ∠ACB = ∠BAC = 60°  [∵ ΔABC is an equilateral triangle]

Also, ∠BAC + ∠BDC = 180°

⇒ 60° + ∠BDC = 180°

⇒ ∠BDC = 180° - 60° = 120°

Also, ∠CBD + ∠BDC + ∠BCD = 180°

⇒ 40° + 120° + ∠BCD = 180°

⇒ ∠BCD = 180° - 40° - 120° = 20°

∴ The value of ∠BCD is 20°

Let radius of the cone be r cm

According to the question

(1/3) × π × r² × 12 = (4/3) × π × 12 × 12 × 12

⇒ r² = 12 × 12 × 4

⇒ r = 12 × 2

∴ r = 24 cm

Minor arc AC will create angle CBA

∠APC = 62º = ∠APB

∠BAP = 90° (diameter perpendicular to tangent)

In Δ APB,

∠APB + ∠BAP + ∠PBA = 180° 

⇒ ∠PBA = 180° - (90° + 62°)

⇒ ∠PBA = 28° 

∴ The measure of minor arc AC is 28° 

Volume of carton = volume of cuboid = Length × Breadth × Height 

⇒ volume of carton = 48 × 27 × Height

∵ 1 foot = 12 inches, then 22.5 cubic feet = 22.5 × 12 × 12 ×12

⇒ 22.5 × 12 × 12 × 12 = 48 × 27 × Height     

⇒ 38,880 = 1,296 × Height 

⇒ Height = 30 inches.

∴ The height of each cartoon is 30 inches.    

From the figure, we can see that the diagonal of the square inscribed in innermost circle is the diameter of the innermost circle.
Side = √32
Or Diagonal = √32 * √2 = 8 cm
Since the circles are spaced at 1.25 cm, the distance between innermost and outermost circles = 1.25 * 4 = 5 cm
Therefore the diameter of the outermost circle = 5 cm + diagonal of inscribed square + 5cm = 18 cm
Now this 18 cm will be the diagonal of the square that needs to be inscribed in outermost circle.
Or a √2 = 18
a = 9 √2 cm
Area = 9 √2 x 9 √2 = 162 sq. cm

Let the area of backyard be x² this year and y² last year
∴ X² - Y² = 131
⇒ (X + Y)*(X - Y) = 131
Now, 131 is a prime number (a unique one too. Check out its properties on Google). Also, always identify the prime number given in a question. Might be helpful in cracking the solution.
⇒ (X + Y) x (X - Y) = 131 x 1
⇒ X + Y = 131
X - Y = 1
⇒ 2X = 132 ⇒ X = 66
and Y = 65
∴ Number of tomatoes produced this year = 66² = 4356

The question is "How many tomatoes did Anil produce this year?"

Hence, the answer is 4356