Mensuration CAT Previous Year Questions with Answer PDF

AB = AC = 8cm
Since BC is the diameter, angle A has to right-angled and angle B = angle C = 45 degrees.
Therefore, the radius of the circle centred at O is 4√ 2.
The radius of the circle centred at Aa is 8cm.
The common area between them = (half of the smaller circle) + (and the minor segment created by the chord BC in the bigger circle)

All the sides of the rhombus are equal.
The area of a rhombus is 12 cm².
Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.
The area of a rhombus is 
d1 x d2 = 24.
The length of the side of a rhombus is given by  This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

Hence, 


or
In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.
The area of a Rhombus is : 
ab = 6.
The length of each side is : 

2a is longer diagonal which is equal to 

We are given that :
Let DP =x
So AB =x
Now DP=CP
So CD = 2x
Now let height of trapezium be h
we can say A(Parallelogram ABPD ) = xh
And A (BPC) = 1/2xh
Now by condition xh - 1/2xh = 10
xh / 2 = 10
so xh =20
Now therefore area of trapezium ABCD = 1/2 (x + 2x)h = 3/2xh = 30

Let A fill the tank at x liters/hour and B drain it at y liters/hour
Now as per Condition 1 :
We get Volume filled till 10pm = 8x - 7y       (1) .
Here A operates for 8 hours and B operates for 7 hours .
As per condition 2
We get Volume filled till 6pm = 4x - 2y        (2)
Here A operates for 4 hours and B operates for 2 hours .
Now equating (1) and (2)
we get 8x - 7y =4x - 2y
so we get 4x = 5y
y = 4x/5
So volume of tank = 
So time taken by A alone to fill the tank 
= 144 minutes

Let us draw the rectangle.

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.
In this question, we are going to assume that the L is greater than B.
Hence, the one side painted at Rs 200/ft should be B to minimise costs.
Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L
Now, L x B = 60000
B = 60000/L
Hence, total cost = 300B + 200L = 18000000/L + 200L
To minimise this cost, we can use AM > = GM,

Hence, minimum cost = Rs 120000.

We need to find out p.
Angle ADE = 180 - 2x
Angle BDF = 180 - 2y
Now, 180 - 2y + p + 180 - 2x = 180 [Straight line = 180 deg]
p = 2x + 2y - 180
Also, x + y + 50 = 180 [Sum of the angles of triangle = 180]
x + y = 130
p = 260 - 180 = 80 degrees.

Applying cosine rule in triangle ACD,
100 + X² − 2 × 10 × Xcos30 = 400

Solving we get X = 
Hence, area = 10Xsin 30 = 
= 

Let the length and the breadth of the rectangle be l and b respectively.
As the circle touches the two opposite sides, its diameter will be same as the breadth of the rectangle. Given, lb = 135 and 
⇒ 
From this 
∴ Required perimeter:

Given the circle is inscribed in the rhombus of diagonals 12 and 16 . Let O be the point of intersection of the diagonals of the rhombus. Then, OE (radius) ⊥ DC.
Also 
As area of ΔODC should be the same, we have, 1/2 x 6 x 8 = 6π/25

As the cone is cut at one-third height from the top (the vertex), the total volume is proportional to the cubes of the heights of the two parts.
Ratio of volumes two parts =   = 1 : 27
Hence the bottom part will have volume of 27 - 1 i.e., 26 parts.
Given ( 26 - 1) i.e., 25 parts -225 cc.
Hence the required answer is 27 parts = 27 x 255 / 25 = 243 cc.

PD + PE + PF = s
Area of = 
As AB = BC = CA, we've
= 
Now 
⇒ 
Required Value = 

Let the breadth of the rectangle be b.
Length of the rectangle = 3b
Let a be the side of the equilateral triangle.
Given,

Given,
2(4b) + 3a = 90
⇒ 8 (a² / 4) + 3a - 90 = 0
⇒ 2a² + 3a - 90 = 0
⇒ (a - 6)(2a + 15) = 0
⇒ a = 6
∴ b = 9 ⇒ 3b = 27

Time taken by each of them to complete one round
= Circumference of the circle / speed
Time taken for Ram to cover one round = 
Time taken for Rahim to cover one round = 
Time taken by Ram and Rahim meet each other for the first time = LCM of 48π and 28.8π = 144π
∴ Number of rounds made by Ram before he meets Rahim for the first time = 144π / 48π
= 3

OT = 2.4M
In ΔOTQ
OQ² = OT² + TQ² ⇒ TQ² = OQ² - OT²
= (5)² - (2.4)²
= 19.24
⇒ 
∴ PQ = 2TQ = 8.8m

Time taken by Anil and Sunil meet at the starting point 
= 3/5
Distance run by Ravi in 3 / 5 hours = (3 / 5) x  8 = 24 / 5 = 4.8 km

From the given data, we know
l² + b² = 9k ...(1)
b² + h² = 12k ...(2)
l² + h² = 15k ...(3)
Adding (1) , (2) and (3) we get
2(l² + b² + h²) = 36 k
l² + b² + h² = 18k ...(4)
Subtracting (1), (2) , (3) from (4) we get
h² = 9k
l² = 6k
b² = 3k
Thus, ratio of shortest edge to longest edge = √3 : 3 = 1 : √3.

It is given that the diagram is a square pyramid and all 4 sides of the triangle are of equal sides.
Now, we need to plug in a right triangle and apply Pythagoras theorem to find the vertical height of the triangle.
Draw a vertical from the apex of the pyramid and draw a line to the base along the equilateral triangle. Connect the points in the base to form a right triangle.
The hypotenuse would be the slant height of the pyramid, which is the altitude of the equilateral triangle.
Length of the base = 10 cms
Hypotenuse = Altitude of the equilateral triangle = x 20 = 10cms
Therefore, (Vertical height)² = (10√3)² - 10² = 200
Vertical height = √200 = 10√2 cms

Calculation:

A man makes cylinder having equal volume by complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper.

Thus the number of cylinder is to be kept at a minimum,

Volume of each cylinder = HCF of 405, 783, 351 = 27

Then, the volume of each cylinder = 27

⇒ πr²h = 27

⇒ π × 3² × h = 27

⇒ h = 27/9π = 3/π 

The volume of each cylinder = 27

The number of cylinder = (405/27) + (783/27) + (351/27)

⇒ 15 + 29 + 13 = 57

The total surface area of cylinder = 2πrh + 2πr²

The total surface area of all the cylinder = 57 × [(2 × π × 3 × 3/π ) + 2 × π × 3²]

⇒ 57 × [18 + 18π] = 1026(1 + π)

∴The total surface area of all the cylinder is 1026(1 + π).

Given Area (ABCD) = 72 sq cm, CD = 9cms, AD = 16cms
Area of Parallelogram = Base × Height
CD × 16 = 72
CD = (72/16) cms
We extend CD till P such that ∠APD = 90°
So, again by applying the area formula considering CD as base
CD × AP = Area of Parallelogram ABCD
9 × AP = 72
AP = 8 cms
We have AP = 8 cms, AD = 16 cms We observe that the ratio of the sides form a 1 : √3 : 2 triangle Therefore DP = 8√3 cms Area (△APD) = 1/2 × AD × DP = 1/2 × 8 × 8√3 sq cms Area (△APD) = 32 √3 sq cms

Given that ABCD be the rectangle of area of 768 sq cm. A semicircular part with diameter AB and area 72π sq cm is removed
From the area given we can find the radius r as

r² = 144
r = 12 cm

The diameter AB is 24 since r = 12
So we have a rectangle ABCD
AB × BC = 768
24 × BC = 768
BC = 768/24
BC = 32
Now we have to find the perimeter of the left over portion (shaded area)
Perimeter = AD + DC + BC + 1/2 perimeter of the circle
Perimeter = 32 + 24 + 32 + 2πr/2 (r = 12 cm)
Perimeter = 88 + 12π cm
The perimeter of the leftover portion, in cm, is 88 + 12π

Given that a parallelogram ABCD has an area 48 sq cm. The length of CD is 8 cm and that of AD is s cm
Since the area given here is 48 sq.cm the height can be found as 6 sq.cm
We have to find which of the following options is necessarily true

Since this (shaded portion) is a right angled triangle with 6 as its height and s as its hypotenuse so we can assure that s ≥ 6 and s cannot be less than 6 so we can eliminate the other options.
By this we can come to the conclusion that s ≥ 6

Given that a solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. We have to find the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube.
Let us take the volume of the 5 cubes to be 1x³ , 1x³ , 8x³ , 27x³ , 27x³
Volume of the original cube = x³(1 + 1 + 8 + 27 + 27)
Volume of the original cube = 64x3
Sides of the original cube = ∛64x³ = 4x
Similarly, sides of the 5 smaller cubes = x , x , 2x , 3x , 3x .
Surface Area of a cube = 6a²
Surface Area of the original cube = (4x) × (4x) = 16x²
Surface area of the smaller cubes = x² , x² , 4x² , 9x² , 9x²
Sum of the surface areas of the smaller cubes = x²(1 + 1 + 4 + 9 + 9) = 24x²
Change in surface area = 24x² – 16x² = 8x²
%change = 
Hence the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to 50%.