CAT Previous Year Questions: Mensuration Banking Exams Notes | EduRev

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

The question is "A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is"
Hence, the answer 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%.