SSC CGL Previous Year Questions: Mensuration - 1 - SSC CGL MCQ

Given, Diagonal of a rhombus is 48 cm

Side of rhombus is 26 cm

The diagonal in a rhombus are perpendicular and bisect of other diagonal.
Given length side 26 cm
One diagonal = 48 cm = DB
Now, OB = OD = 24 cm
Now consider triangle AOB this is a right angled  triangle
and we know AB = 26 cm and OB = 24 cm
Now using the Pythagorus theorem we can find out
length of OA as
OA² = AB² – OB²
         = 676 – 576
OA² = 100
OA = 10 cm
So the length of diagonal AC is 2 × 10 = 20 cm.

Edge of cube = 8 cm
The total surface area of cube is = 6a²
a = 8 cm
S.A. = 6 × 8 × 8 = 64 × 6 = 384 cm²

Hence the radius of circle is = 7 cm
Circumference of the circle is = 2πr

When we join 5 cubes by end to end
Then length = 5 x 3 = 15 cm
breadth = 3 cm
height = 3 cm
total surface Area = 2(15 x 3 + 15 x 3 + 3 x 3) = 198 cm²

Let initially length was l and breadth was b.
Area (A) = l × b.

Total cost of repairing = 1.15R × 1.188 A = 1.3662 R.A
Percentage increase in cost

Volume of cylinder (V) = πr².h, where r = radius h = height

We know that diagonal of rhombus bisect perpendicularly. Let ABCD is rhombus and diagonal bisect at point 'O'.

As, (8)² + (15)² = 64 + 225 = 289 = (17)²
Hence, triangle is right angle triangle.
Sum of angle of any triangle is 180°.
Sum of area of all the three sectors

So, remaining area = 60 – 19.25 = 40.75 cm²

Number of sphere of radius 2 cm

After joining 6 cubes of edge length 4 cm each, Length of resulting cuboid = 6 × 4 = 24 cm

Breadth  = 4 cm Height  = 4 cm
Surface Area of the cuboid
= 2(lb +bh + hl)
= 2(24 × 4 + 4 × 4 + 4 × 24)
= 416 cm²

By Pythagorean theorem,
(20)² = (12)² + (x²)
∴ x² = 400 – 144 = 256
∴ x = √256 = 16 cm
∴ Diagonal of rhombus = 2x = 2 × 16 = 32 cm and other diagonal = 24 cm

Here,
Radius of cylinder = 7 cm Height of cylinder = 20 cm
∵ Total surface area  of cylinder = 2πrh + 2πr²

= 880 + 308 = 1188 cm²
When cylinder is cuting along height, two new cylinders are generated–radius of new cylinder = 7cm height of new cylinder = 10cm
∵ Total surface area of new cylinder = 2πrh + 2πr²

⇒ 440 + 308 = 748 cm²
∴ Total surface area of two new cylinders = 748 × 2 = 1496 cm²
∴ Percentage increase in surface area

Let radius of hemisphere = 2x
Radius of sphere = x
∴ Ratio of total surface area of hemisphere and sphere

Here,
The ratio of curved surface area of two cones = 1 : 4
Let curved surface area of first cone be x and curved surface area of second cone be 4x.
Let slant height of first cone be 2x and slant height of second cone be x.
According to question,

∴ The ratio of the radius of the two cones = 1 : 8

Let side of rhombus be x

By pythagorean theorem,
x² = (15)² + (8)²
x² = 225 + 64 = 289
∴ x = 289= 17 cm
∴ Perimeter of rhombus = 4 × side
= 4 × 17 = 68 cm

Total surface area of sphere = 9π cm²
∵ 4πr² = 9π cm²

Here,
Total surface area of hemisphere = 166.32 sq cm.
r = ?
∵ 3πr² = 166.32

Volume of the cylinder = πr²h
Since, radius is increased by 25%
Then, increased radius 
Let h₁ be the height of cylinder whose radius is increased.
Since, volume remains same

So, height must be reduced by 36%.