Mensuration - Problems | SSC CGL Tier 2 - Study Material, Online Tests, Previous Year PDF Download

Problem 1

The area of a rectangle is thrice that of a square. The length of the rectangle is 20 cm and the breadth 3/2 times the length of a side of the square. The side length of the square (in cm) is,
a) 10
b) 20
c) 60
d) 30

Ans: a

By given statement area of the rectangle,

A_r=20 × breadth = 20×3/2a, where a is length of a side of the square, =30a
=3A_s, where A_s, is the area of the square.
So,

A_s=10a=a², by definition.
or, a=10.

Problem 2

The perimeter of a rhombus is 40 cm and its height is 5 cm. Its area in (cm2) is,
a) 45
b) 60
c) 55
d) 50

Ans: d

Let us see how a rhombus behaves and what are its basic characteristics. In the figure below ABCD is a rhombus all four sides of which are of equal length a,

As length of all sides of a rhombus is equal in length, its side length = 40/4 = 10 cm.

In a rhombus, its area = base side × height = a × h. If you push the rhombus from right to left holding its corner A and keeping its base CD fixed, it adjusts its shape to the rectangle shape of CDGF without losing or adding any area. The area of the triangles △AGD and △BFC are same.

As area of the rectangle = a × h, we get the area of the rhombus as = base × height.

Thus area of the rhombus in our problem = side length × height = 50 cm.

Problem 3

The sum of the length, breadth and height of a rectangular parallelepiped is 24 cm and the length of the diagonal is 15 cm. Then its total surface area is,
a) 351 cm²
b) 256 cm²
c) 265 cm²
d) 315 cm²

Ans: a

The following is the figure corresponding to the problem.

Let the length, breadth and height be, L, B and H. The base diagonal,

P²=L²+B².
Again, Main diagonal,
D²=P²+H²
=L²+B²+H²=225.

Given L+B+H=24 cm.
Knowing that the total surface area is, 2(LB+BH+HL) using the square of three term sum expression,
(L+B+H)²
=L²+B²+H²+2(LB+BH+HL), we get total surface area as,
2(LB+BH+HL)

=242−225

=576−225

=351 cm².

Problem 4

In the figure given below two shaded regions are formed in a circle of radius a.

The area of the shaded region is,

a)
b) a² (π -1)
c)
d)

Ans: d

The triangle formed by the horizontal diameter is a right triangle (as all diameters subtend a 900 angle at the periphery) with two inclined sides equal and the vertical radius as the perpendicular bisector of the base.
Area of this triangle = a².
Area of the semi-circle = 1/2πa²
So, area of the shaded region

Problem 5

Three circles of radii a, b and c touch each other externally. The area of the triangle formed by the three centres is,
a) ab+bc+ca
b)
c)
d) none of the above

Ans: c

The following figure represents the solution. 

The three sides of the triangle are,
k=a+b
l=b+c, and
m=c+a.
The half perimeter of the triangle is then,

Using the relation of area of triangle with its half perimeter, area

where x, y and z are the three side lengths.

In our case the area of the triangle is then,

Problem 6

The circumference of a circle is 11 cm. The area of a sector of the circle subtending an angle of 600 at the periphery (take π = 22/7) is,

a)
b)
c)
d)

Ans: a
The visualized figure for the problem is as below.

The circumference is,
2πr=11, where r is radius
or,  2×22r=7×11,
or, r = 4/7 cm
Area subtended by 60º sector is one-sixth of the total area (60º is one-sixth of 360º the whole angle covering the circle).
Thus area of the sector

Problem 7

If the difference between the areas of the circumcircle and incircle of an equilateral triangle is 44 cm², then the area of the triangle (in cm², take π=22/7), is,
a) 28 cm²
b) 21 cm²
c) 7√3 cm²
d) 14√3 cm²

Ans: d
The corresponding figure is as below.

The circumcentre and incentre of the equilateral triangle is the same point P and the perpendicular bisector CD and the other two such bisectors from vertices A and B intersect at P which in this case is also the centroid and CD is a median.

Let's assume side length of the triangle as a.
So,

CD  being the median, PD is one-third of CD, that is,

Similarly, CP is two-third of CD, that is,

Thus difference in the areas of the circles,

Or, a²=56.

The area of the equilateral triangle with side length  a is,

Problem 8

If area of an equilateral triangle is A and its height is b, the value of b²/4 is,
a) 1/3
b) 3
c) √3
d) 1/√3

Ans: c
The following figure depicts the problem situation.

If the side length of an equilateral triangle is a, its height is,

And its area is,

So desired expression,

Problem 9

ABCD is a parallelogram. P and Q are the mid-points of sides BC and CD respectively. If the area of the △ABC is 12 cm², the area of △APQ is,
a) 9 cm²
b) 12 cm²
c) 10 cm²
d) 8 cm²

Ans: a
The following figure corresponds to visualization of the problem.
P being midpoint of BC, it divides the △ABC into two equal parts (height being same, base being half of larger base), that is,
Area of △APC = 1/2 of area of △ABC=6 cm²

Similarly, area of △AQC is half of area of △ACD=6 cm².

So area of quadrilateral APCQ=12 cm².

Now we only have to find the area of △PCQ and subtract it from this area of the quadrilateral to get the desired area of △APQ.

The other diagonal BD also divides the area of the parallelogram into two equal parts and so, area of the △BCD=12 cm².

Again BD||PQ and P and Q are the midpoints of the other two sides BC and CD of the △BCD. Thus these two triangles △PCQ and △BCD are similar and each side including the height of the smaller triangle is half its corresponding side and the height of the larger triangle.

This makes the area of the △PCQ = 1/4th of the area of △BCD=3 cm².
Finally then, the area of △APQ=12−3=9 cm².

Problem 10

A wire when bent in the form of a square encloses an area of 484 sq cm. What will be the enclosed area when the same wire is bent into the form of a circle?
a) 616 sq cm
b) 693 sq cm
c) 462 sq cm
d) 539 sq cm

Ans: a

If the side of the square is a cm,

a²=484=22².

So, a=22 cm.

Periphery of the square is then = 4×22=88 cm.

This peripheral length of the wire will then enclose a circular area.

So if the radius of the circular area is r cm,

2πr=88,

Or, r=14 cm.

The area of the circular area is then,