## 26th SSC CGL level Solution Set, 1st on topic Mensuration

This is the 26th solution set of 10 practice problem exercise for SSC CGL exam and 1st on topic Mensuration. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set.

We will repeat here the method of taking a 10 problem test if you have not gone through it already. And if you have not taken the test yet, you can refer to * SSC CGL level Question Set 26, Mensuration 1*, and then after taking the test come back to this solution.

### Method for taking this 10 problem test and get the best results from the test set:

**Before start,**go through the tutorials**Basic concepts on Geometry 1**,**, Basic concepts on Geometry 2**or any other short but good material to refresh your concepts if you so require.**Basic concepts on Geometry 3****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 15 minutes.**When the time limit of 15 minutes is over,**mark up to which you have answered,**but go on to complete the set.****At the end,**refer to the answers to the questions to mark your score at 15 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.**Identify and analyze**the problems that**you couldn't do**to learn how to solve those problems.**Identify and analyze**the problems that**you solved incorrectly**. Identify the reasons behind the errors. If it is because of**your shortcoming in topic knowledge**improve it by referring to**only that part of concept**from the best source you get hold of. You might google it. If it is because of**your method of answering,**analyze and improve those aspects specifically.**Identify and analyze**the**problems that posed difficulties for you and delayed you**. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.**Give a gap**before you take a 10 problem practice test again.

Important:bothandpractice testsmust be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.mock tests

**Resources that should be useful for you**

**You may refer to:**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Tutorials that you should refer to**

**If you like, **you may * subscribe* to get latest content from this place.

### 26th solution set- 10 problems for SSC CGL exam: topic Mensuration - Answering time 15 mins

**Problem 1.**

Three circles of radii 5.5 cm, 4.5 cm and 3.5 cm touch each other externally. The perimeter of the triangle formed by joining the centres of the circles (in cm) is,

- $27\pi$
- $27$
- $13.5$
- $\pi(3.5^2 + 4.5^2 + 5.5^2)$

**Solution 1 - Problem analysis and Visualization**

As the circles touch each other externally, the two radii of first circle (say - any first) will be perpendicular to the tangents passing through the touching points with the other two circles. This is by the definition of circles touching. But at the same time, the radii of the two touching circles will be also perpendicular to the tangents at the same points.

When two circles touch each other at one point (otherwise they will intersect at two points, if at all), they have a common tangent passing through the single touching point.

The following is the figure.

#### Solution 1 - Problem solving execution

Thus any two perpendiculars from two centres to the same tangent will actually be one single line forming one side of the triangle that will comprise of one part radius of one circle and second part radius of second circle. The same is true for the other two sides of the triangles.

Side $AB = AK + KB$

$\hspace{17mm}= r_1 + r_2 $

$\hspace{17mm}= 3.5 + 4.5$

$\hspace{17mm}= 8$ cm,

Side $AC = AM + MC $

$\hspace{17mm}=r_1 + r_3$

$\hspace{17mm}=3.5 + 5.5 $

$\hspace{17mm}= 9$ cm, and

Side $CB = CL + LB $

$\hspace{17mm}= r_3 + r_2 $

$\hspace{17mm}= 5.5 + 4.5$

$\hspace{17mm}= 10$ cm, that is,

a total of $10+9+8=27$ cm.

Otherwise, conceptually the perimeter

$ = 2(\text{radius 1 + radius 2 + radius 3}) $

$= 2(3.5 + 4.5 + 5.5) $

$= 27$ cm.

**Answer:** Option b: 27.

**Key concepts used:** Visualization -- Basic Geometry concepts of circles and tangents.

**Problem 2.**

Three circles of diameter 10 cm each are bound together by a rubber band as shown below,

The length of the rubber band (in cm) in stretched condition as shown, will be,

- $30 + \pi$
- $30$
- $10\pi$
- $60 + 20\pi$

**Solution 2 - Problem analysis and Visualization**

Let us reproduce the figure in enriched condition with dashed perpendiculars drawn to the three pairs of tangent points.

We can see that the total length of the rubber band will comprise of three pairs of,

- length of tangent section between a pair of circles colored green, say, PQ
- length of sector of one circle held by the angle between two perpendiculars to the tangents colored red, say, QR.

#### Solution 2 - Reasoning

* Regarding length of tangent section:* As PQ is a tangent to the two circles each of equal diameter 10cm, ABQP form a rectangle and $PQ=AB=10$cm.

Being perpendiculars to the common tangent $PQ$, the sides $AP||BQ$ and also being radii of same length 5 cm, $AP=BQ$. Thus $AB||PQ$ and $AB=PQ$ forming rectangle ABQP.

There are three such tangent sections in the total length, that total up to 30 cm.

* Regarding the arc length of arc QR:* three sides of $\triangle ABC$ being equal, it is an equilateral triangle and $\angle ABC = 60^0$. So,

$\angle QBR = 360^0 - 2\times{90^0} - 60^0 = 120^0$, which is one third of $360^0$ and so the arc length $QR$ is one-third of perimeter of one circle, that is $\frac{1}{3}\times{10\pi}$. Three such arcs total up to $10\pi$.

Thus total length of the rubber band is, $30 + 10\pi$.

**Answer:** Option a : $30 + 10\pi$.

**Key concepts used:** Basic concepts of circles, tangents, triangles, rectangles and arcs.

**Note:** With a good conceptual base, drawing of the second figure is not necessary, as with quick visualization and simple reasoning, the solution can be reached in a short time.

**Problem 3.**

A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in a ratio of 6 : 5. The smaller side of the rectangle is, (take $\pi =\frac{22}{7}$)

- 30 cm
- 60 cm
- 36 cm
- 25 cm

**Solution 3 - Problem analysis and elegant solution**

The perimeter of the circle will form the perimeter of the rectangle which will comprise of twice sum of length and breadth.

The perimeter of the circle, that is, the rectangle is,

$P =2\pi{r} $

$\hspace{5mm}= 2\times{\displaystyle\frac{22}{7}}\times{42}$

$\hspace{5mm}= 12\times{22}$ cm.

Let us assume the actual length and breadth of the rectangle be, $6x$ and $5x$ using ratio concepts.

So perimeter will be,

$P= 2(6x+5x) = 22x = 12\times{22}$.

So $x = 12$.

The smaller side or breadth is then $=5x = 60$ cm.

**Answer:** b: 60cm.

**Key concepts used:** Circle perimeter to rectangle perimeter transformation -- application of ratio concepts effectively.

**Note:** Try solving this problem using a different method and compare the methods.

**Problem 4.**

Two equal maximum sized circular plates are cut-off from a circular paper-sheet of circumference 352 cm. The circumference of each circular plate is,

- 176 cm
- 180 cm
- 165 cm
- 150 cm

**Solution 4 - Problem analysis and elegant solution**

To cut-off two equal and maximum sized circles out of the large circle as shown in the figure, the diameter of the larger circle must be shared equally by the two smaller internal circles, that is, the diameter of each of the smaller plates will be half of the diameter of the larger circle.

As the perimeter of a circle with diameter $d$, is $P= \pi{d}$, the perimeter of each of the smaller circular paper plate with half the diameter of the larger circular paper will be half the perimeter of the larger circular paper, which in this case will be,

$=\displaystyle\frac{352}{2}=176$ cm.

**Answer:** Option a: 176 cm.

**Key concepts used:** Visualization -- use of concept of proportionality between diameter and perimeter of a circle for quick elegant solution.

**Note:** Try solving this problem in a different method and compare methods for increasing clarity on concepts and methods.

**Problem 5.**

If length and perimeter of a rectangle are in the ratio 5 : 16, then its length and breadth will be in the ratio,

- 5 : 11
- 5 : 4
- 5 : 3
- 5 : 8

**Solution 5 - Problem analysis and elegant solution**

Using ratio concepts let the actual length and perimeter be $5x$ and $16x$ respectively.

As the perimeter contains twice the length or $10x$ plus twice the breadth, reducing this amount of length, we have twice the breadth as, $16x -10x = 6x$. So length and breadth ratio is, $10x : 6x = 5 : 3$.

**Answer:** Option c: 5 : 3.

**Key concept used:** Use of canceled common factor introduction technique in ratio quickly gives us the solution as finally also ratio was desired.

**Note:** Try solving this problem also using a different method for comparing methods using the * Many ways technique* for

**problem solving skill improvement.****Problem 6.**

From a point in the interior of an equilateral triangle, the perpendicular distances of the sides are, $\sqrt{3}$ cm, $2\sqrt{3}$ cm and $5\sqrt{3}$ cm. The perimeter (in cm) of the triangle is,

- 48
- 64
- 24
- 32

**Solution 6 - Problem analysis and visualization**

The visualized figure for the problem is as below.

We are given three heights from a single internal point. This urges us to get the three areas and sum up to get the area of the whole triangle. In equilateral triangle, as the side lengths are same, from area we can find out the side length and then the perimeter.

#### Solution 6 - Problem solving execution

Let us assume side length of the equilateral triangle as $a$.

Area of $\triangle ABC$ = Area of $\triangle PBC$ + Area of $\triangle PBA$ + Area of $\triangle PAC$

$\hspace{20mm} = \displaystyle\frac{1}{2}\left(PS\times{BC} + PQ\times{AB} + PR\times{AC}\right)$

$\hspace{20mm}=\displaystyle\frac{a}{2}\left(PS + PQ + PR\right)$

$\hspace{20mm}=\displaystyle\frac{a}{2}\times{8\sqrt{3}}$

$\hspace{20mm}=4a\sqrt{3}$.

But again, area of the equilateral triangle = $\frac{\sqrt{3}}{4}a^2 = 4a\sqrt{3}$,

Or, $a = 16$, and so,

Perimeter = 48.

**Answer:** Option a : 48.

**Key concepts used:** Converting the heights to areas then summing to get the area of the equilateral triangle -- equating this area with standard formula gives us the side length and then the perimeter.

**Problem 7.**

Through each vertex of a triangle, a line parallel to the opposite side is drawn. The ratio of the perimeter of the new triangle thus formed with the original triangle is,

- 3 : 2
- 4 : 1
- 5 : 3
- 2 : 1

** Solution 7 - Problem analysis**

The corresponding figure is as below.

As $AC||BE$ and $AB||CE$, $ABEC$ forms a parallelogram where $AC = BE$ and $AB=CE$.

Similarly, as $AC||DB$ and $BC||AD$, $ADBC$ forms a second parallelogram where $AC = BD$ and $AD=BC$.

The important poin that we can conclude from these two relations is that $B$ is the mid-point of $DE$ and $AC$ is half of DE.

Because of symmetry this will be true for the other two sides $FD$ and $EF$ also.

In short, each of the three sides of the original triangle is half the length of a corresponding side of the new triangle.

Thus ratio of perimeter of new triangle to that of original triangle is 2 : 1.

**Answer: d: **2 : 1.

**Key concepts used:** Concepts of bounding pairs of parallel lines creating a parallelogram -- relating length of sides of two triangles -- context visualization.

**Problem 8.**

A right triangle with side lengths 3 cm, 4 cm and 5 cm is rotated about the side 3 cm as rotation axis, to form a cone. The volume of the cone so formed is,

- $16\pi$ cm$^3$
- $20\pi$ cm$^3$
- $15\pi$ cm$^3$
- $12\pi$ cm$^3$

** Solution 8 - Problem analysis and solving**

The following figure depicts the problem situation.

The height of the right triangle is the side 3 cm while the side 5 cm is the hypotenuse. Naturally the base is the side 4 cm and it forms the radius of the cone base, $r=4$ cm.

With its height, $h=3$ cm the volume of the cone is,

$V =\displaystyle\frac{1}{3}{\pi}r^2h=\displaystyle\frac{1}{3}{\pi}\times{16}\times{3} = 16\pi$ cm$^3$.

**Answer:** Option a: $16\pi$ cm$^3$

**Key concepts used:** Visualization of the cone formation -- cone volume concept.

**Problem 9.**

An equilateral triangle of side 6 cm has its corners cut-off to form a regular hexagon. Area (in cm$^2$) of this hexagon will be,

- $3\sqrt{3}$
- $6\sqrt{3}$
- $5\displaystyle\frac{\sqrt{3}}{2}$
- $3\sqrt{6}$

**Solution 9 - Problem analysis **

The following figure corresponds to visualization of the problem.

In a regular hexagon thus formed, each internal angle, say, $\angle BCD = 120^0$ so that its corresponding external angle, $\angle BCQ = 60^0$. This necessarily makes the third angle of the $\triangle BQC$, $\angle QBC = 60^0$ as already $\angle BQC = 60^0$ in the equilateral $\triangle PQR$.

We can thus visualize the small corner triangles cut-off as also equilateral triangles, and as in the regular hexagon $BC=CD$, $QC$ is one-third of side length of $\triangle PQR$.

**Solution 9 - Problem solving execution**

In a general equilateral triangle of side length $a$, the area of the triangle,

$A = \displaystyle\frac{1}{2}\times{a}\times{height} $

$\hspace{5mm}= \displaystyle\frac{1}{2}\times{a}\times{\displaystyle\frac{\sqrt{3}}{2}a}$, as $height=a\sin 60^0$.

Or, $A = \displaystyle\frac{\sqrt{3}}{4}a^2$.

If side length of larger triangle is $a$, side length of smaller triangle is $\displaystyle\frac{1}{3}a$ and its area will be,

$A_s =\displaystyle\frac{\sqrt{3}}{36}a^2$ and the area of the hexagon will be area of the larger triangle minus three times the area of the smaller triangle,

$A_h = A - 3A_s $

$\hspace{7mm}= \displaystyle\frac{\sqrt{3}}{4}a^2 - \displaystyle\frac{\sqrt{3}}{12}a^2 $

$\hspace{7mm}= 9\sqrt{3} - 3\sqrt{3} = 6\sqrt{3}$.

We have delayed calculations as late as possible.

**Answer:** Option b: $6\sqrt{3}$.

**Key concepts used:** Visualization of hexagon formation -- discovery of the fact that side length of a smaller triangle is one-third of the side-length of the larger triangle -- concept of area of equilateral triangles.

** Problem 10.**

The ratio of sides of a triangle is 3 : 4 : 5 and area of the triangle is 72 square units. Then the area of an equilateral triangle with same perimeter as the first triangle (in square units) is,

- $48\sqrt{3}$
- $96$
- $60\sqrt{3}$
- $32\sqrt{3}$

**Solution 10 - Problem analysis and visualization**

Converting the terms of the ratio of side lengths to actual side lengths by multiplying with canceled out HCF $x$ we have the transformed ratio as, $3x : 4x : 5x$. By the amounts it is obvious that the triangle is a right triangle,

$(5x)^2 = (3x)^2 + (4x)^2$.

Its area is then,

$\frac{1}{2}\times{4}\times{3}x^2 = 72$ square units,

Or, $x^2 = 12$ square units,

Or, $x = 2\sqrt{3}$ unit.

So Perimeter of the triangle $= (3x + 4x + 5x) = 12x = 24\sqrt{3}$ unit.

If this equals the perimeter of an equilateral triangle, its side length is one-third the perimeter, or $8\sqrt{3}$ unit.

This gives us the area of the equilateral triangle as,

$A = \frac{\sqrt{3}}{4}(64 \times{3}) = 48\sqrt{3}$ square units.

**Answer:** a: $48\sqrt{3}$.

**Key concepts used:** Ratio concepts -- area and perimeter concepts of right triangle and equilateral triangle.

### Other related question set and solution set on SSC CGL mensuration

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Tutorials that you should refer to**

*Basic concepts on Geometry 1 lines points and triangles*

*Basic concepts on Geometry 2 quadrilaterals polygons squares*

*Basic and rich concepts on Geometry 3 Circles*

*Basic and rich Geometry concepts part 4 Arc angle subtending concept proof*

#### Question and Solution Sets on Mensuration

**SSC CGL level Solution Set 88 on Mensuration 8**

**SSC CGL level Question Set 88 on Mensuration 8**

**SSC CGL level Solution Set 87 on Mensuration 7**

**SSC CGL level Question Set 87 on Mensuration 7**

**SSC CGL level Solution Set 86 on Mensuration 6**

**SSC CGL level Question Set 86 on Mensuration 6**

**SSC CGL level Solution Set 43 on Mensuration 5**

**SSC CGL level Question Set 43 on Mensuration 5**

**SSC CGL level Solution Set 42 on Mensuration 4**

**SSC CGL level Question Set 42 on Mensuration 4**

**SSC CGL level Solution Set 41 on Mensuration 3**

**SSC CGL level Question Sety 41 on Mensuration 3**

**SSC CGL level Solution Set 27 on Mensuration 2**

**SSC CGL level Question Set 27 on Mensuration 2**

**SSC CGL level Question Set 26 on Mensuration 1**