## 88th SSC CGL level Solution Set, 8th on topic Mensuration

This is the 88th solution set of 10 practice problem exercise for SSC CGL exam and 8th on topic Mensuration. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource. This can be used for other competitive tests with Mensuration in syllabus also.

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 should refer to * SSC CGL level Question Set 88, Mensuration 8*, and then after taking the test come back to this solution.

The *10 problem question* set can effectively be used as a *mini-mock test on SSC CGL Mensuration*.

### 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 at the end of this question set 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**

*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*

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### 88th solution set - 10 problems for SSC CGL exam: topic Mensuration 8 - Answering time 15 mins

**Problem 1.**

The volume of the metal of a cylindrical pipe is 748 $\text{cm}^3$. The length of the pipe is 14 cm and its external radius is 9 cm. Its thickness is, (Take $\pi=\frac{22}{7}$)

- 23 cm
- 3.7 cm
- 5.2 cm
- 1 cm

**Solution 1 - Problem analysis and execution**

The volume of the cylindrical pipe would be volume of a cylinder with cross-section radius equal to the outer radius of the pipe minus the volume of the inner cylinder with cross-section radius equal to the inner radius,

$V_{cylinder}=V_{outer}-V_{inner}$,

Or, $748=\pi \times{9^2}\times{14}-V_{inner}=81\times{44}-V_{inner}$,

So,

$V_{inner}=81\times{44}-17\times{44}$

$=\pi r_i^2 14=44r_i^2$, where $r_i$ is the inner radius.

So the square of inner radius is,

$r_i^2=81-17=64$.

The inner cylinder radius is then 8 cm and thickness of the metal cylinder is,

$9-8=1$ cm.

If you can take out the factor of 44 from 748, you can solve this problem easily in mind.

Mark that we haven't evaluated the product of 44 and 81, expecting 748 also to contain a factor of 44, more so because we could see that the third term also contains a factor of 44. This is applying *delayed evaluation technique* on anticipation based on *favorable indications.*

**Answer:** Option d: 1 cm.

**Key concepts used:** Volume of a cylindrical pipe* -- Delayed evaluation technique -- Solving in mind.*

**Problem 2.**

The sum of areas of two circles touching externally is $130\pi$ sq cm and the distance between their centers is 14 cm. The radius of the larger circle is, (Take $\pi=\frac{22}{7}$)

- 22 cm
- 44 cm
- 11 cm
- 33 cm

**Solution 2 - Problem analysis and execution**

Assume $r_1$ and $r_2$ as the radii of the two circles.

From the given conditions,

$r_1+r_2=14$, and

$130\pi=\pi(r_1^2+r_2^2)$.

Then $r_1^2+r_2^2=130$.

Squaring the first equation,

$196=(r_1+r_2)^2=r_1^2+2r_1r_2+r_2^2=130+2r_1r_2$.

So, $2r_1r_2=196-130=66$.

Subtracting this from $r_1^2+r_2^2=130$, you get,

$(r_1-r_2)^2=130-66=64$,

Or, $r_1-r_2=8$.

And the radius of the larger circle $r_1$,

$r_1=\displaystyle\frac{8+14}{2}=11$ cm.

You have added $(r_1+r_2)=14$, $(r_1-r_2)=8$ and divided by 2. $r_1$ is the radius of the larger circle here.

**Answer:** Option c : 11 cm.

**Key concepts used:** Two circles touching externally -- Area of circle -- Basic algebraic relations -- Solving in mind.

Using a bit of basic algebra you can easily solve this problem.

**Problem 3.**

The sides of a triangle are in ratio 2 : 3 : 4. The perimeter of the triangle is 18 cm. The area (in $\text{cm}^2$) of the triangle is,

- $9$
- $36$
- $3\sqrt{15}$
- $\sqrt{42}$

**Solution 3 - Problem solving using basic ratio concepts and area of a triangle in terms of its semi-perimeter and three side lengths**

This again is an algebra based problem where you have to use basic ratio concepts, perimeter as sum of all three sides and area of the triangle formula in terms of its semi-perimeter (half of perimeter) and lengths of its three sides.

Assuming the three side lengths to be $2x$, $3x$ and $4x$ where $x$ is the reintroduced cancelled out HCF in the ratio, the perimeter of the triangle is,

$18=2x+3x+4x=9x$,

Or $x=2$,

And lengths of three sides, 4 cm, 6 cm and 8 cm respectively.

The area of a triangle in terms of its semi-perimeter (half of perimeter) and lengths of three sides is,

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{9(9-4)(9-6)(9-8)}$, as semi-perimeter $s=9$ cm

$=\sqrt{9\times{5}\times{3}\times{1}}$

$=3\sqrt{15}\text{ cm}^2$.

**Answer:** Option c: $3\sqrt{15}$.

**Key concepts used:** Basic ratio concepts -- HCF reintroduction technique -- Area of a triangle in terms of its semi-perimeter and the lengths of three sides -- Basic algebra concepts -- Solving in mind.

**Problem 4.**

The inradius of a triangle is 4 cm and its area is 34 sq cm. The perimeter of the triangle is,

- 20 cm
- 17 cm
- 34 cm
- 8.5 cm

#### Solution 4 - Problem analysis and execution

Let us first solve the problem mentally without drawing the diagram.

**Let's assume, **$r$ to be the inradius and the three side lengths to be $a$, $b$ and $c$.

If you imagine three lines from the three vertices to be drawn to the incenter, effectively you divide the triangle into three component triangles height of each of which is the inradius.

So the total area of the triangle would be,

$A=\frac{1}{2}r(a+b+c)$,

Or, $p=\displaystyle\frac{2A}{r}$, where perimeter $p=a+b+c$.

So, $p=\displaystyle\frac{2\times{34}}{4}=17$ cm

Solved mentally within maximum 20 seconds.

Let us show you the diagram for clear visualization.

The three colored triangles $\triangle OBC$, $\triangle OCA$ and $\triangle OAB$ make up the whole area of $\triangle ABC$. The heights of these three triangles is the inradius of the triangle,

$OF=OD=OE=r=4$ cm (given),

And the bases are $a$, $b$ and $c$ respectively.

Perimeter of the triangle is,

$p=(a+b+c)=\displaystyle\frac{2A}{r}=\displaystyle\frac{2\times{34}}{4}=17$ cm.

Not difficult at all. Isn't it?

**Answer:** Option b: 17 cm.

**Key concepts used:** Inradius of a triangle -- Incenter dividing a triangle into three component triangles height of each of which is the inradius -- Perimeter concept -- Visualization -- Solving in mind.

**Problem 5.**

The sum of radii of two spheres is $10\text{ cm}$ and sum of their volumes is $880\text{ cm}^3$. What will be the product of their radii?

- $33\frac{1}{3}$
- $26\frac{1}{3}$
- $21$
- $70$

**Solution 5 - Problem analysis and execution**

Here also you have to depend on a bit of algebra.

Assume the two radii to be $r_1$ and $r_2$.

So sum of the two radii is,

$r_1+r_2=10$ cm.

And sum of the volumes of the two spheres is,

$V_s=\frac{4}{3}\pi(r_1^3+r_2^3)=880$,

Or, $r_1^3+r_2^3=210$.

Now we will use a bit of algebra to express the sum of cubes of radii in terms of sum and product of the radii.

$r_1^3+r_2^3=(r_1+r_2)(r_1^2-r_1r_2+r_2^2)$

$=(r_1+r_2)\left[(r_1+r_2)^2-3r_1r_2\right]$,

Or, $210=10(100-3r_1r_2)$,

Or, $r_1r_2=\displaystyle\frac{79}{3}=26\frac{1}{3}$.

**Answer:** Option b: $26\frac{1}{3}$.

**Key concept used:** Sphere volume -- Basic algebra concepts -- Sum of cubes of radii two-factor enhanced expansion in terms of sum of radii and product of radii -- Solving in mind.

**Problem 6.**

If the volume of a sphere is numerically equal to its surface area, then its diameter is,

- 6 cm
- 2 cm
- 3 cm
- 4 cm

**Solution 6 - Problem analysis and execution**

As the volume of the sphere is numerically equal to its surface area, assuming $r$ to be its radius,

$\frac{4}{3}\pi r^3=4\pi r^2$.

Simplifying you get,

$r=3$ cm.

And diameter equal to 6 cm.

**Answer:** Option a : 6 cm.

**Key concepts used:** Volume of sphere -- Surface area of sphere -- Solving in mind.

**Problem 7.**

In $\triangle ABC$, AD and AE are bisectors of $\angle BAC$ and $\angle BAD$ respectively. If $\angle BAE=30^0$, AE = 9 cm and EC = 15 cm, what is the area (in $\text{cm}^2$) of $\triangle AEC$.

- 72
- 54
- 36
- 216

** Solution 7 - Problem visualization and solution**

This problem is more Geometry dependent and we'll solve it first mentally using the technique of imagining the figure in mind, that is, visualization.

Assume $\angle BAC$ to be the apex angle and $\angle ABC$ to be the left base angle of $\triangle ABC$.

**First** a line segment AD bisects the apex angle $\angle BAC$ so that $\angle BAD=\angle DAC$.

Don't bother about the location of point D—* just concentrate on the angles*. D may lie inside or outside the triangle.

**Second time again** $\angle BAD$ is bisected by a second line segment AE.

*What really happens by these two consecutive bisections?*

The angle $\angle BAC$ is simply divided into four equal parts, one of which $\angle BAE=30^0$. As a consequence, $\angle EAC$ turns out to be $120^0-30^0=90^0$ and $\triangle EAC$ a right triangle.

The hypotenuse of this triangle EC is 15 cm and the base AE is 9 cm.

That gives the height (by using Pythagoras theorem),

$AC=15^2-9^2=12^2$.

And the area of the right $\triangle AEC$ becomes simply half of base multiplied by its height,

$A_{AEC}=\frac{1}{2}\times{9}\times{12}=54 \text{ cm}^2$.

Let us show you the diagram.

As you can see AD bisects $\angle BAC$, and next, AE further bisect one half of the bisected angle $\angle BAD$. $\angle BAE$ being $30^0$, $\triangle EAC$ turns out to be a right triangle with right angle at $\angle EAC$.

With the given lengths of the hypotenuse EC and base AE, you can perceive now that E must lie outside the triangle. But where D or E lies is not important. Key information is in formation of the right riangle with its hypotenuse and base lengths given.

You don't really need to draw the diagram to solve this awkward looking problem.

**Answer: **Option b**: **54.

**Key concepts used:** Bisectors of angles -- Visualization -- Pythagoras theorem -- Right triangle formation -- Area of right triangle -- Solving in mind.

Yes, you can easily solve the problem in mind with clear visualization and step by step reasoning of what actually happens.

**Problem 8.**

If O is the centroid and AD, BE and CF are the three medians of $\triangle ABC$ with an area of 96 sq cm, then the area of $\triangle BOD$ (in $\text{cm}^2$) is,

- 8
- 12
- 24
- 16

** Solution 8 - Problem analysis and execution**

The representative figure is shown below.

By **median property**, median AD divides the triangle into two equal area triangles. Area of $\triangle BAD$ is then $48\text{ cm}^2$.

This area is again half of the base multipled by the height of the triangle,

$48=\frac{1}{2}\times{BD}\times{AQ}$.

You have to find the area of the $\triangle BOD$ which is,

$A_{BOD}=\frac{1}{2}\times{BD}\times{OP}$

$=48\times{\displaystyle\frac{OP}{AQ}}$, you have substituted BD from the first equation.

Now your task is specific—you have to find the value of the ratio $\displaystyle\frac{OP}{AQ}$.

#### Second stage of solution of problem 8: using equal corresponding side ratio in two similar triangles

The two triangles $\triangle DOP$ and $\triangle DAQ$ are similar as PO is parallel to QA (being perpendicular to same base line). The ratios of corresponding sides of these two triangles (sides opposite to equal angles) are equal. This results in,

$\displaystyle\frac{OP}{AQ}=\frac{OD}{AD}$.

Now we use the concept of median segmentation at centroid in a ratio of 2 : 1—centroid O divides AD in a ratio of 2 : 1 so that OD : AD = 1 : 3.

So,

$A_{BOD}=48\times{\frac{1}{3}}=16\text{ cm}^2$.

**Answer:** Option d: 16.

**Key concepts used:** Property of median dividing a triangle into two equal area parts -- Area of triangle -- Similar triangle condition -- Equal side ratio property of two similar triangles -- Median section ratio at centroid -- Solving in mind.

**Problem 9.**

The area and length of one of the diagonals of a rhombus are 84 sq cm and 7 cm respectively. Find the length ( in cm) of the other diagonal.

- 36
- 12
- 24
- 48

**Solution 9 - Problem analysis and execution**

Without drawing the diagram we'll first solve the problem using rhombus properties.

In a rhombus,

- all four sides are of equal lengths but internal angles are not $90^0$.
- the diagonals are perpendicular bisectors to each other.

Because of the second property of the diagonals, one diagonal divides the area of the rhombus exactly into two equal halves, and the second diagonal divides each half exactly into two more equal halves.

In effect, the two mutually bisecting diagonals form four triangles of equal areas, each one-fourh the area of the rhombus, height half of a diagonal and base half of second diagonal.

Given area of rhombus being 84 sq cm, area of each of the four component triangles is 21 sq cm,

$21=\frac{1}{2}\times{\frac{1}{4}}\times{7}\times{d_2}$, the first diagonal being 7 cm, and the second $d_2$ cm, say

So, $d_2=24$ cm.

Can easily be solved in mind.

Still for clarity the figure is shown below.

Diagonal AC and BD are perpendicular bisectors of each other and divide the area of rhombus ABCD into four triangles of equal areas.

Area of each such triangle, say of $\triangle AED$ is,

$A_{AED}=\frac{1}{2}ED\times{AE}$

$=\frac{1}{8}AC\times{BD}$

$=\frac{1}{4}A_{ABCD}$

$=21$.

AC being 7 cm, $BD=3\times{8}=24$ cm.

**Answer:** Option c: 24.

**Key concepts used:** Properties of rhombus diagonals -- Diagonals of a rhombus segments area of rhombus into four equal area triangles -- Solving in mind.

** Problem 10.**

The smaller diagonal of a rhombus is equal to its length of its sides. If the length of each side is 6 cm, what is the area ( in sq cm) of an equilateral triangle whose side is the larger diagonal?

- $27\sqrt{3}$
- $18\sqrt{3}$
- $36\sqrt{3}$
- $32\sqrt{3}$

**Solution 10 - Problem analysis and quick solution by problem definition and cylinder volume concepts**

As the smaller diagonal of the rhombus is equal in length to the side length 6 cm, it divides the area of the rhombus into two equilateral triangles each with internal angle $60^0$.

Because the two diagonals of a rhombus are perpendicular bisectors to each other, the longer diagonal divides the shorter one into two 3 cm segments.

Considering a right triangle with this 3 cm segment as height and 6 cm side as hypotenuse, the base length of half the longer diagonal is,

$\frac{1}{2}d=\sqrt{6^2-3^2}=3\sqrt{3}$ cm.

So the longer diagonal length is, $d=6\sqrt{3}$ cm.

Area of an equilateral triangle with this side length is,

$A=\displaystyle\frac{\sqrt{3}}{4}d^2=27\sqrt{3}$.

For clarity, the rhombus is shown below.

**Answer:** Option a: $27\sqrt{3}$.

**Key concepts used:** Visualization -- Properties of a rhombus -- Properties of rhombus diagonals -- Pythagoras theorem -- Area of an equilateral triangle -- Solving in mind.

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