## 38th SSC CGL level Solution Set, 6th on topic Geometry

This is the 38th solution set of 10 practice problem exercise for SSC CGL exam and 6th on topic Geometry.

If you have not taken the corresponding test yet, you may refer to * SSC CGL level Question Set 38, Geometry 6* and then continue with this solution.

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

**Before start,**go through the**tutorials on****Geometry basic concepts part 1 on points lines triangles,****Geometry basic concepts part 2 on Quadrilaterals Squares Rectangles,****Geometry basic and rich concepts part 3 on Circles,**or any other short but good material to refresh your concepts if you so require.**Basic and rich Geometry concepts part 4 on proof of arc angle subtending concept****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 given at the end 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

### 38th solution set- 10 problems for SSC CGL exam: 6th on Geometry - answering time 15 mins

#### Problem 1.

If I be the incentre of $\triangle ABC$, $\angle ABC=60^0$ and $\angle ACB=50^0$, then $\angle BIC$ is,

- $70^0$
- $65^0$
- $125^0$
- $55^0$

#### Solution 1.

The following figure describes the problem,

##### Problem analysis and solving

I being the incentre, BI and CI are the bisectors of the angles $\angle ABC=60^0$ and $\angle ACB=50^0$. So,

$\angle CBI=30^0$, and

$\angle BCI=25^0$.

Thus in the $\triangle BIC$,

$\angle BIC=180^0 - (\angle CBI + \angle BCI)$

$\hspace{14mm}= 180^0 - 55^0$

$\hspace{14mm}=125^0$.

**Answer:** c: $125^0$.

**Key concepts used:** Basic property of incentre as * intersection of angle bisectors* --

*.*

**incentre concepts**#### Problem 2.

If ratio of number of sides of two polygons is 5 : 6, and ratio of their internal angles is 24 : 25, the number of sides of the two polygons are,

- 20, 24
- 5, 6
- 15, 18
- 10, 12

#### Solution 2.

Using the ratio concepts, let's assume the actual values of the number of sides be, $5x$ and $6x$ and the internal angles be, $24y$ and $25y$.

We know that for a polygon with number of sides as $n$ the relation between $n$ and the total internal angles is,

$(n-2)\pi=nI$, where I is the internal angle.

So for the two polygons we have,

$(5x-2)\pi = 5x\times{24y}=120xy$, and

$(6x-2)\pi=6x\times{25y}=150xy$.

Taking a ratio of the two,

$\displaystyle\frac{6x - 2}{5x-2}=\frac{150}{120}=\frac{5}{4}$,

Or, $24x - 8=25x - 10$,

Or, $x=2$.

So the number of sides are,

$5x=10$, and

$6x=12$.

**Answer:** d: 10, 12.

**Key concepts used:** * Basic Ratio concepts* --

**polygon side angle relation concept.**#### Problem 3.

If the inradius of an equilateral triangle is 3 cm, its side length is,

- $3\sqrt{3}$ cm
- $6$ cm
- $9\sqrt{3}$ cm
- $6\sqrt{3}$ cm

#### Solution 3.

The problem is depicted in the figure below.

##### Problem analysis and solving

In our * SSC CGL level solution set 36 Geometry 4*, we have shown how in an equilateral triangle all the three important points in a triangle, the incentre, the circumcentre and the centroid coincide. This is shown in the figure above.

In our problem this time, the inradius OD=3 cm is given and legnth of side BC is to be found out.

In $\triangle ABC$, AOD being the median, the centroid divides it in a ratio of 2 : 1. So the circumradius $AO=2\times{3}=6$ cm.

Thus in right $\triangle ODC$,

$CD^2 = OC^2 - OD^2 = 6^2 - 3^2 = 27$.

Finally then,

$CD = 3\sqrt{3}$, and length of a side,

$BC = 2\times{CD} = 6\sqrt{3}$ cm.

**Answer:** d: $6\sqrt{3}$ cm.

**Key concepts used:** Rich concept of * three point coincidence* of incentre, circumcentre and centroid

*-- use of concept of*

**in an equilateral triangle***to get circumradius --*

**centroid dividing median in a ratio of 2 : 1***.*

**pythagoras theorem**#### Problem 4.

ABC is an isosceles triangle such that AB=AC and AD is the median to the base BC with $\angle ABC=35^0$. Then $\angle BAD$ is,

- $35^0$
- $55^0$
- $110^0$
- $70^0$

#### Solution 4.

The following figure depicts the given problem.

##### Problem analysis and solving

The $\triangle ABC$ being isosceles and AD being the median, in the two triangles $\triangle ABD$ and $\triangle ACD$, the two sides AB=AC, BD=DC and the third side AD is common. So the SSS test for congruency of the two triangles is satisfied.

Thus $\angle BDA=\angle CDA=90^0$.

So finally in $\triangle ABD$,

$\angle BAD=180^0 - (\angle ABD + \angle BDA) $

$\hspace{14mm}= 180^0 - (35^0 + 90^0)$

$\hspace{14mm}=55^0$.

**Answer:** b: $55^0$

**Key concepts used:** Properties of isosceles triangles -- * Congruency test of triangles rich concepts* -- total angle in a triangle concept -- concept of equal adjacent angles at intersection of two lines are $90^0$,

*.*

**perpendicular line condition**#### Problem 5.

Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the larger circle which is outside the inner circle is of length,

- $2\sqrt{2}$ cm
- $4\sqrt{2}$ cm
- $2\sqrt{3}$ cm
- $3\sqrt{2}$ cm

#### Solution 5.

The figure depicting the problem is shown below.

##### Problem analysis and relevant concepts

In the figure above the length of any chord of the larger circle that is outside the smaller circle will be closely related to the angle it subtends on any point on the major arc. The angle and correspondingly the length of the chord will increase as the chord approaches the centre of the larger circle. If it were allowed to reach the centre its length would have been maximum, that is, equal to the the length of the diameter of the larger circle.

In this case, the chord can't reach the centre and the nearest it can reach the centre is the point D at which it touches the smaller circle. The shortest or perpendicular distance of the chord from the centre of the larger circle will be minimum at this point.

If the chord moves further towards the centre P, some of its portion will be inside the smaller circle which is prohibited.

The * second basic concept* that makes this position of the chord as the maximum length position under the given conditions is,

A chord of same length will subtend an angle that is equal at all points on the complementary arc.

##### Problem solving

In the figure above,

PC=3 cm, CD=4 cm, being twice the radius of the smaller circle. So,

DP = 1 cm, and

AP = 3 cm.

Thus in right $\triangle APD$,

$AD^2 = AP^2 - DP^2 = 9 -1 = 8$,

Or, $AD = 2\sqrt{2}$ cm,

Or, chord length $AB = 4\sqrt{2}$ cm.

**Answer:** Option b: $4\sqrt{2}$ cm.

**Key concepts used:** Visualization -- **Internally *** touching *circles concept -- chord angle subtending concepts -- maximum chord length concept -- tangent concept -- Pythagoras theorem.

#### Problem 6.

In a right $\triangle ABC$, point D lies on side BC. If AB be the hypotenuse, then,

- $AB^2 + CD^2=BC^2 + AD^2$
- $AB^2=AD^2+BD^2$
- $AD^2 + AC^2=2AD^2$
- $CD^2+BD^2=2AD^2$

#### Solution 6.

The following figure represents the problem.

##### Problem analysis

On examining the desired relations we understand that we need to use some amount of algebraic manipulations in the right triangles involved.

##### Problem solving

In right $\triangle ABC$ by Pythagoras theorem,

$AB^2 = AC^2 + BC^2$,

Or, $AC^2 = AB^2 - BC^2$,

Similarly in right $\triangle ACD$,

$AC^2 = AD^2 - CD^2 = AB^2 - BC^2$, using the earlier result.

So we finally have,

$AB^2 + CD^2 = BC^2 + AD^2$

**Answer:** a: $AB^2 + CD^2 = BC^2 + AD^2$.

**Key concepts used:** Pythagoras theorem -- * basic algebraic concepts* --

*.*

**deductive reasoning**#### Problem 7.

If each angle of a triangle is less than the sum of the other two angles, then the triangle is,

- Obtuse angled
- Right angled
- Acute angled
- Not a triangle

#### Solution 7.

##### Problem analysis and solving

By our previous knowledge we know that the triangle must be acute angled, but here we will establish the mechanism behind this result.

In a triangle if the angles are A, B and C, we have their sum,

$A + B + C = 180^0$.

Again by the inequalities we have first,

$A \lt B + C$,

Or, $A \lt 180^0 - A$,

Or, $2A \lt 180^0$,

Or, $A \lt 90^0$.

In exactly the same way,

$B \lt 90^0$, and

$C \lt 90^0$.

All three angles being less than $90^0$ the triangle is necessarily an acute angled triangle.

**Answer:** c: Acute angled.

**Key concepts used:** Concept of total angle in a triangle -- target driven * inequality analysis* -- result is a basic triangle concept but if unable to recall, can be quickly established from the basic concepts.

#### Problem 8.

The side AB of a parallelogram ABCD is extended to E such that BE=AB and DE intersects BC at Q. Then the point Q divides BC in the ratio,

- 1 : 2
- 2 : 1
- 1 : 1
- 2 : 3

#### Solution 8.

The following is a depiction of the problem graphically.

As BE is an extension of AB, AE || CD (as AB || CD in the parallelogram). So, BE || CD. Thus,

$\angle BCD = \angle CBE$,

$\angle CDE = \angle DEB$, and at the intersection Q,

$\angle EQB = \angle CQD$.

This makes the two triangles similar. Moreover as CD=AB=EB, the triangles are actually congruent (corresponding side ratio equal and 1, that means corresponding sides are equal which is the SSS test satisfaction for congruency ), so that, CQ=QB, and the desired ratio is 1 : 1.

**Answer:** c: 1 : 1.

**Key concepts used:** * Basic parallelogram concepts* -- basic

*--*

**intersected parallel lines concepts**

**congruent**

*concepts -- deductive reasoning.***triangles**#### Problem 9.

A cyclic quadrilateral ABCD is such that AB=BC, AD=DC, AC perpendicular to BD and $\angle CAD=\theta$. Then the $\angle ABC=$,

- $\theta$
- $\displaystyle\frac{\theta}{2}$
- $3\theta$
- $2\theta$

#### Solution 9.

The following figure represents the problem description.

##### Problem analysis

In $\triangle ABC$, AB=BC so that, $\angle BAC= \angle BCA$. Similarly in $\triangle ADC$, AD=DC so that $\angle ACD= \angle CAD=\theta$. Furthermore, opposite angles summing up to $180^0$, we also have $\angle ABC + \angle ADC = 180^0$.

##### Problem solving

In $\triangle ADC$ then,

$\angle ADC = 180^0 -2\theta$

Again opposite angles summing up to $180^0$ in the cyclic quadrilateral, we have,

$\angle ADC = 180^0 - \angle ABC = 180^0 -2\theta$.

So,

$\angle ABC = 2\theta$.

**Answer:** d: $2\theta$.

**Key concepts used:** Basic concepts of isosceles triangle -- total angle in a triangle concept -- **rich concept of cyclic quadrilateral.**

**Note:** AB=BC and AC perpendicular to BD are superfluous information.

#### Problem 10.

ABCD is a rhombus whose side AB = 4 cm and $\angle ABC = 120^0$. Then the length of the diagonal BD is,

- 4 cm
- 2 cm
- 1 cm
- 3 cm

#### Solution 10.

The following is the figure relevant to the problem.

##### Problem analysis and relevant concepts

Among other concepts the most relevant concept that needs to be used here is the concept that the two diagonals of a rhombus intersect each other perpendicularly and also these are angle bisectors. Let us see how this happens.

#### Concept of diagonals of a rhombus are perpendicular bisectors to each other as well as angle bisectors, the mechanism

In a rhombus the diagonals are perpendicular bisectors to each other as well as bisectors of corresponding angles of the rhombus.

Let us see how it happens.

In the figure above, as $AD=DC$ in $\triangle ADC$, $\angle CAD = \angle DCA$.

Again as AB || DC, $\angle DCA =\angle CAB$.

So,

$\angle CAD = \angle CAB$, and so AC bisects $\angle BAD$.

In the same way other three angles of the rhombus are also bisected by the diagonals. Let now see how the second property of perpendicular bisectors happens.

In $\triangle ABP$ and $\triangle ADP$ as AD=AB, $\angle ABP =\angle ADP$ and the $\angle BAD$ is bisected.

Thus in these two triangles two pairs of angles being equal, the third pair of angles are also equal making the two triangles similar and as corresponding sides AD=AB, the triangles are actually congruent. So,

$BP = DP$ and $\angle APB =\angle APD = 90^0$.

The same is true for the other diagonal. Thus the special property of diagonals in a rhombus holds good. This is** rhombus diagonals property.**

Continuing with our problem solving, diagonal BD bisecting $\angle ABC = 120^0$, in right $\triangle ABP$, the $\angle ABP = 60^0$ and,

$\cos 60^0 = \displaystyle\frac{BP}{AB}$,

Or, $\displaystyle\frac{1}{2}=\displaystyle\frac{BP}{4}$,

Or, $BP = 2$ cm,

Or, $BD = 2\times{2} = 4$ cm

**Answer:** a: 4 cm.

**Key concepts used:** * Rhombus diagonals property* --

**basic trigonometry concepts.****Related 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.**

**Concept tutorials for SSC CGL and other competitive exams on Geometry**

**Basic and rich Geometry concepts part 7, Laws of sines and cosines**

**Basic and rich Geometry concepts part 6, proof of triangle area from medians**

**Basic and rich Geometry concepts part 5, proof of median relations**

**Basic and rich Geometry concepts part 4, proof of arc angle subtending concept**

**Geometry, basic and rich concepts part 3, Circles**

**Geometry, basic concepts part 2, Quadrilaterals polygons and squares**

**Geometry, basic concepts part 1, points lines and triangles**

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