## Twentieth SSC CGL level Solution Set, 2nd on topic Geometry

This is the twentieth solution set of 10 practice problem exercise for SSC CGL exam and 2nd on topic Geometry.

We will repeat here the method of taking the test if you have not gone through it already. And if you have not taken the test yet, you can refer to * SSC CGL Question Set 20, Geometry 2*, and then after taking the test come back to this solution.

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

**Before start,**go through the tutorials,**Geometry basic concepts part 1, points lines and triangles**or any other short but good material to refresh your concepts if you so require. This question set is in fact the set of exercise problems at the end of the first tutorial. Don't do the exercise as you are preparing for a hard competitive test. For you, taking the test will involve a different more stringent method.**Geometry basic concepts part 2, Quadrilaterals squares,****Geometry basic concepts part 3, Circle****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.**When the time limit of 12 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 12 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

### Twentieth solution set- 10 problems for SSC CGL exam: 2nd on Geometry - answering time 12 mins

#### Problem 1.

In a cyclic quadrilateral $ABCD$, side $AB$ is extended to $E$ so that $BE = BC$. If $\angle ADC=70^0$ and $\angle BAD=95^0$ then $\angle DCE$ is,

- $140^0$
- $165^0$
- $120^0$
- $110^0$

#### Solution 1.

As $\angle ADC=70^0$, its opposite angle in the cyclic quadrilateral is,

$\angle ABC = 180^0 - 70^0 = 110^0 $,

because in a circumscribed cyclic quadrilateral opposite angles sum to $180^0$.

Being the external angle in the triangle $\triangle BEC$ again,

$\angle ABC= \angle BEC + \angle BCE = 2\angle BCE$, as $BC=BE$ and so, $\triangle BCE$ is isosceles.

Thus $\angle BCE = 55^0$, half of $110^0$.

On the other hand as it is opposite to $\angle BAD = 95^0$,

$\angle DCB =180^0 - 95^0 = 85^0$.

Finally then,

$\angle DCE = 85^0 + 55^0 = 140^0$.

**Answer:** a: $140^0$.

#### Problem 2.

The ratio between the number of sides of two regular polygons is $1 : 2$ and the ratio between their interior angles is $2 : 3$. The number of sides of the polygons are respectively,

- 5, 10
- 6, 12
- 7, 14
- 4, 8

#### Solution 2.

For a polygon with number of sides $n$ its total of internal angles = $\pi(n - 2)$.

For the two polygons with sides $n_1$ and $n_2$ the ratio of sides, $n_1 : n_2 = 1 : 2$.

So, $n_2 = 2n_1$.

Again, ratio of their internal angles is, $ a_1:a_2 = 2:3$, or, $a_2 =\frac{3}{2} a_1$.

From its number of sides, total internal angle of the second triangle is,

$ TI_2=\pi(n_2 - 2) = \pi(2{n_1} - 2) $

Again it has $n_2$ number of $a_2$ internal angles and so, total internal angle for the second polygon is again is,

$TI_2 = n_2a_2=\frac{3}{2} n_2(a_1)=3n_1a_1 $.

Or, $n_1a_1 = \frac{2}{3}\pi(n_1 - 1)$

This is for the second polygon.

For the first polygon, its total internal angle is,

$n_1a_1 = \pi(n_1 - 2)$.

Taking the ratio of the two,

$3(n_1 - 2) = 2(n_1 - 1)$,

Or, $n_1 = 4$,

and $n_2 = 8$.

Answer: d: 4, 8.

#### Problem 3.

The length of the diagonal $BD$ of the parallelogram $ABCD$ is 18cm. If $P$ and $Q$ are the centroids of $\triangle ABC$ and $\triangle ADC$ respectively, length of $PQ$ is,

- 4cm
- 12cm
- 6cm
- 9cm

#### Solution 3.

Being a parallelogram its diagonals bisect each other and so $BD$ is a median to both the triangles $\triangle ABC$ and $\triangle ADC$. $AE$ and $AF$ are the two other medians drawn to opposite sides intersecting the other medians at $P$ and $Q$ respectively which are then the centroids of the two triangles.

Now $BD = 18$cm and half of it is 9cm. This is the length of the median divided by $P$ and $Q$ in ratio 2:1 from vertices. Between vertices then, out of 18cm, 2 portions out of 6 is the length of $PQ$, which is, $PQ=\frac{2}{6}\times{18} =6$cm.

**Answer:** c: 6cm

#### Problem 4.

In rhombus $ABCD$, a straight line through $C$ cuts extended $AD$ at $P$ and extended $AB$ at $Q$. If $DP =\displaystyle\frac{1}{2}AB$ the ratio of the lengths of $BQ$ and $AB$ is,

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

AS $AP||BC$ in triangle $\triangle APQ$ the two triangles $\triangle APQ$ and $\triangle BCQ$ are similar. The base $AP = AD + DP = 3$ portions while, the base $BC$ of the $\triangle BCQ$ is 2 portions. As the corresponding sides in the two similar triangles are in equal ratio, $AQ : BQ = 3:2$, Or, $BQ: AB = 2:1$.

Answer: b: 2 : 1.

#### Problem 5.

$A$, $B$ and $C$ are three points on the circumference of a circle. If $AB = AC = 5\sqrt{2}$cm and $\angle BAC=90^0$ the length of radius is,

- 15cm
- 5cm
- 10cm
- 20cm

#### Solution 5.

As subtended angle at the periphery by the chord $BC$ is $90^0$, the chord is a diameter of the circle, and forms the hypotenuse of the right angled isosceles $\triangle ABC$.

By Pythagoras theorem then,

$BC^2 = AB^2 + AC^2 = 50 + 50=100$.

So, diameter $BC=10$, or, radius=5cm.

**Answer:** b: 5cm.

#### Problem 6.

If $PN$ is the perpendicular from a point $P$ on the circumference of a circle of radius 7cm to its diameter $AB$ and the length of the chord $PB$ is 12cm, the length of $BN$ is,

- $6\displaystyle\frac{5}{7}$ cm
- $3\displaystyle\frac{5}{7}$ cm
- $12\displaystyle\frac{2}{7}$ cm
- $10\displaystyle\frac{2}{7}$ cm

#### Solution 6.

In two triangles, $\triangle APB$ and $\triangle PNB$, apart from the equal right angles (diameter subtends an angle of $90^0$ at peripheral point $P$), the $\angle B$ is common to both triangles. So the third angles are also same and the triangles are similar.

The ratio of corresponding sides in these two similar triangles, $ BN : PB = PB : AB$,

Or, $BN = \displaystyle\frac{PB^2}{AB} = \displaystyle\frac{144}{14} = 10\displaystyle\frac{2}{7}$cm.

**Answer:** d: $10\displaystyle\frac{2}{7}$cm.

#### Problem 7.

If the angle subtended by a chord at its center is $60^0$, the ratio between the lengths of the chord and the radius is,

- $1 : 1$
- $2 : 1$
- $\sqrt{2} : 1$
- $1 : 2$

#### Solution 7.

Radii $AO=BO$ makes,

$\angle OAB = \angle OBA$

$\hspace{15mm}= \frac{1}{2}(180^0 - 60^0)$

$\hspace{15mm}= 60^0$.

So the triangle is equilateral and the ratio between the chord and the radius is $1 : 1$.

**Answer:** a: $1 : 1$

#### Problem 8.

$AB$ and $CD$ are two parallel chords of respective lengths 8cm and 6cm on the same side of the center of a circle. The distance between them is 1cm. Then the radius of the circle is,

- 4cm
- 5cm
- 3cm
- 2cm

#### Solution 8.

In the two triangles $\triangle APO$ and $\triangle CQO$ all values of the sides are known except the portion $OP$ which we assume here for this reason as unknown $x$. From the two triangles we get two equations by applying Pythagoras theorem,

$r^2 = AP^2 + x^2 = x^2 + 16$, and

$r^2 = CQ^2 + (x+1)^2$

$\hspace{5mm}= 9 + x^2 + 2x + 1$

$\hspace{5mm}= x^2 + 2x + 10$.

Subtracting the second from the first,

$2x = 10$,

Or, $x=5$cm.

**Answer:** b: 5cm.

#### Problem 9.

$ABCD$ is a trapezium whose side $AD$ is parallel to $BC$. Diagonals $AC$ and $BD$ intersect at $O$. If $AO = 3$, $CO = x - 3$, $BO = 3x - 19$ and $DO = x - 5$, the value(s) of $x$ will be,

- 7, 10
- 7, 6
- 12, 6
- 8, 9

#### Solution 9.

The sides $AD || BC$ so that, $\angle ADO = \angle CBO$ and $\angle DAO = \angle BCO$ and so the triangle $\triangle AOD$ is similar to $\triangle BOC$.

In a pair of similar triangles ratio of corresponding sides are equal. Applying this rule in this case we have,

$\displaystyle\frac{AO}{CO} = \displaystyle\frac{DO}{BO}$,

Or, $\displaystyle\frac{3}{x-3} = \displaystyle\frac{x-5}{3x-19}$,

Or, $3(3x - 19) = (x - 3)(x-5)$

Or, $x^2 - 17x + 72 = 0$

Or, $(x - 8)(x - 9) = 0$.

So, $x = 8, 9$

**Answer:** d: 8, 9.

#### Problem 10.

A quadrilateral $ABCD$ circumscribes a circle and $AB=6$cm, $CD=5$cm and $AD=7$cm. The length of side $BC$ is,

- 3cm
- 4cm
- 5cm
- 6cm

The sides of the circumscribed quadrilateral are all tangents to the inscribed circle. The tangent points are respectively, $P$, $Q$, $R$ and $S$. By property of two tangents from a single external point to a circle, the tangent segment lengths are equal. In our case for example, $BP=BS$.

Using this property we will arrive at the solution.

Let us assume the two parts of side $BC$ length of which is to be found out, are, $CS=x$ and $BS=y$, so that its length is $BC=x+y$.

For the adjacent side $AB$, $BP = y$, and $PA = 6 - y$; for the side $CD$, $CR=x$ and $RD = 5 - x$.

Finally reaching the side $AD$, $DQ=5-x$ and $AQ = 6-y$. Their sum is,

$5 - x + 6 - y = 7$,

Or, $x + y = 11 - 7=4$.

This is the desired length of the fourth side.

**Answer:** b: 4cm.

**Note:** Do not think that in the test venue you have to draw such a precise detailed diagram. In fact, the underlying principle is so simple, even without a drawing just by visualizing the solution can be obtained.

*Furthermore, in MCQ based Geometry problems in competitive exams, deductions generally are minimal. If you can visualize on the basis of a quick sketch of a diagram, you can solve most problems in a minute.*

**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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