## Twentyfirst SSC CGL level Solution Set, 3rd on topic Geometry

This is the twentyfirst solution set of 10 practice problem exercise for SSC CGL exam and 3rd on topic Geometry.

If you have not taken the corresponding test yet, you may refer to * SSC CGL level Question Set 21, Geometry 3* and then continue with this solution.

### Twentyfirst solution set- 10 problems for SSC CGL exam: 3rd on Geometry - answering time 12 mins

#### Problem 1.

In a right triangle $\triangle ABC$, the $\angle A = 90^0$ and $AD$ is perpendicular to $BC$. If areas of the triangles $\triangle ABC = 40cm^2$ and $\triangle ACD = 10cm^2$ with $AC = 9cm$, the length of $BC$ is,

- 4cm
- 12cm
- 18cm
- 6cm

#### Solution 1.

The following figure describes the problem,

In two right-angled triangles $\triangle ABC$ and $\triangle ACD$, apart from the equal right angles, the angle $\angle C$ is common. So the two triangles are similar to each other. The equal ratios of corresponding sides are (first triangle taken as the $\triangle ACD$),

$\displaystyle\frac{AC_1}{BC_2}=\frac{AD_1}{AB_2}=\frac{CD_1}{AC_2}$

Ratio of their areas is,

$\displaystyle\frac{CD_1.AD_1}{AB_2.AC_2}=\displaystyle\frac{10}{40}=\frac{1}{4}$.

In this ratio from equal side ratios because of triangle similarity,

$\displaystyle\frac{CD_1}{AC_2} = \displaystyle\frac{AC_1}{BC_2} = \displaystyle\frac{AC}{BC}$, and

$\displaystyle\frac{AD_1}{AB_2} = \displaystyle\frac{AC_1}{BC_2} = \displaystyle\frac{AC}{BC}$.

So, the ratio of areas = $\displaystyle\frac{1}{4} = \frac{AC^2}{BC^2}$,

Or, $BC^2 = 4\times{9^2}$,

Or, $BC = 2\times{9} = 18$cm.

**Answer:** c: 18cm.

#### Special note

You may refer to a detailed treatment of this problem at **How to solve intriguing SSC CGL level Geometry problem in a few steps 4.**

#### Problem 2.

If the ratio of an external angle and an internal angle of a regular polygon is $1 : 17$, the number of sides of the regular polygon is,

- 12
- 36
- 20
- 18

#### Solution 2.

The total of internal angles of a regular polygon is,

$\pi(n - 2)=na_i$, and an individual internal angle is,

$a_i = \displaystyle\frac{\pi(n-2)}{n}$.

On the other hand the total of external angles is $na_e=2\pi$, and an individual external angle is,

$a_e=\displaystyle\frac{2\pi}{n}$.

The ratio of an external angle to an internal angle is then,

$\displaystyle\frac{a_e}{a_i} = \frac{1}{17}=\frac{2\pi}{\pi(n-2)}$

Or, $(n - 2) = 34$.

Or, $n=36$.

Answer: b: 36.

#### Problem 3.

In an isosceles $\triangle ABC$, $AB=AC$. A circle passing through $B$ and touching AC at its middle point, intersects $AB$ at $P$. Then $AP : AB$ is,

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

#### Solution 3.

The line $APB$ cutting the circle at two points is called * a secant of the circle* and it

*to the circle,*

**forms a rich conceptual relationship with the section $AD$ to the tangent point**$AD^2 = AP\times{AB}$.

You may refer to this concept in our tutorial **Geometry basic concepts part 3, Circles.**

Otherwise also, it is easy to see how this relationship comes about by detecting that the two triangles $\triangle ABD$ and $\triangle APD$ are similar because, $\angle A$ is common and external angle $\angle ADP = \angle PBD$, the angle subtended by the chord on the opposite section of the periphery. This again is a rich concept and you may refer to the tutorial just mentioned to understand how it happens. We will not explain it here.

With this similarity of triangles, $\triangle APD$ (we will assume it as first) and $\triangle ABD$ (we will assume it as second) the side equality relationships are,

$\displaystyle\frac{AD_1}{AB_2}=\frac{PD_1}{BD_2}=\frac{AP_1}{AD_2}$,

Or, $AD_1\times{AD_2} = AP_1\times{AB_2}$.

In our case as the $\triangle ABC$ is isosceles, $AB = AC = 2AD$. Thus from the secant - tangent section relationship we have,

$AP\times{AB} = AD^2$,

Or, $\displaystyle\frac{AP}{AB} = \displaystyle\frac{AD^2}{AB^2}$

$\hspace{10mm}= \displaystyle\frac{AD^2}{AC^2}$

$\hspace{10mm}=\displaystyle\frac{1}{4}$.

**Answer:** c: 1 : 4.

#### Problem 4.

In $\triangle ABC$, $\angle C$ is an obtuse angle. The bisectors of exterior angles at $A$ and $B$ meet extended $BC$ and $AC$ at $D$ and $E$ respectively. If $AB = AD=BE$ then $\angle ACB$ is

- $105^0$
- $110^0$
- $135^0$
- $108^0$

#### Solution 4.

The following figure depicts the given problem.

Let $x$ and $y$ be the external angles after bisection by $AD$ and $BE$, and $\alpha$ and $\beta$ be the equal angles in triangles $\triangle ABE$ for $AB=BE$ and in $\triangle ABD$ for $AB=AD$.

We have, external angle,

$2x = \angle C + (180^0 - 2y)$,

Or, $\angle C = 2(x + y) - 180^0$ -----------------(E1).

Again from triangles $\triangle ABD$ and $\triangle ABE$,

$\alpha + x = 180^0 - 2\beta$

$\beta + y = 180^0 - 2\alpha$.

Adding the two,

$x + y = 360^0 - 3(\alpha + \beta) = 360^0 - 3(180^0 - \angle C)=3\angle C - 180^0$.

Substituting this value of $x + y$ in $E1$,

$\angle C = 2(3\angle C - 180^0) - 180^0$

Or, $5\angle C = 3\pi$,

Or, $\angle C = 108^0$.

**Answer:** d: $108^0$

#### Problem 5.

How many triangles can be formed by taking any three from the four line segments of lengths, 2cm, 3cm, 5cm and 6cm?

- 1
- 2
- 3
- 4

#### Solution 5.

We have to select three values out of the four length values given so that sum of any two values in the chosen set is larger than the third value.

This is Choosing three out of four combination problem and we can choose in $^4C_3=4$ ways. But because of the triangle formation constraint, sides 2cm, 3cm can't be taken together in any choice. This reduces number of combinations by 2, [2, 3, 5] and [2, 3, 6], leaving only two possibilities, [2, 5, 6] and [3, 5, 6].

**Answer:** b: 2.

**Note:** You had a very slight need to use concepts from a different topic area namely permutation and combination. This is not unusual - concepts from different topic areas may very well be mixed together in forming more demanding problems.

Here there was no real need to use concepts on permutation and combination but try out a little more complex problem,

**Modified problem: **

How many triangles can be formed by taking any three from the five line segments of lengths, 2cm, 3cm, 4cm, 5cm, and 6cm?

You may refer to our tutorial on * Permutation and Combination *for gaining a basic idea on this general topic.

#### Problem 6.

In right $\triangle ABC$, $BL$ and $CM$ are two medians with right angle at $\angle A$ and $BC=5cm$. If $BL = \displaystyle\frac{3\sqrt{5}}{2}$, then the length of $CM$ is,

- $2\sqrt{5}$cm
- $10\sqrt{2}$cm
- $5\sqrt{2}$cm
- $4\sqrt{5}$ cm

#### Solution 6.

The following figure represents the problem.

**Problem analysis:**

As we are dealing with medians and right triangles, we may reach the length of second median through stage by stage application of Pyhthagoras relationship.

**Problem solving:**

From the outer triangle,

$4a^2 + 4b^2 = 25$,

Or, $a^2 + b^2 = \frac{25}{4}$.

Fron $\triangle ABL$,

$4a^2 + b^2 = \frac{45}{4}$.

From $\triangle ACM$,

$CM^2 = 4b^2 + a^2$.

Adding the last two equations,

$CM^2 + \frac{45}{4} = 5(a^2 + b^2)$.

Substituting from the first equation,

$CM^2 + \frac{45}{4} = \frac{125}{4}$,

Or, $CM^2 = \frac{80}{4} = 20$,

Or, $CM = 2\sqrt{5}$cm.

**Answer:** a: $2\sqrt{5}$cm.

#### Problem 7.

In $\triangle ABC$, $D$ and $E$ are two points on the sides $AC$ and $BC$ respectively, such that $DE=18$cm, $CE=5$cm and $\angle DEC = 90^0$. If $tan \angle ABC = 3.6$, then $AC : CD$ is

- $2BC : CE$
- $BC : 2CE$
- $CE : 2BC$
- $2CE : BC$

#### Solution 7.

The following figure represents the problem.

A useful pattern recognition is,

$\tan C = \displaystyle\frac{18}{5} = 3.6 = \tan B$.

This makes $\angle B = \angle C$ and is the starting breakthrough in the problem.

By this break, the $\triangle ABC$ is transformed to an isosceles triangle with $AB=AC$.

The second stratagem we employ is to draw a perpendicular $AF$ on $BC$. The $\triangle ABC$ being an isosceles triangle, $AF$ is also a median and bisects BC.

But more importantly, $DE$ and $AF$ both being perpendiculars on the same line $BC$ these are parallel to each other which results in similarity of two triangles $\triangle AFC$ (first) and $\triangle DEC$ (second).

By equality of ratio of corresponding sides we have then,

$\displaystyle\frac{AC}{CD} = \frac{AF}{CE}$.

But as, $AF = \displaystyle\frac{1}{2}BC$, we have finally,

$\displaystyle\frac{AC}{CD}=\frac{BC}{2CE}$.

**Answer:** b: $\displaystyle\frac{BC}{2CE}$.

#### Problem 8.

$AB$ is a chord to a circle and $TAP$ is the tangent to the circle at $A$. If $\angle BAT = 75^0$ and $\angle BAC= 45^0$, where $C$ is a point on the circle, then $\angle ABC$ is,

- $40^0$
- $60^0$
- $70^0$
- $45^0$

#### Solution 8.

The following is a depiction of the problem graphically.

With $\angle BAT + \angle BAC = 75^0 + 45^0 = 120^0$ consumed out of $180^0$, the leftover $\angle CAP=60^0$.

But this angle being an external angle, it will be equal to any angle subtended by the chord $AC$ on the periphery.

Thus, $\angle ABC=60^0$.

**Answer:** b: $60^0$.

#### Problem 9.

If radii of two circles be 6cm and 3cm and the length of the transverse common tangent be 8cm, then the distance between the centers is,

- $\sqrt {140}$cm
- $\sqrt {145}$cm
- $\sqrt {135}$cm
- $\sqrt {150}$cm

#### Solution 9.

The following figure represents the problem description.

Given, transverse common tangent length, $AB = CD = 8$cm, radii $PC=6$cm and $QB=3$cm. Needed, distance between centers, $PQ = d = d_1 + d_2$.

Right triangles $\triangle QBX$ and $\triangle PAX$ are similar, as one more angle $\angle BXQ = \angle AXP$. The corresponding side ratios are then equal. As radii form a pair in ratio 2 : 1, the other two pairs of sides are also in ratio 2: 1. Specifically, $AX : BX = 2 : 1$. As also $AX + BX = 8cm$,

$AX = \displaystyle\frac{8}{3}$cm,

$BX = \displaystyle\frac{16}{3}$cm.

Using these values in $\triangle PAX$,

$d_1^2 = 6^2 + AX^2 = 36 + \displaystyle\frac{256}{9} = \displaystyle\frac{580}{9}$,

Or, $d_1 = \frac{2}{3}\sqrt{145}$.

As $d_2$ is half of $d_1$,

$d_2 = \frac{1}{3}\sqrt{145}$.

Thus $d = d_1 + d_2 = \sqrt{145}$.

**Answer:** b: $\sqrt{145}$.

**Note:** If you manage to remember, you can use a formula relating the length of the common transverse tangent and the distance between the centers. Length of the common transverse tangent,

$AB = \sqrt{d^2 - (r_1+ r_2)^2}$,

Or, $64 = d^2 - 81$,

Or, $d = \sqrt{145}$.

This relationship can be easily derived with a little bit of algebraic manipulation.

#### Problem 10.

Two circles with centers at $P$ and$Q$ intersect each other at $B$ and $C$. Two points on both sides of $C$ are placed such on the circles that $ACD$ are collinear with $A$ on the first circle. If $\angle APB = 130^0$, find $\angle BQD$.

- $130^0$
- $165^0$
- $65^0$
- $15^0$

#### Solution 10.

The following is the figure relevant to the problem.

The chord $AB$ subtends an angle of $130^0$ at the center and so will subtend an angle half of $130^0$, that is, $65^0$ at the periphery opposite to the center.

So $\angle ACB = 65^0$.

Moving into the second circle, we get the internal angle of the cyclic quadrilateral, $BCDE$ as, $\angle BCD = 180^0 - 65^0 = 115^0$.

In a cyclic quadrilateral, the opposite angles sum to $180^0$, So, opposite angle $\angle BED = 180^0 - 115^0 = 65^0$.

This is the angle subtended by the chord $BD$ at the periphery and so the same chord will subtend double the peripheral angle, that is, $2\times{65^0} = 130^0$ at the center.

Thus $\angle BQD = 130^0$.

**Answer:** a: $130^0$.

**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**

**How to solve difficult Geometry problems quickly in a few steps**

**How to solve intriguing SSC CGL level Geometry problem in a few steps 4**

**How to solve difficult SSC CGL Geometry problems in a few steps 3**

**How to solve difficult SSC CGL Geometry problems in a few steps 2**

**How to solve difficult SSC CGL Geometry problems in a few steps 1**

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