## Difficult geometry questions for SSC CGL Solved question set 95

Solve 10 difficult geometry questions for SSC CGL in 15 minutes in Set 95. Verify correctness from answers and learn to solve quickly from solutions.

The solved question set contains,

**Difficult geometry questions for SSC CGL**to be answered in 15 minutes (10 chosen questions)**Answers to the questions**, and,**Quick solutions**to the questions.

For best results, take the test first and then learn to solve quickly from solutions.

### 10 Difficult geometry questions for SSC CGL Set 95 - answering time 15 mins

**Q1.** In the given figure, $\angle BAC=70^0$, $\angle ACB=45^0$ and $\angle DEA=140^0$. What is the value of $\angle BDE$?

- $10^0$
- $25^0$
- $15^0$
- $20^0$

**Q2.** In $\triangle PQR$, the sides $PQ$ and $PR$ are produced to $A$ and $B$ respectively. The bisectors of $\angle AQR$ and $\angle BRQ$ intersect at point $O$. If $\angle QOR=50^0$, what is the value (in degrees) of $\angle QPR$?

- $50$
- $60$
- $100$
- $80$

**Q3.** In $\triangle ABC$, $AB=a-b$, $AC=\sqrt{a^2+b^2}$ and $BC=\sqrt{2ab}$. Find $\angle B$.

- $90^0$
- $60^0$
- $45^0$
- $30^0$

**Q4.** Sides of a right triangle $\triangle ABC$ are, $a$, $b$ and $c$, where $c$ is the hypotenuse. What will be the radius of the incircle of this triangle?

- $\displaystyle\frac{(a+c-b)}{2}$
- $\displaystyle\frac{(a+b+c)}{2}$
- $\displaystyle\frac{(b+c-a)}{2}$
- $\displaystyle\frac{(a+b-c)}{2}$

**Q5.** G is the centroid of the $\triangle ABC$, where AB, BC and CA are 7 cm, 24 cm and 25 cm respectively. Then BG is,

- $8\frac{1}{3}$ cm
- $6\frac{1}{3}$ cm
- $5\frac{1}{2}$ cm
- $4\frac{1}{6}$ cm

**Q6.** The area of an equilateral $\triangle ABC$ is $18\sqrt{3}$ sq. units. Find the value of the side length (in units) of the triangle.

- $\sqrt{2}$
- $6\sqrt{2}$
- $3\sqrt{2}$
- $2\sqrt{2}$

**Q7.** How many diagonals are there in an octagon?

- $24$
- $12$
- $20$
- $14$

**Q8.** If the side of a square is increased by $20\text{%}$, then what will be the percentage increase in the perimeter?

- $100$
- $20$
- $40$
- $50$

**Q9.** Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the greater circle which is outside the inner circle is of length,

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

**Q10.** In the given figure, ABCD is a rhombus and BCE is an isosceles triangle with $BC=CE$, $\angle CBE=84^0$, and $\angle ADC=78^0$. What is the value (in degrees) of $\angle DEC$?

- $20$
- $28$
- $36$
- $33$

### Answers to the 10 difficult geometry questions for SSC CGL Set 95

**Q1. Answer:** Option b: $25^0$.

**Q2. Answer:** Option d: $80$.

**Q3. Answer:** Option a: $90^0$.

**Q4. Answer:** Option d: $\displaystyle\frac{(a+b-c)}{2}$.

**Q5. Answer:** Option a: $8\frac{1}{3}$ cm.

**Q6. Answer:** Option b: $6\sqrt{2}$.

**Q7. Answer:** Option c: $20$.

**Q8. Answer:** Option b: $20$.

**Q9. Answer:** Option c: $4\sqrt{2}$ cm.

**Q10. Answer:** Option d: $33$.

### Solutions to 10 difficult geometry questions for SSC CGL Set 95 - answering time was 15 mins

**Q1.** In the given figure, $\angle BAC=70^0$, $\angle ACB=45^0$ and $\angle DEA=140^0$. What is the value of $\angle BDE$?

- $10^0$
- $25^0$
- $15^0$
- $20^0$

#### Solution 1: Quickest solution by external angle properties of a triangle

**Problem analysis and reasoning to know requirement:**

Given $\angle DEA=140^0$ being external to $\triangle BDE$, it will be equal to the sum of two internal opposite angles, one of which is our target $\angle BDE$.

So we need to find $\angle DBE$.

**External angle property of a triangle** states,

An external angle of a triangle is equal to the sum of two opposite internal angles.

**Problem solving:**

$\angle DBE$ is the external angle to the $\triangle ABC$. So,

$\angle DBE =70^0+45^0=115^0$.

And target angle,

$\angle BDE=140^0-115^0=25^0$.

This is the quickest solution based on basic concepts, problem analysis and requirement specification.

**Answer:** Option b: $25^0$.

**Key concepts used:** **Problem analysis -- Solution requirement specification-- External angle in a triangle -- Solving in mind.**

**Q2.** In $\triangle PQR$, the sides $PQ$ and $PR$ are produced to $A$ and $B$ respectively. The bisectors of $\angle AQR$ and $\angle BRQ$ intersect at point $O$. If $\angle QOR=50^0$, what is the value (in degrees) of $\angle QPR$?

- $50$
- $60$
- $100$
- $80$

#### Solution 2: Quickest solution by problem analysis, requirement specification and deductive reasoning using concept of angles at point of incidences on a straight line

The following graphic describes the problem.

**Problem analysis and requirement specification:**

There are many angles involved, So we need to get a clear idea about *exactly which angles are required to be found out to find target angle $\angle QPR$.*

We'll **reason backwards starting from the end requirement.**

To find the value of $\angle QPR$, in $\triangle PQR$, you have to find the value of the sum of other two angles $\angle x+\angle y$.

Finding individual values of $\angle x$ and $\angle y$ are not required, **only their sum is required.**

**Problem solving following the question thread, how to find $(\angle x+\angle y)$ starting now from what is given:**

By deductive reasoning, now explore how to find this sum *from given value of $\angle QOR$.*

Try to connect now from what is given to finding $(\angle x+\angle y)$.

The **link** is easy to see:

1. At incidence point Q on line PQA, $(\angle x+2\angle \alpha)=180^0$, and,

2. At incidence point R on line PRB, $(\angle y+2\angle \beta)=180^0$.

3. Add the two to get a relation between $(\angle x+\angle y)$ and $(\angle \alpha+\angle \beta)$.

4. From $\triangle OQR$ get the value of $(\angle \alpha+\angle \beta)$, hence the value of $(\angle x+\angle y)$ and finally value of target $\angle QPR$.

This is the * deductive reasoning chain in mind*, and the solution comes out to be $80^0$ in no time.

**Actual Step details**

Adding the two angle relations at incidence points Q and R,

$(\angle x+\angle y)=2[\pi-(\angle \alpha+\angle \beta)]$.

In $\triangle QOR$, $(\angle \alpha+\angle \beta)=\pi-50^0$.

Substituting,

$(\angle x+\angle y)=2\times{50^0}=100^0$, and

$\angle QPR=180^0-100^0=80^0$.

**Answer:** Option d: $80$.

**Key concepts used:** **Problem analysis -- Requirement specification -- Deductive reasoning -- Angles at point of incidence on a straight line -- Total of angles in a triangle -- Solving in mind.**

**Q3.** In $\triangle ABC$, $AB=a-b$, $AC=\sqrt{a^2+b^2}$ and $BC=\sqrt{2ab}$. Find $\angle B$.

- $90^0$
- $60^0$
- $45^0$
- $30^0$

#### Solution 3: Immediate solution by mapping well-known algebraic relation $(a-b)^2=a^2-2ab+b^2$ onto Pythagoras relation in a right triangle

The following graphic describes the problem.

Though the figure is not really required, it will help visualization.

By well known expansion of $(a-b)^2$,

$(a-b)^2=a^2-2ab+b^2=(a^2+b^2)-2ab$

$=(\sqrt{a^2+b^2})^2-(\sqrt{2ab})^2$

Or, $(\sqrt{a^2+b^2})^2=(a-b)^2+(\sqrt{2ab})^2$.

Relating the given values AB, AC and BC,

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

This is the relation between three sides by Pythagoras theorem in the right $\triangle ABC$ with AC as hypotenuse and right angle at $\angle B$.

**Answer:** Option a: $90^0$.

**Key concepts used:** **Domain mapping -- Relating algebraic expansion of $(a-b)^2$ with side lengths and hence with Pythagoras relation -- Solving in mind.**

**Q4.** Sides of a right triangle $\triangle ABC$ are, $a$, $b$ and $c$, where $c$ is the hypotenuse. What will be the radius of the incircle of this triangle?

- $\displaystyle\frac{(a+c-b)}{2}$
- $\displaystyle\frac{(a+b+c)}{2}$
- $\displaystyle\frac{(b+c-a)}{2}$
- $\displaystyle\frac{(a+b-c)}{2}$

#### Solution 4: Quick solution by equating sum of areas of three intriangles formed by incentre and the vertices with the whole triangle area and then Algebraic simplification aided by Pythagoras relation

We'll use the following graphic for explaining the solution.

#### Problem analysis, Key pattern identification and deductive reasoning

The very first problem that is to be overcome is,

To get a relation between inradius $r$ and the sides $a$, $b$ and $c$.

Inradius being the radius of the circle inscribed in the triangle,

All three sides are tangents to the circle, and theradii from centre I are perpendicular to the sides at tangent points.

This *knowledge visualization helps to identify the key pattern,*

In all three intriangles (as we call them), $\triangle ICB$, $\triangle IBA$ and $\triangle IAC$, radius $r$ is the height. As area of a triangle is half of base multiplied with height, if we sum up the areas of these three intriangles we'll get the area of the main $\triangle ABC$ and

a relation between $r$ and the three sides.

**Problem solving stage 1:**

As thought out, equating *sum of areas of three component intriangles to the area of the whole* $\triangle ABC$,

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

Or, $r=\displaystyle\frac{ab}{a+b+c}$.

But **this value is very dissimilar in form with any of the choice values.**

**Reasoning says**,

This value of inradius is certainly correct, and it must be equal to one of the choice values.

**Question is**, how to convert this inradius value to one of the choice values?

As the answer to the question, **reasoning asks,**

What more resources do we have in terms of relations among the sides?

Surely we have the **resource of** side relation, $a^2+b^2=c^2$ from **Pythagoras theorem** that has not been used yet.

This is the **second key pattern** that we'll use now to solve the problem in the final stage.

**Final solution:**

In right $\triangle ABC$ by Pythagoras theorem,

$a^2+b^2=c^2$,

Or, $(a^2+b^2-c^2)=0$.

Now we need **Algebraic simplification concepts,**

$(a+b+c)(a+b-c)=(a+b)^2-c^2$

$=(a^2+b^2-c^2)+2ab$

$=2ab$, as $(a^2+b^2-c^2)=0$,

Or, $\displaystyle\frac{ab}{(a+b+c)}=\displaystyle\frac{(a+b-c)}{2}$.

The fourth choice value is the answer then,

$r=\displaystyle\frac{ab}{a+b+c}=\frac{a+b-c}{2}$.

**Answer:** Option d: $\displaystyle\frac{(a+b-c)}{2}$.

**Key concepts used:** *Problem analysis -- Deductive reasoning -- Key pattern identification -- **Incentre concepts --** Mapping sum of three internal triangle areas to get a relation between inradius and the three sides -- Algebraic simplification -- Pythagoras theorem -- Solving in mind.*

Not so easy a problem that can nevertheless be solved quickly by problem analysis, mapping sum of three internal triangle areas to get inradius to sides relation and finally by algebraic simplification aided by Pythagoras relation.

In itself none of the ideas are difficult if you are able to use them* in conjunction by deductive reasoning.*

**Q5.** G is the centroid of the $\triangle ABC$, where AB, BC and CA are 7 cm, 24 cm and 25 cm respectively. Then BG is,

- $8\frac{1}{3}$ cm
- $6\frac{1}{3}$ cm
- $5\frac{1}{2}$ cm
- $4\frac{1}{6}$ cm

#### Solution 5: Quick solution by Key pattern identification of the median BGF in a right triangle dividing the triangle into two isosceles triangles and concept of segmentation of median at centroid

The following graphic helps to make the explanations clear.

First identify that line segment FE joining two mid-points of two sides of the $\triangle ABC$, the two triangles $\triangle ABC$ and $\triangle FEC$ **must be similar.**

#### Reason why $\triangle ABC$ is similar to $\triangle FEC$

In two triangles, $\triangle ABC$ and $\triangle FEC$, ratio of two pairs of sides are equal, as well as, common angles at vertex C are equal. This is SAS condition for similarity of two triangles.

$\displaystyle\frac{AC}{FC}=\frac{BC}{EC}=2$, and $\angle ACB=\angle FCE$.

Thus $\triangle ABC$ and $\triangle FEC$ are similar with all corresponding pairs of angles equal,

$\angle BAC=\angle EFC$, and $\angle CBA=\angle CEF=90^0$.

So **FE is perpendicular to BC at its mid-point E.**

This makes the* two triangles $\triangle FBE$ and $\triangle FCE$ congruent with three corresponding pairs of angles equal* (congruent because two pairs of sides equal in two right triangles, making the third side also equal, SSS condition for congruency).

And hence, $\triangle FBC$ is isosceles as $FB=FC$.

This is the **Primary breakthrough** in the problem.

So,

$BF=FC=AF=\frac{1}{2}AC=\displaystyle\frac{25}{2}$.

Final solution is now only a step away.

By the well-known concept of **segmentation of a median at the centroid** in the ratio of 2 : 1 with larger segment towards the vertex,

$BG=\frac{2}{3}BF=\displaystyle\frac{25}{3}=8\frac{1}{3}$ cm.

**Answer:** Option a: $8\frac{1}{3}$ cm.

**Key concepts used:** **Key pattern identification of median from right angle vertex dividing the right triangle into two isosceles triangles, this is a rich concept -- ***Triangle similarity conditions --*** Similarity of triangles -- Isosceles triangles -- ***Rich concept of median from right angle vertex --***Median section ratio at centroid**** -- Solving in mind.**

**Q6.** The area of an equilateral $\triangle ABC$ is $18\sqrt{3}$ sq. units. Find the value of the side length (in units) of the triangle.

- $\sqrt{2}$
- $6\sqrt{2}$
- $3\sqrt{2}$
- $2\sqrt{2}$

#### Solution 6: Quick solution by equilateral triangle area to side relation

Though not necessary for ease of visualization, the problem graphic is shown below.

AD is the height of the equilateral triangle perpendicular to base BC and bisecting it at D.

In right $\triangle ADB$ by Pythagoras relation,

$AD^2=a^2-\left(\displaystyle\frac{a}{2}\right)^2=\displaystyle\frac{3}{4}a^2$.

So, $AD=\displaystyle\frac{\sqrt{3}}{2}a$, and,

Area $A=\frac{1}{2}AD\times{a}=\displaystyle\frac{\sqrt{3}}{4}a^2=18\sqrt{3}$ sq units,

Or, $a^2=4\times{18}$,

Or, $a=6\sqrt{2}$ sq units.

**Answer:** Option b: $6\sqrt{2}$.

**Key concepts used:** **Area of equilateral triangle by side -- Pythagoras theorem -- Solving in mind.**

**Q7.** How many diagonals are there in an octagon?

- $24$
- $12$
- $20$
- $14$

#### Solution 7: Quick solution by forming the special pattern of diagonals in an octagon using common sense reasoning and visualization

If you know the formula it takes just a little time to get the answer, but we are not expected to remember all special types of formulas.

In this case we will derive the answer quickly enough using common sense experience and deductive reasoning. Together we call it common sense reasoning that most people can use effectively when the need arises.

The following graphic will help visualization and explanation.

Visualize in your mind's eye that you have selected one vertex A to draw all diagonals from it that can be drawn.

With 8 vertices of an octagon, you can connect one vertex to at most 7 other vertices. Would all of these 7 be diagonals?

By common sense reasoning you can easily deduces that,

Connections to two of its adjacent vertices are in fact edges and are not diagonals.

So total number of diagonals that you can draw from any vertex of an octagon is,

$7-2=5$.

For eight vertices, there will be then,

$8\times{5}=40$ diagonals.

Apply your common sense reasoning the second time,

A diagonal from A to F is counted twice for vertex A to F and also for vertex F to A.

So, actual number of diagonals in an octagon will be,

$\displaystyle\frac{40}{2}=20$.

**Answer:** Option c: $20$.

**Key concepts used:** **Visualization -- ****Key pattern identification -- Key method creation -- Common experience or Common sense -- Deductive reasoning -- Common sense reasoning -- ***Number of diagonals in a polygon -- ***Solving in mind.**

**Note:** This problem touches the interesting topic of Graph theory. With the ideas that you have gained, you can create and solve many interesting puzzles.

In general, if you take a polygon of $n$ vertices, each vertex may have $(n-2)$ diagonals.

So total number of diagonals will be,

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

You have just derived the formula for maximum number of diagonals possible to be drawn in a polygon of $n$ vertices from scratch.

**Q8.** If the side of a square is increased by $20\text{%}$, then what will be the percentage increase in the perimeter?

- $100$
- $20$
- $40$
- $50$

#### Solution 8: Immediate solution by perimeter of a square concept and percentage increase concept

The following diagram will help explanation.

Assume the side length of the square to be 5 cm. With 20% increase it increased one-fifth by 1 cm.

So each side increasing by 1 cm, perimeter increased by 4 cm. As the perimeter was 4 times 5 cm, 4 cm increase is equivalent to 1 cm increase for every 5 cm, just a 20% increase.

That's the perimeter increase concept of a square,

Whatever be the percentage increase of the side length of a square, its perimeter will also be increased by the same percentage.

**Answer:** Option b: $20$.

**Key concepts used:** **Perimeter of square -- ***Properties of a square -- ***Basic percentage concepts**** -- Solving in mind.**

**Q9.** Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the greater circle which is outside the inner circle is of length,

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

#### Solution 9: Quick solution by key pattern identification of excess overlapping radii, concept of biggest chord outside inner circle in a pair of internally touching circles and chord bisection property

The graphic below will help explanation. Ignore the dotted line segments for the time being.

Visualize the key pattern that the radii of the two circles overlap with an excess.

Diameter of the smaller internally touching circle $CD=4$ cm and radius of outer bigger circle is, $CQ=3$ cm.

So segment length exceeding overlap portion,

$QD=CD-QC=4-3=1$ cm.

The **biggest chord outside inner circle in a pair of internally touching circles** must fulfill the condition that,

It must be the tangent to the inner circle at the point on the common diameter of the two circles.

Chord AB fulfilling this condition is the biggest such chord in this problem.

By chord bisection property, as QD from centre Q of the bigger circle is perpendicular bisector of the chord, in right $\triangle QBD$ by Pythagoras theorem,

$BD^2=BQ^2-QD^2=9-1=8$,

Or, $BD=2\sqrt{2}$.

So, $AB=2BD=4\sqrt{2}$.

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

**Key concepts used:** *Key pattern identification of excess of overlapping radii of the two internally touching circles -- Chord length to perpendicular bisector from centre relation -- Biggest chord in internally touching circles -- Chord bisection property -- Pythagoras theorem.*

#### Mechanism of condition for biggest chord outside inner circle in a system of two internally touching circles

The biggest chord outside the internally touching circle must be nearest to the circle to be the biggest. That means, it must be a tangent to the internal circle.

But there may be infinite number of such tangents.

Which one should be the biggest?

Appreciate that,

The chord length of outer circle increases as the perpendicular bisector QD of the chord from its centre decreases and vice versa.

As QD gets smaller and smaller approaching zero, the length of the biggest chord approaches its maximum value equal to the diameter of the circle.

And in a system of a pair of internally touching circles, QD with D lying on the common diameter would be of smallest perpendicular bisector length so that the corresponding chord will be the longest such chord.

A counter-example is shown as tangent point M not lying on the common diameter. The perpendicular bisecting point L has shifted in this case away from D and went outside the inner circle.

It is much further away than D from centre Q so that it is much longer than QD and so the corresponding chord KN is much smaller than AB.

In fine,

Tangent point D, the nearest point to the inner circle lying on the common diameter is also nearest to centre Q of the outer circle resulting in the longest chord at D outside the inner circle.

**Q10.** In the given figure, ABCD is a rhombus and BCE is an isosceles triangle with $BC=CE$, $\angle CBE=84^0$, and $\angle ADC=78^0$. What is the value (in degrees) of $\angle DEC$?

- $20$
- $28$
- $36$
- $33$

#### Solution 10: Quick solution by identification of two isosceles triangles and rhombus angle properties

We'll use the same question graphic for explanation.

In a rhombus opposite corner angles being equal, $\angle ABC=\angle ADC=78^0$.

Value of each of the other two equal angles will be,

$\angle BCD=\frac{1}{2}(2\pi-2\times{78^0})=102^0$.

In isosceles $\triangle BCE$, with $BC=CE$ and $\angle CBE=84^0$,

$\angle BEC=84^0$, and,

$\angle BCE=180^0-2\times{84^0}=12^0$.

Adding,

$\angle DCE=102^0+12^0=114^0$.

But $\triangle DCE$ is also an isosceles one with $DC=BC=CE$. So the two base angles are equal,

$\angle DEC=\angle CDE=\frac{1}{2}(180^0-114^0)=33^0$.

**Answer:** Option d: $33$.

**Key concepts used:** **Identification of two isosceles triangles -- Total of three angles in a triangle -- Rhombus angle properties -- Solving in mind.**

### End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

**Concept based pattern and method formation**, and,**Identification of the key pattern and use of the method**associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the concept based pattern identification and use of quick problem solving method.

**Recommendation:** In Geometry problem solving, first represent the problem figure quickly as a draft figure. It need not be perfect at all. Then analyze the problem to identify the useful patterns and apply suitable methods. Identify what element value is required to be found out as final answer, and then work backwards.

If necessary, introduce new elements such as an angle, a side or a line segment. Or, if you find it faster, use trigonometric, algebraic or arithmetic concepts such ratios, basic algebraic relations without any hesitation. Your goal is to arrive at the correct answer quickly and accurately.

And one more important point to remember, you must not depend on guesswork in this math exam.

**Keep using the basic concepts** and **use the advanced concepts** or rules **only when** you require it.

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