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SSC CGL level Solution Set 20, Geometry 2

Learn to solve Geometry questions for SSC CGL Set 20

Solution Set SSC CGL 20 Geometry questions on Cicle Cyclic quadrilateral

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SSC CGL Question Set 20, Geometry 2.

Learn to solve Geometry questions for SSC CGL Set 20 - answering time was 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,

  1. $140^0$
  2. $165^0$
  3. $120^0$
  4. $110^0$

Solution 1.

SSC CGL Solution Set 20 geometry2 q1

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,

  1. 5, 10
  2. 6, 12
  3. 7, 14
  4. 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,

  1. 4cm
  2. 12cm
  3. 6cm
  4. 9cm

Solution 3.

SSC CGL Solution Set 20 geometry2 q3

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. 1 : 2
  2. 2 : 1
  3. 3 : 1
  4. 1 : 1

SSC CGL Solution Set 20 geometry2 q4

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,

  1. 15cm
  2. 5cm
  3. 10cm
  4. 20cm

Solution 5.

SSC CGL Solution Set 20 geometry2 q5

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,

  1. $6\displaystyle\frac{5}{7}$ cm
  2. $3\displaystyle\frac{5}{7}$ cm
  3. $12\displaystyle\frac{2}{7}$ cm
  4. $10\displaystyle\frac{2}{7}$ cm

Solution 6.

SSC CGL Solution Set 20 geometry2 q6

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. $1 : 1$
  2. $2 : 1$
  3. $\sqrt{2} : 1$
  4. $1 : 2$

SSC CGL Solution Set 20 geometry2 q7

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,

  1. 4cm
  2. 5cm
  3. 3cm
  4. 2cm

Solution 8.

SSC CGL Solution Set 20 geometry2 q8

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,

  1. 7, 10
  2. 7, 6
  3. 12, 6
  4. 8, 9

Solution 9.

SSC CGL Solution Set 20 geometry2 q9

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,

  1. 3cm
  2. 4cm
  3. 5cm
  4. 6cm

SSC CGL Solution Set 20 geometry2 q10

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.

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