SSC CGL Geometry Questions with Detailed Solutions

Given:

Distance between the centers of the two circles = 36 cm

Radius of the first circle = 15 cm

Radius of the second circle = 9 cm

Concept Used:

Length of the direct common tangent

​$$L_{\text{direct}} = \sqrt{d^2 - (r_1 - r_2)^2}$$​

Length of the transverse common tangent

​$$L_{\text{transverse}} = \sqrt{d^2 - (r_1 + r_2)^2}$$​

Solution:

Direct common tangent:

$$L_{\text{direct}} = \sqrt{36^2 - (15 - 9)^2}\\ \ \\L_{\text{direct}} = \sqrt{1296 - 6^2}\\ \ \\L_{\text{direct}} = \sqrt{1296 - 36}\\ \ \\L_{\text{direct}} = \sqrt{1260}$$ 

Transverse common tangent:

$$L_{\text{transverse}} = \sqrt{36^2 - (15 + 9)^2}\\ \ \\L_{\text{transverse}} = \sqrt{1296 - 24^2}\\ \ \\L_{\text{transverse}} = \sqrt{1296 - 576}\\ \ \\L_{\text{transverse}} = \sqrt{720}$$ 

Sum of the lengths:

$$\text{Sum} = L_{\text{direct}} + L_{\text{transverse}} = \sqrt{1260}+\sqrt{720}\\ \ \\\sqrt{2\times2\times3\times3\times5\times7}+\sqrt{2\times2\times2\times2\times3\times3\times5}$$ 

​$$=6\sqrt{7\times5}+12\sqrt5\\ \ \\=6\sqrt{5}(\sqrt7+2)$$​​

​$$\textbf{Given:} \\ \space \, \\ \text{Median PM length} = 15 \\ \space \, \\ \text{G is Centroid.} \\ \space \, \\ \textbf{Concept Used:} \\ \space \, \\ \text{Centroid divides median in 2:1 ratio (Vertex to Base).} \\ \space \, \\ \textbf{Solution:}$$​​

​$$PG = \frac{2}{3} \times PM \\ \space \, \\ PG = \frac{2}{3} \times 15 \\ \space \, \\ PG = 10 \text{ cm} \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ 10 \text{ cm}$$​​

Given:

Length of the chord on the smaller circle = 6 cm

Length of the chord on the larger circle = 18 cm

Concept Used:

For a chord of a circle, the distance from the center to the chord (perpendicular distance) and the radius of the circle are related by the Pythagorean theorem:

​$$R^2 - d^2 = \left(\frac{l}{2}\right)^2$$​​

where:

R is the radius of the circle

d is the perpendicular distance from the center to the chord

l is the length of the chord

Solution: 

For the smaller circle:

​$$r_1^2 - d^2 = \left(\frac{6}{2}\right)^2 = 9$$​​

For the larger circle:

​$$r_2^2 - d^2 = \left(\frac{18}{2}\right)^2 = 81$$​​

Subtract the equation for the smaller circle from the equation for the larger circle:

​$$r_2^2 - d^2 - (r_1^2 - d^2) = 81 - 9$$​​

​$$r_2^2 - r_1^2 = 72$$​​

The difference in the squares of the radii of the two circles is 72.

Given:

Two parallel chords ( AB) and ( CD) in a circle with center ( O)

Distance between the chords = 21 cm.

Length of ( AB) = 24 cm.

Length of ( CD) = 18 cm.

Solution:

d$$_1 = \sqrt{r^2 - \left(\frac{AB}{2}\right)^2} = \sqrt{r^2 - 12^2} = \sqrt{r^2 - 144}$$​​

d$$_2 = \sqrt{r^2 - \left(\frac{CD}{2}\right)^2} = \sqrt{r^2 - 9^2} = \sqrt{r^2 - 81}$$​​

d$$_1 - d_2 = 21$$​​

d$$_1 + d_2 = 21$$​​

​$$\sqrt{r^2 - 144} + \sqrt{r^2 - 81} = 21$$​​

a + b = 21

( a$$^2 - b^2 = (r^2 - 144) - (r^2 - 81) = -63)$$​​

(a - b)(a + b) = -63 $$\implies (a - b) \times 21 = -63 \implies$$​ a - b = -3

2a = 18 $$\implies$$​ a = 9

9 + b = 21 $$\implies$$​ b = 12

a = $$\sqrt{r^2 - 144} = 9 \implies r^2 - 144 = 81 \implies r^2 = 225 \implies$$​ r = 15 cm

Alternate Method:  

Radius = CO 

CO = $$\sqrt{(\frac{CD}2)^2+ (\frac{AB}2)^2}$$​​

CO = $$\sqrt{(\frac{18}2)^2+ (\frac{24}2)^2}$$ 

CO = $$\sqrt{(9)^2+(12)^2}$$ 

CO = $$\sqrt{81+144}$$​​

CO = $$\sqrt{225}$$​

CO = 15

Given:
Radius of the circumscribed circle (r) = 10 cm.
Formula Used:
For a regular hexagon inscribed in a circle, the side length of the hexagon (a) is equal to the radius of the circle (r).
Area of a regular hexagon = $$\frac{3\sqrt{3}}{2} \times a^{2}$$​​
​$$\sqrt{3} \approx 1.732$$​​
Solution:
Side length of the hexagon a = 10 cm.
Substitute a into the area formula:
​$$\text{Area} = \frac{3\sqrt{3}}{2} \times 10^{2}\\ \ \\\text{Area} = \frac{3\sqrt{3}}{2} \times 100\\ \ \\\text{Area} = 150\sqrt{3}\\ \ \\\text{Area} \approx 150 \times 1.73205\\ \ \\\text{Area}\approx 259.8075$$​​
Rounding to two decimal places, we get 259.81$$\text{ cm}^{2}$$​​
So the correct answer is (d).

​$$\textbf{Given:} \\ \space \, \\ PT:TQ=3:4, RT:TS=6:5. PQ=p, RS=q. \\ \space \, \\ \textbf{Concept Used:} \\ \space \, \\ \text{Intersecting Chords Theorem: } PT \times TQ = RT \times TS. \\ \space \, \\ \textbf{Solution:} \\ \space \, \\ 12x^2 = 30y^2 \implies x^2/y^2 = 2.5. \\ \space \, \\ p = 7x, q = 11y. \\ \space \, \\ p^2/q^2 = 49x^2 / 121y^2 = (49/121) \times 2.5 = 245/242. \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ 245/242$$​​

​$$\textbf{Given:}\\ \, \\\triangle MNO\\ \, \\LK \parallel NO\\ \, \\\text{Area of } \triangle MLK : \text{Area of trapezoid } LKNO = 9 : 40\\ \, \\\textbf{Concept Used:}\\ \, \\\text{Ratio of areas of similar triangles equals square of ratio of sides}\\ \, \\\textbf{Formula Used:}\\ \, \\\frac{\text{Area of smaller triangle}}{\text{Area of whole triangle}} = \left(\frac{\text{Corresponding side}}{\text{Whole side}}\right)^2\\ \, \\\textbf{Solution:}$$

​$$​​\text{Let area of } \triangle MLK = 9x\\ \, \\\text{Area of trapezoid } LKNO = 40x\\ \, \\\text{Area of } \triangle MNO = 49x\\ \, \\\frac{\text{Area of } \triangle MLK}{\text{Area of } \triangle MNO} = \frac{9}{49}\\ \, \\\left(\frac{ML}{MN}\right)^2 = \frac{9}{49}\\ \, \\\frac{ML}{MN} = \frac{3}{7}\\ \, \\LN = MN - ML\\ \, \\ML : LN = 3 : 4\\ \, \\\textbf{Final Answer:}\\ \, \\3 : 4$$​​

​Given :

In trapezoid ABCD, AB||CD.
Diagonals AC and BD intersect at E.

Formula Used :

Triangles formed between parallel sides and intersecting diagonals are similar.

For similar triangles:
​$$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{Corresponding side}_1}{\text{Corresponding side}_2}\right)^2$$​​

Solution :

Since AB$$\parallel C$$​D, triangles $$\triangle$$​ABE and $$\triangle$$​CDE are similar.

Corresponding bases are AB and CD.

Therefore,
​$$\frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle CDE}$$​​

​$$= \left(\frac{AB}{CD}\right)^2$$​​

Given:

( TP) is a tangent to the circle at point ( P)

( TQR ) is a secant intersecting the circle at points ( Q) and ( R)

​$$( \angle PTQ = 27^\circ )$$​ and $$( \angle TPQ = 55^\circ)$$  

Concept Used: 

Alternate Segment Theorem:

This theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment

Exterior Angle Theorem (Triangle):

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Tangent-Secant Theorem (Power of a Point):

For a point outside the circle with a secant and a tangent:

Solution: 

​$$\angle PRQ = \angle TPQ$$ (Angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment)​

$$\angle RQP = \angle QTP + \angle QPT \\\angle RQP = 55 + 27 = 82\\\angle QPR = 180 - (\angle RQP+\angle PRQ)\\\angle QPR = 180 - (82 + 55) \\\angle QPR = 180 - (137) \\\angle QPR = 43 \\2\angle QPR = \angle QOR \\ \angle QOR = 2 \times 43 \\\angle QOR = 86^\circ$$ 

Alternate Method: 

In  $$\triangle TPR$$ 

$$\angle T + \angle + P + \angle R = 180 \\ \ \\27 + (55 + A) + 55 = 180 \\ \ \\27 + 55 + A + 55 = 180 \\ \ \\A = 180 - 137\\ \ \\A = 43 \\ \ \\2A = 2 \times43 \\ \ \\2A = 86$$​

Given:

Radius of the circle = 7 units

Chords PQ = QR = 7 units 

Concept Used:

A Quadrilateral is rhombus if it has four equal side

Formula Used: ​

For a Rhombus ;

​$${D_1}^2 +{D_2}^2 = 4a^2$$ 

where, $$D_1 \ , \ \ D_2$$ are diagonals, a is side

Solution:   

PQ = QR = 7  unit

OP = OR = OQ = 7 unit

Four sides are equal , PQRO is a rhombus

OQ and PR is diagonal 

So, OQ² + PR² = 4 side² 

7² + PR² = 4×  7² 

PR² = 4× 7² - 7² ​

PR² =  7² ( 4 - 1)

 PR² =  7² × 3 

PR = $$7\sqrt3$$ = $$\frac{21}{\sqrt3}$$ unit​

​

Given:
Radius of the circle r = 20 cm
Length of chord  AB = 32 cm
Concept Used:​

​The tangent at any point of a circle is perpendicular to the radius through the point of contact.

The line joining the center of the circle to the point of intersection of the tangents bisects the chord.

Pythagorean theorem.

Solution:

Let O be the center of the circle. Let M be the midpoint of chord AB. OM is perpendicular to AB.

AM = MB = $$\frac{AB}{2} = \frac{32}{2} = 16$$​ cm.

In right triangle OMA, using the Pythagorean theorem:

OM² + AM² = OA²

OM² + 16² = 20²

OM² = 400 - 256 = 14

OM = 12 cm

Triangles OMA and OTA are similar. Therefore:

​$$\frac{AM}{TA} = \frac{OM}{OA}$$​​

​$$\frac{16}{TA} = \frac{12}{20}$$​​

TA = $$16 \times \frac{ 20}{ 12}$$​​

TA = $$\frac{80}{3}$$​= 26.666... cm $$\approx 26.67$$ cm​

Given:

Triangle ABC with side BC produced to point D.

AC = BC.

∠BAC = 70°.

Concept Used:

Properties of isosceles triangles: If two sides of a triangle are equal, the angles opposite those sides are also equal.

Angle sum property of a triangle: The sum of the angles in a triangle is 180°.

Linear pair: Angles on a straight line add up to 180°.

Solution:

Triangle ABC is isosceles:

Since AC = BC, ∆ABC is an isosceles triangle.

Therefore, ∠ABC = ∠BAC = 70°.

Using the angle sum property of a triangle:

∠BAC + ∠ABC + ∠ACB = 180°

70° + 70° + ∠ACB = 180°

∠ACB = 180° - 140° = 40°.

∠ACB and ∠ACD form a linear pair.

∠ACB + ∠ACD = 180°

40° + ∠ACD = 180°

∠ACD = 180° - 40° = 140°.

2.5∠ACD - 1.5∠ABC = (2.5 $$\times$$​ 140°) - (1.5 $$\times$$ 70°)​

= 350 - 105

= 245°.

The value of 2.5∠ACD - 1.5∠ABC is 245°.

Given:

Kite ABCD, where BD is the longer diagonal.

∠ABD = 54°

∠ADB = 50°

Solution:

Diagonals of a kite are perpendicular bisectors → BD ⊥ AC.

​$$\angle ABC = \angle ADC$$ That's Why $$\angle BDC = 54$$ , $$\angle DBC = 50$$ 

$$\angle ADC +\angle ABC = 2(50)+2(54)$$ =  100 + 108 = 208 ​

In △ABD, sum of interior angles is 180°:

∠ABD + ∠ADB + ∠BAD = 180°

50 + 54 + $$\angle BAD$$ = 180

$$\angle BAD$$ = 180 - 104 = 76 

ABCD = 360$$^\circ$$​​

​$$\angle A +\angle B+ \angle C +\angle D = 360$$ 

76 + 108 + $$\angle C$$ + 100 = 360 

​$$\angle C$$ = 360 - 284

76$$^\circ$$​​

Given:

In ΔPQR, PT is the median, meaning T is the midpoint of QR.

Concept Used: 

The Apollonius Theorem states that in any triangle, if a median is drawn, then:

PQ$$^2 + PR^2 = 2(PT^2 + QT^2)$$​

Thus, the correct Option (b) 

Proof of Apollonius' Theorem:

Coordinate System Approach

Let the coordinates of the points be:

P(0, h)

Q(-a, 0)

R(a, 0)

Since T is the midpoint of QR, its coordinates are: T(0, 0)

Distance Calculations

Using the distance formula:

$$PQ = \sqrt{(0 + a)^2 + (h - 0)^2} = \sqrt{a^2 + h^2}\\ \ \\PR = \sqrt{(0 - a)^2 + (h - 0)^2} = \sqrt{a^2 + h^2}\\ \ \\PT = \sqrt{(0 - 0)^2 + (h - 0)^2} = h\\ \ \\QT = TR = \frac{QR}{2} = \frac{2a}{2} = a\\ \ \\PQ^2 = a^2 + h^2\\ \ \\PR^2 = a^2 + h^2\\ \ \\PT^2 = h^2\\ \ \\QT^2 = a^2$$​

Adding both sides:

$$PQ^2 + PR^2 = (a^2 + h^2) + (a^2 + h^2) = 2a^2 + 2h^2 2\\ \ \\(PT^2 + QT^2) = 2(h^2 + a^2) = 2a^2 + 2h^2$$​

Since both sides are equal, we get:

​$$PQ^2 + PR^2 = 2(PT^2 + QT^2)$$​

Given:

AOC is the diameter of the circle, meaning ∠AOC = 180$$^\circ$$​​

B is a point on the circumference such that arc AB is $$\frac{1}{5}$$​ of arc BC

Concept Used:

​The central angle subtended by an arc is equal to the degree measure of the arc.

Solution: 

$$\angle$$AOC = 180(straight line angle)

​$$\angle AOC = 180^\circ =$$ 5x + x

6x = 180 

x = 30 

So, $$\angle BOC = 5x$$​​

5x = $$5 \times 30=$$​150$$^\circ$$