Geometry: Solved Examples- 1

# Geometry: Solved Examples- 1 | CSAT Preparation - UPSC PDF Download

Question 1:  x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.
A. 2
B. 3
C. 1
D. More than one possible value of z exists

Explanation. xy = 4
xy could be 2 * 2 or 4 * 1
221
222 These are the possible triangles.
223
441
22x will be a triangle if x is 1, 2 or 3 (trial and error).
44x is a triangle only if x is 1.
1. 221 is acute. 12 + 22 > 22
2. 222 is equilateral. So acute.
3. 223 is obtuse. 22 + 22 < 32
4. 144 is acute. 12 + 42 > 42
The question is " x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z."
And so, the third side has to be 3
Choice B is the correct answer.

Question 2How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?
A. 11
B. 6
C. 17
D. 23

Explanation.
Two possibilities: Two equal sides could add up to 12 or sum of 2 unequal sides = 12.
i.e. Sum of 2 equal sides = 12
Sum of 2 unequal sides = 12
⇒ If sum of two equal sides were 12, sides of the triangle should be 6, 6, x.
What are the values x can take?
x could range from 1 to 11.
11 integer values exist.
Now 2 unequal sides adding to 12. This could be 1 + 11, 2 + 10, 3 + 9, 4 + 8 or 5 + 7.
How many isosceles triangles are possible with the above combinations?
Isosceles triangles with the above combination:

 1, 11, 11    ☑ 1, 1, 11     ☒ 2, 10, 10    ☑ 2, 2, 10     ☒ 3, 9, 9        ☑ 3, 3, 9        ☒ 4, 8, 8        ☑ 4, 4, 8        ☒ 5, 7, 7        ☑ 5, 5, 7        ☑

6 possibilities. Triplets such as (1, 1, 11), (2, 2, 10), etc are eliminated as sum of the two smaller values is less than the largest value. These cannot form a triangle.
The question is " How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?"
11 + 6 = 17 possibilities totally.
Choice C is the correct answer.

Question 3: Sides of a triangle are 6, 10 and x for what value of x is the area of the △ the maximum?

A. 8 cms

B. 9 cms

C. 12 cms

D. None of these

Explanation.
Side of a triangle are. 6, 10, x.
Area = 1/2 * 6 * 10 * sin∠BAC.
Area is maximum, when ∠BAC = 90◦
x = √(100+36) = √136
There is a more algebraic method using hero’s formula. Try that also.
The question is "Sides of a triangle are 6, 10 and x for what value of x is the area of the △ the maximum?"
Hence, the answer is None of these.
Choice D is the correct answer.

Question 4Two circles are placed in an equilateral triangle as shown in the figure. What is the ratio of the area of the smaller circle to that of the equilateral triangle

A. π:36√3
B. π:18√3
C. π:27√3
D. π:42√3

Explanation.

In-radius of equilateral triangle of side a
Diameter of larger circle
Let us say common tangent PQ touches the two circle at R, center of smaller circle is I.
Now, PQ is parallel to BC. AR is perpendicular to PQ. Triangle PQR is also an equilateral triangle and AORID is a straight line. (Try to establish each of these observations. Just to maintain the rigour.)

Area of smaller circle = πr2

The question is " What is the ratio of the area of the smaller circle to that of the equilateral triangle?"
Choice C is the correct answer.

Question 5: Perimeter of a △ with integer sides is equal to 15. How many such triangles are possible?
A. 7
B. 6
C. 8
D. 5

Explanation.
Let us assume a ≤ b ≤ c.
a = 1, Possible triangle 1, 7, 7
a = 2, possible triangle 2, 6, 7
a = 3, possible triangles 3, 6, 6 and 3, 5, 7
a = 4, possible triangles 4, 4, 7 and 4, 5, 6
a = 5, possible triangle is 5, 5, 5
There are totally 7 triangles possible.
The question is "Perimeter of a △ with integer sides is equal to 15. How many such triangles are possible?"
Choice A is the correct answer.

Question 6There is an equilateral triangle with a square inscribed inside it. One of the sides of the square lies on a side of the equilateral △. What is the ratio of the area of the square to that of the equilateral triangle?
A. 12 : 12 + 7√3
B. 24 : 24 + 7√3
C. 18 : 12 + 15√3
D. 6 : 6 + 5√3

Explanation.

APQ is an equilateral △. As PQ is parallel to BC.
Let side of the square be ‘a’
AP = a = AQ
△QRC has angles 30 – 60 – 90.

Area of equilateral

Ratio of area of square to that of equilateral △ is
= 12 : 12 + 7√3
The question is "What is the ratio of the area of the square to that of the equilateral triangle?"
Hence, the answer is 12 : 12 + 7√3.
Choice A is the correct answer.

Question 7: △ABC has integer sides x, y, z such that xz = 12. How many such triangles are possible?
A. 8
B. 6
C. 9
D. 12

Explanation.
xz = 12
x,z can be 1, 12 or 2, 6 or 3, 4
Possible triangles
1 - 12 - 12
2 - 6 - 5; 2 - 6 - 6; 2 - 6 - 7
3 - 4 - 2; 3 - 4 - 3; 3 - 4 - 4; 3 - 4 - 5; 3 - 4 - 6.
The question is "△ABC has integer sides x, y, z such that xz = 12. How many such triangles are possible?"
Hence, the answer is 9 triangles.
Choice C is the correct answer.

Question 8: ABCDE is a regular pentagon. O is a point inside the pentagon such that AOB is an equilateral triangle. What is ∠OEA?

A. 66°
B. 48°
C. 54°
D. 72
°

Explanation.

Join OE and OD.
Internal angle of regular pentagon = 108°
∠EAB = ∠EDC = 108°
∠OAB = 60°
∠EAO = 48°
AO = OB = AB as the triangle is equilateral.
AB = AE as this is a regular pentagon.
Triangle AEO is isosceles as AO = EA.
∠AEO = ∠AOE = x (say)
In triangle AEO,
∠OAE + 2x = 180°
48° + 2x = 180°
2x = 132°, or x = 66°
The question is " What is ∠OEA?"
∠OEA = 66°
Choice A is the correct answer.

Question 9△ has sides a2, b2 and c2. Then the triangle with sides a, b, c has to be:
A. Right-angled
B. Acute-angled
C. Obtuse-angled
D. Can be any of these three

Explanation.
Assume a ≤ b ≤ c,
we have a2 + b2 > c2.
This implies the triangle with sides a, b, c has to be acute-angled.
The question is "△ has sides a2, b2 and c2. Then the triangle with sides a, b, c has to be:"
Hence, the answer is Acute-angled .
Choice B is the correct answer.

Question 10Consider a right–angled triangle with inradius 2 cm and circumradius of 7 cm. What is the area of the triangle?
A. 32 sq cms
B. 31.5 sq cms
C. 32.5 sq cms
D. 33 sq cms

Explanation.
r = 2
R = 7 (Half of hypotenuse)
Hypotenuse = 14
r = (a + b - h)/2
2 = (a + b - 14)/2
a + b - 14 = 4
a + b = 18
a2 + b2 = 14 2
a2 + (18 - a)2 = 14 2
a2 + 324 + a2 - 36a = 196
2a2 - 36a + 128 = 0
a2 - 18a + 64 = 0
Now the 2 roots to this equation will effectively be a, 18 – a. Product of the roots = 64.
Area = 1/2 * product of roots (How? Come on, you can figure this out.)
= 32 sq. cms
The question is "What is the area of the triangle?"
Hence, the answer is 32 sq. cms
Choice A is the correct answer.

Question 11: What is the ratio of longest diagonal to the shortest diagonal in a regular octagon?

A. √3 : 1

B. 2 : 1

C. 2 : √3

D. √2 : 1

Explanation.

Consider regular octagon ABCDEFGH
Its longest diagonal would be AE or BF or CG or DH.
Let us try to find out AE.
PQ = a
AP = QD
a2 = BP2 + AP2 => a2 = 2 AP2 {since BP=AP}
a = √2AP => AP = a/(√2)
AD =AP + PQ + QD = a/(√2) + a + a/(√2)
⇒ a + a√2
AE2 = (a + a√2) 2 + a2
AE2 = (a2 + 2 * a * √2 + 2a2) + a2
AE2 = a2 (1 + 2√2 + 2) + a2
⇒ a2 (4 + 2√2)
Shortest diagonal = AC or CE
AC2 = AB2 + BC2 – 2AB × BC cos135 degree
(Alternatively, we can deduce this using AC2 = AQ2 + QC2. We use cosine rule just to get some practice on a different method.)
= a2 + a2 – 2a2 * ((−1)/√2)
= 2a2 + √2a2
= a2 (2 + √2)
AE2 = a2 (4 + 2√2)
AE2/AC2 = a2 (4 + 2√2) / a2 (2 + 2√2) = 2
AE / AC = √2
Remember, for a regular octagon.
Each internal angle = 135 degrees.
Each external angle = 45 degrees.
So, we get a bunch of squares and isosceles right–angled △s if we draw diagonals.
A regular hexagon breaks into equilateral triangles. A regular octagon breaks into isosceles right angled triangles.

The question is "What is the ratio of longest diagonal to the shortest diagonal in a regular octagon?"

##### Hence, the answer is √2 : 1

Choice D is the correct answer.

Question 12: Find the altitude to side AC of triangle with side AB = 20 cm, AC = 20 cm, BC = 30 cm.

A. 10√7

B. 8√7

C. 7.5√7

D. 15√7

Explanation.

This is an isosceles triangle. So, let us find the altitude to BC first.

{RHS congruence; AD is common, AB = AC}
DC = 15
AD2 + DC 2 = AC2
AD2 + 15 2 = 202
Area of the triangle

h = 7.5√7

The question is " Find the altitude to side AC of triangle with side AB = 20 cm, AC = 20 cm, BC = 30 cm."

##### Hence, the answer is 7.5√7.

Choice C is the correct answer.

Question 13: ABCDEF is a regular hexagon inscribed inside a circle. If the shortest diagonal of the hexagon is of length 3 units, what is the area of the shaded region.

A. 1/6(3π − (9√3)/2)

B. 1/6(2π − (6√3)/2)

C. 1/6(3π − (8√3)/2)

D. 1/6(6π − (15√3)/2)

Explanation.

Let side of regular hexagon be a.
The shortest diagonal will be of length a√3. Why?
A regular hexagon is just 6 equilateral triangles around a point. The shortest diagonal is FD.
FD = FP + PD
△FOE is equilateral and so is △ EOD.
Diagonal FD can be broken as FP + PD, both of which are altitude of equilateral s.
FP = (√3a)/2
FD = √3a = shortest diagonal
The question tells us that the shortest diagonal measures 3 cm.
√3a = 3 ⇒ a = √3
Area of hexagon = (√3a2)/4 * 6
Area of circle – area of hexagon = π (√3)2 − √3/4 * (√3)2 * 6
= 3π − (9√3)/2
Area of shaded region = 1/(6) (area(circle) – area(hexagon))
= 1/(6)(3π − (9√3)/2)

The question is "what is the area of the shaded region?"

##### Hence, the answer is 1/(6)(3π − (9√3)/2)

Choice A is the correct answer.

Question 14: A circle of radius 5 cm has chord RS at a distance of 3 units from it. Chord PQ intersects with chord RS at T such that TS = 1/3 of RT. Find minimum value of PQ.

A. 6√3

B. 4√3

C. 8√3

D. 2√3

Explanation.

OM = 3, OS = 5
MS = 4 = RM {Using Pythagoras theorem}
⇒ RS = 8 cms
TS = 1/3 of RT
TS = 1/4 of RS
TS = 2 cms
RT * TS = PT * TQ
{Intersecting Chords theorem: When there are two intersecting chords, the product of the rectangle formed by the segments of one chord is equal to the product of the rectangle formed by the segments of the other.}
6 × 2 = PT * TQ
PT * TQ = 12
By AM – GM inequality, (PT+TQ)/2 ≥ √(PT * TQ)
(PT+TQ)/2 ≥ √12
PT + TQ ≥ 2√12
⇒ PQ ≥ 2√12
Or PQ ≥ 4√3
Minimum PQ = 4√3

The question is " Find minimum value of PQ."

##### Hence, the answer is 4√3.

Choice B is the correct answer.

Question 15: Triangle has perimeter of 6 + 2√3 . One of the angles in the triangle is equal to the exterior angle of a regular hexagon another angle is equal to the exterior angle of a regular 12-sided polygon. Find area of the triangle.

A. 2√3

B. √3

C. √3/2

D. 3

Explanation.

Given, Perimeter = 6 + 2√3
One of the angles in the triangle is equal to the exterior angle of a regular hexagon which is equal to 60°
Another angle is equal to the exterior angle of a regular 12-sided polygon = 30°.
From this we can deduce that the other angle is equal to 90°.
The property of a 60-30-90 triangle is that, the sides are in the ratio √3x, x and 2x.
Therefore, Perimeter is sum of all sides = x(3+ √3)= 6 + 2√3 .
⇒ x = (6 + 2√3)/(3+ √3) = 2.
Therefore, the sides are 2√3, 2 and 4.
Area of a Right Triangle = 1/2 * Product of Perpendicular sides = 1/2 * 2 * 2√3 = 2√3 .

The question is "Perimeter of a △ with integer sides is equal to 15. How many such triangles are possible?"

##### Hence, the answer is 2√3.

Choice A is the correct answer.

Question 16: Area of a Rhombus of perimeter 56 cms is 100 sq cms. Find the sum of the lengths of its diagonals.

A. 33.40

B. 34.40

C. 31.20

D. 32.30

Explanation.

Given, Perimeter = 56 and area = 100.
Let the side of the rhombus be “a”, then 4a = 56 => a = 14.
Area of Rhombus = Half the product of its diagonals. Let the diagonals be d1 and d2 respectively.
1/2 * d1 * d2 = 100 => d1 * d2 = 200. By Pythagoras theorem, (d1)2 + (d2)2 = 4a2
⇒ (d1)2 + (d2)2 = 4*196 = 784.
(d1)2 + (d2)2 + 2d1 * d2 = (d1+ d2)2 = 784 +2*200 = 1184 => (d1+ d2) = √1184 = 34.40
Therefore, sum of the diagonals is equal to 34.40 cm .

The question is " Find the sum of the lengths of its diagonals."

##### Hence, the answer is 34.40.

Choice B is the correct answer.

Question 17Rhombus has a perimeter of 12 and one angle = 120°. Find its area.

A. 9 * (√3)/2

B. 3 * (√3)/2

C. 9 * √3

D. 18 * √3

Explanation.

Perimeter = 12.
Let the side of the rhombus be “a”, then 4a = 12 => a = 3.
One angle = 120°.
Adjacent angles of a rhombus are supplementary. Therefore, the other angle = 60°.
Diagonals of a rhombus bisect each other, therefore, ∠DAC = 60° ∠DCA =60° ∠BAC =60°
Therefore, Triangle DAC and BAC are equilateral triangles.
Therefore, Area of Rhombus = 2 * Area of the Equilateral Triangle
= 2 * (√3/4) * a2 = (√3/2) * 9 = 9 * (√3/2).

The question is " Find the area of the rhombus"

##### Hence, the answer is 9 * (√3/2).

Choice A is the correct answer.

Question 18: Circle with center O and radius 25 cms has a chord AB of length of 14 cms in it. Find the area of triangle AOB

A. 144 cm2

B. 121 cm2

C. 156 cm2

D. 168 cm2

Explanation.

Given :
Radius of the circle = 25 cm
Center is O
Length of Chord AB = 14 cm
Therefore, △ AOB will be an isosceles triangle with 2 sides OA and OB as 25 cm and side AB as 14 cm.
For an isosceles triangle with the equal sides as 'a' and base as 'b' then

Hence Area

The question is "Find the area of triangle AOB."

##### Hence, the answer is 168 cm2.

Choice D is the correct answer.

Question 19: Two mutually perpendicular chords AB and CD intersect at P. AP = 4, PB = 6, CP = 3. Find radius of the circle.

A. 31.25(1/2)

B. 37.5(1/2)

C. 26(1/2)

D. 52(1/2)

Explanation.

When 2 chords AB and CD intersect at P then AP * PB = CP * PD
Hence 4 * 6 = 3 * PD
Thus, PD = 8
Now AB = AP + PB = 10
And CD = CP + PD
Thus, CD = 11
Consider the circle with center O.
Drop a perpendicular from O to chord AB and CD.
This will bisect the chords at X and Y i.e AX=XB and CY = YD.
Here AX = AP + PX i.e 5 = 4 + PX
PX = 1
Similarly, since CD = 11, PY+CP+YD = 11,
⇒ PY = 11-3-5.5 = 2.5.
PY = OX and PX = OY.
So, PXOY will from a rectangle as seen in the figure.
Now consider the triangle BOX, it is a right triangle where OB is the radius.
XB = 5, OX = 2.5
Then OB = ( OX2 + XB2 )1/2 OB = 31.251/2 Thus radius = 31.251/2 The radius can also be found out using the triangle YOD.

The question is "Find radius of the circle."

##### Hence, the answer is 31.251/2.

Choice A is the correct answer.

Question 20: Triangle ABC has angles A = 60° and B = 70°. The incenter of this triangle is at I. Find angle BIC.

A. 90°

B. 130°

C. 80°

D. 120°

Explanation.

A + B + C = 180 in a triangle
Hence C = 180 – 130 = 50°
Incenter is the meeting point of angle bisectors. BIC will form a triangle.
In this triangle B = 35, C = 25
Hence BIC or I = 180 – (35 + 25) = 120°

The question is "Find angle BIC."

##### Hence, the answer is 120°.

Choice D is the correct answer.

Question 21: Rhombus of side 6 cm has an angle equal to the external angle of a regular octagon. Find the area of the rhombus.

A. 18√2 cm2

B. 9√2 cm2

C. 15√2 cm2

D. 12√2 cm2

Explanation.

Side of the rhombus 's' = 6 cm
Angle between sides 'a' = exterior angle of an octagon
Therefore a = 45 degrees
Area of rhombus = sin a * s2
Area = (1/√2) * 36
Area = 18√2 cm2

The question is "Find the area of the rhombus."

##### Hence, the answer is 18√2 cm2.

Choice A is the correct answer.

Question 22: A circle inscribed in a square of side 2 has an equilateral triangle inscribed inside it. What is the ratio of areas of the equilateral triangle to that of the square?

A. 9√3 : 16

B. 3√3 : 4

C. 9√3 : 4

D. 3√3 : 16

Explanation

The radius of the circle is 1 unit.
If we considered the side of the equilateral triangle to be ‘a’, the circumradius of an equilateral triangle is

Therefore, a = √3
So, ratio of areas of the equilateral triangle to that of the square is

3√3 : 16

The question is "What is the ratio of areas of the equilateral triangle to that of the square? "

##### Hence, the answer is 3√3 : 16 Units

Choice D is the correct answer.

Question 23: An acute-angled isosceles triangle has two of its sides equal to 10 and 16. Find the area of this triangle.

A. √231 units

B. 12√66 units

C. 24 units

D. 5√231 units

Explanation.

The third side is either 10 or 16 and the sum of the squares of the two smaller sides must be greater than the square of the third side.
102 + 102 < 162
Therefore, the third side of the triangle is 16.
Semiperimeter, S = (10 + 16 + 16)/2 = 21
Area = √(21(21−16)(21−16)(21−10))
= √(21 * 5 * 5 * 11)
= 5√231 units

The question is "Find the area of this triangle."

##### Hence, the answer is 5√231 units.

Choice D is the correct answer.

Question 24: Three equal circles are placed inside an equilateral triangle such that any circle is tangential to two sides of the equilateral triangle and to two other circles. What is the ratio of the areas of one circle to that of the triangle

A. π : (6+4√3)

B. 3π : (6+4√3)

C. 2π : (6+4√3)

D. π : (6+2√3)

Explanation.

Let radii of each circle be ‘a’.
Distance PQ = 2a.
Draw PM perpendicular to AB
∠MAP = 30°

MP = a , AM = a3
After this we are done.
Area of circle = πa2
Area of equilateral triangle

= a2 (6+4√3)
Ratio = π : (6+4√3)

The question is "What is the ratio of the areas of one circle to that of the triangle?"

##### Hence, the answer is π : (6+4√3)

Choice A is the correct answer.

Question 25: There is an equilateral triangle with a square inscribed inside it. One of the sides of the square lies on a side of the equilateral △. What is the ratio of the area of the square to that of the equilateral triangle?

A. √3 : (5 + 4√3)

B. 2√3 : (7 + 4√3)

C. 4√3 : (7 + 4√3)

D. 4√3 : (5 + 2√3)

Answer. 4√3 : (7 + 4√3)

Explanation.

Let side of square be ‘a’
MN = LP = a
∠ABC = 60°

ML = a. BL  (Use some trigonometry for this)

Area of square = a2
Area of equilateral triangle

Ratio = 4√3 : (7 + 4√3)

The question is "What is the ratio of the area of the square to that of the equilateral triangle?"

##### Hence, the answer is 4√3 : (7 + 4√3)

Choice C is the correct answer.

Question 26: Consider Square S inscribed in circle C, what is the ratio of the areas of S and Q? And, Consider Circle C inscribed in Square S, what is the ratio of the areas of S and Q?

A. 2:π, 4:π

B. 4:π, 2:π

C. 1:π, 4:π

D. 2:π, 1:π

Explanation.

If square is inside the circle, ratio of areas of square to that of circle is 2 : π.

If circle is inside square, ratio of areas of square to that of circle is 4 : π.
Remember, circle area goes with π, square area goes with number.

If square is inside the circle, ratio of areas of square to that of circle is 2 : π.
If circle is inside square, ratio of areas of square to that of circle is 4 : π.

The question is "What is the ratio of the areas of S and Q? and what is the ratio of the areas of S and Q?"

##### Hence, the answer is 2:π, 4:π.

Choice A is the correct answer.

Question 27: Consider equilateral triangle T inscribed in circle C, what is ratio of the areas of T and C? Consider Circle C inscribed in equilateral triangle T, what is ratio of the areas of T and C?

A. 3√3:π , 3√3:16π

B. 3√3:4π , 3√3:π

C. √3:π , 3√3:4π

D. √3:π , √3:16π

Explanation.

For any equilateral triangle of side ‘a’

Area of equilateral triangle

Area of smaller circle

Area of larger circle

When Equilateral triangle T inscribed in circle C,
Ratio of Areas of T and C :: 3√3:4π
When Circle C inscribed in equilateral triangle T,
Ratio of Areas of T and C :: 3√3:π

The question is "Consider equilateral triangle T inscribed in circle C, what is ratio of the areas of T and C? Consider Circle C inscribed in equilateral triangle T, what is ratio of the areas of T and C? "

##### Hence, the answer is 3√3:4π , 3√3:π

Choice B is the correct answer.

Question 28: Consider Regular Hexagon H inscribed in circle C, what is ratio of the areas of H and C? Consider Circle C inscribed in Regular Hexagon H, what is ratio of the areas of H and C?

A. 2√3 : 3π , 3√3 : 4π

B. 3√3 : π , 3√3 : 4π

C. 3√3 : 2π, 2√3 : π

D. √3 : π , √3 : 4π

Answer. 3√3 : 2π, 2√3 : π

Explanation.

Inradius = altitude of equilateral triangle of side a = √3a/2

Area of regular hexagon = 6 * √3a2/4
Area of smaller circle = π ∗ √3a/2 ∗ √3a/2 = 3πa2/4
Area of larger circle = πa2
When Hexagon H inscribed in circle C,
Ratio of Areas of H and C :: 3√3 : 2π
When Circle C inscribed in Regular Hexagon H,
Ratio of Areas of H and C :: 2√3 : π

The question is "Consider Regular Hexagon H inscribed in circle C, what is ratio of the areas of H and C? Consider Circle C inscribed in Regular Hexagon H, what is ratio of the areas of H and C?"

##### Hence, the answer is 3√3 : 2π, 2√3 : π.

Choice C is the correct answer.

Question 29: What is the distance between the orthocentre and the circumcenter of a triangle who sides measure 24 cm, 26 cm and 10 cm?

A. 13 cm

B. 12 cm

C. 7.5 cm

D. √30 cm

Explanation.

The sides measure 24 cm, 26 cm and 10 cm. 10, 24, 26 is a Pythagorean triple!
So it is a right angled triangle we are talking about.
Draw the perpendicular bisectors to get the circumcenter. Orthocenter is the point where all altitudes meet.
In a right angled triangle it is the vertex that makes 90 deg angle.

Now it is easy to find OC! It is the length of the diagonal of the rectangle formed.
Length = 12 cm, breadth = 5 cm
Diagonal = √(122+52) = 13

The question is "What is the distance between the orthocentre and the circumcenter of a triangle who sides measure 24 cm, 26 cm and 10 cm?"

##### Hence, the answer is 13 cm.

Choice A is the correct answer.

Question 30: Two circles with centres O1 and O2 touch each other externally at a point R. AB is a tangent to both the circles passing through R. P’Q’ is another tangent to the circles touching them at P and Q respectively and also cutting AB at S. PQ measures 6 cm and the point S is at distance of 5 cms and 4 cms from the centres of the circles. What is the area of the triangle SO1O2?

A. 9 cm2

B. 3(4+√7)/2 cm2

C. 27/2 cm2

D. (3√41)/2 cm

Explanation.

From the diagram we see that SP, SR are tangents to circle1 from same point S. Similarly SR, SQ are tangents from same point to circle 2.
SP = SR; SQ = SR implies SP = SQ
Given PQ = 6cm
SP + SQ = 6
Therefor SR = SP = SQ = 3 cm.
SR is the altitude to the triangle SO1O2. We need to find the length of the base O1O2 to determine the area.
O1RS is a right angled triangle with hypotenuse = 5 and one side = 3
Therefore, O1R = √(52 - 32) = 4cm
Similarly, O2RS is a right angled triangle with hypotenuse = 4 and one side = 3
Therefore, O2R = √(42 - 32) = √7cm
O1O2 = O1R + O2R = 4 + √7
Area of the triangle SO1O2 = 1/2 * SR * O1O2 = 1/2 * 3 * (4 + √7) cm2

The question is " What is the area of the triangle SO1O2?"

##### Hence, the answer is 3(4+√7)/2 cm2.

Choice B is the correct answer.

The document Geometry: Solved Examples- 1 | CSAT Preparation - UPSC is a part of the UPSC Course CSAT Preparation.
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## FAQs on Geometry: Solved Examples- 1 - CSAT Preparation - UPSC

 1. What are some common geometry concepts that are frequently asked in exams?
Ans. Some common geometry concepts that are frequently asked in exams include angles, triangles, circles, polygons, and coordinate geometry. These topics often involve properties, measurements, and relationships between various geometric figures and shapes.
 2. How can I improve my problem-solving skills in geometry?
Ans. To improve problem-solving skills in geometry, it is important to practice regularly. Solve a variety of geometry problems, understand the concepts and formulas thoroughly, and try to visualize the given information. Break down complex problems into smaller steps, use diagrams and drawings to aid your understanding, and look for patterns or similarities in different problems. Additionally, seeking guidance from teachers or online resources can also be helpful.
 3. What are some useful strategies for tackling geometry problems in exams?
Ans. When tackling geometry problems in exams, it is essential to read the question carefully and identify the given information. Draw accurate diagrams or figures to visualize the problem better. Apply relevant theorems, formulas, and properties to solve the problem step by step. Break down complex problems into smaller, manageable parts and cross-check your answers to ensure accuracy. Practice solving similar problems beforehand to build confidence and speed.
 4. How important is understanding the properties of angles in geometry exams?
Ans. Understanding the properties of angles is crucial in geometry exams as angles play a significant role in various geometric relationships and calculations. Knowledge of angle types (such as acute, obtuse, right, and straight angles), angle addition and subtraction theorems, and angle properties within polygons is essential. It helps in determining congruence, similarity, and measurements of different geometric figures accurately.
 5. How can coordinate geometry be applied to solve geometry problems?
Ans. Coordinate geometry involves using algebraic techniques to solve geometric problems. It uses the coordinate plane, where points are represented by ordered pairs (x, y). By assigning coordinates to points, various geometric properties and relationships can be analyzed through equations and formulas. Coordinate geometry enables the calculation of distances between points, slopes of lines, equations of lines, and the determination of geometric figures' intersections or symmetries.

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