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CAT | GEOMETRY TestCracker

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Geometry - 1

1. For what values of x in the figure given

below are the lines C-G-D and A-F-B

parallel, given that EG and EF intersect

at E?

Solution:

The lines CD and AB are parallel to each other,

hence ∠EFA = ∠FHD = 10x +5, ∠EHD + ∠FHD =

1800, Hence ∠EHD = 180 – (10x+5), In Triangle

EHG, we know the three angles, ∠E = 300, EGH =

6x-5, ∠EHD = 180 – (10x+5, and sum of all these

angles will be 1800, solve for x.

2. The lines CD and FG are parallel to each

other, what is the value of ∠𝐸𝐹𝐺 ?

Solution:

3. What is the value of x?

Solution:

4. What is the value of x?



Solution:

Extend AD to meet the other line at B,

∠𝐴𝐷𝐶 + ∠𝐷𝐵𝐶 = 180, as they are

interior opposite angles. Hence ∠𝐵𝐷𝐸 =

180 – (80 +30) = 700, hence x+10 + 70 =

1800, therefor x will be 1000

5. Which of the following statements is

false?

a) Two straight lines can intersect at

only one point.

b) Through a given point, only one

straight line can be drawn

c) A line segment can be produced to

any desired length

d) Through two given points, it is

possible to draw one and only one

straight line

Solution:

Option b), many lines can be drawn

from a single point

6. Maximum number of points of

intersection of four lines on a plane is:

a) 4

b) 5

c) 6

d) 8

Solution:

It can also be solved using combinatorics. There

are 4 lines and for a point of intersection, you

need two lines. It is equal to selecting two lines

out of four lines, hence it can be written as 4C2,

which is equal to 6.

7.

In the above figure, if the length of AB is twice

that of length of BC , length of MN =10 units,

what is the length of MO?

Solution:



By Basic proportionality theorem (BPT), 𝐴𝐵

𝐵𝐶=

𝑀𝑁

𝑁𝑂, It is also given that AB = 2 BC,

therefore, NO = 10/2 = 5 units.

8. AB and CD are two parallel line

segments. E is a point between them

such that ∠𝐸𝐴𝐵 is 600 and ∠𝐸𝐶𝐷 is 700.

What is the measure of ∠𝐴𝐸𝐶 if A and C

lie on the same side of E?

Solution:

9. In the figure given below, AB = AC = CD,

If ∠𝐴𝐷𝐵 = 200, what is the value of

∠𝐵𝐴𝐷?

a) 400

b) 600

c) 700

d) 1200

Solution:

10. The two sides of a triangle are 13 cm

and 8 cm, if the third side is an integer,

what is the sum of all values of the third

side?

Solution:

In a triangle, the sum of any two sides

must be greater than the third side and

the difference between any two sides is

less than the remaining side. Let us say

that the third side is c, let a = 13 and b =

8 cm.

Hence, a-b < c < a+b, thus 13-8 < c < 13

+8, therefore 5 < c < 21. Hence c can

take any values from 6 to 20. The

question is about finding the sum of all

values from 6 to 20, there are 15

numbers and their sum will be 195.

11. Consider obtuse angled triangles with

sides 8 cm, 15 cm and x cm. If x is an

integer, how many such triangles exist?

a) 5

b) 21

c) 10

d) 15



Solution:

In an obtuse angled triangle, the square of the

longest side will be greater than the sum of the

squares of the remaining sides. If a,b,c are the

lengths of 3 sides, and if c be the length of the

largest side, then c2 > a2 + b2. Here a = 8 cm, b =

15 cm, and let c = x cm, thus

x2 > 152 + 82

x2 > 289

x > 17

x can take values like 18,19,20,21 and 22. It

cannot take 23, as it will violate the condition in

any triangle which is, the sum of any two sides

must be less than the third side.

In the above scenario we had assumed that x is

the length of the largest side, x can also not be

the largest side. In that case, 152 > x2 + 82, solve

for x, x will be less than 12.69, hence it can take

values like 12,11,10,9 and 8, it cannot take the

value of 7. Thus in total there are 10 triangles

possible.

12. ABC is a triangle, and points D, E and F

are points on AB, AC and BC

respectively. D, E and F divide ABC in

the ratio 1:3, 1:4 and 1:1 respectively.

What fraction of the area of triangle

ABC is the area of triangle DEF?

Solution:

13. If a, b and c are the sides of a triangle,

and a2 + b2 + c2 = ab + bc + ca, then the

triangle is

a) Equilateral

b) Isosceles

c) Right angled

d) Obtuse angled

Solution:

(a+b+c)2 = a2 + b2 + c2 + 2 (ab + bc +ca)

Going by the given condition, (a+b+c)2 =

3(ab+bc+ca). This will be satisfied only if

a = b = c.

a2 + b2 + c2 = ab + bc + ca

Multiply both sides by 2

2 (a2 + b2 + c2) = 2 (ab + bc +ca)

2 (a2 + b2 + c2) - 2 (ab + bc +ca) = 0

This can be written as (a-b)2 + (b-c)2 +

(c-a)2 = 0. Sum of three things will be

zero if each of them are equal to zero.

Thus, (a-b)2 = 0, (b-c)2 = 0, (c-a)2 = 0, this

makes is a=b, b=c, c=a, thus a = b = c

and it is an equilateral triangle



14. The longest side of a triangle has length

20, and another of its sides has length

10, Its area is 80 units, what is the exact

length of the third side?

a) √240

b) √250

c) √260

d) √270

Solution:

15. In the given triangle, DEF, points A, B

and C are taken on DE, DF and EF

respectively such that EC = AC, and CF =

BC. If angle D is 400, what is the

measure of ∠𝐴𝐶𝐵?

a) 1400

b) 700

c) 1000

d) None of these

Solution:

16. In the given triangle, if AB = 3cm, BC =

4cm and AC = 5 cm, what is the length

of the perpendicular DB?



Solution:

17. A piece of paper is in the shape of a

right angled triangle and is cut along a

line that is parallel to the hypotenuse,

leaving a smaller triangle. There was a

35% reduction in the length of the

hypotenuse of the triangle, if the area

of the original triangle is 34 square

inches before the cut, what is the area

of the smaller triangle?

a) 16.665

b) 16.565

c) 15.465

d) 14.365

Solution:

Triangle ABC and Triangle DBE are similar.

Two geometrical objects are called similar if

they both have the same shape, or one has the

same shape as the mirror image of the other.

Two triangles are similar when:

The corresponding angles are equal in size (two

is actually also sufficient, as the third angle

always has to make 180°)

or

The corresponding sides have the same ratio

(scale factor)



For examples on similar triangles, refer to this

link here

Few points about similar triangles

1) Similar triangles are equiangular.

2) In similar triangles corresponding sides are

proportional.

3) Congruent (Congruent means equal) triangles

are similar, but the converse is not always true.

4) Triangles similar to the same triangle are

similar to each other.5) Similar figures have the

same shape, but not necessarily the same size.

If two triangles are similar, the ratio of the areas

of two triangles will be equal to the square of

the ratio of the corresponding sides.

Also, if two triangles are similar, then:

Ratio of corresponding sides will be equal and

the Ratio of sides = Ratio of heights (altitudes)

= Ratio of Medians

= Ratio of Angle Bisectors

= Ratio of inradii

= Ratio of Circumradii

In the above problem, triangle BDE is similar to

triangle ABC.

Thus 𝐵𝐷

𝐴𝐵=

𝐵𝐸

𝐵𝐶=

𝐷𝐸

𝐴𝐶

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐵𝐷𝐸

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶=

𝐷𝐸2

𝐴𝐵2

If the length of the side AB is a, then side DE will

be 0.65 a. Hence the ratio of the areas = 0.652

12

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐵𝐷𝐸

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶=

0.652

12

Triangle BDE = 0.652 x 34

http://www.dr-aart.nl/Geometry-similarity.html



= 14.365

18. In the diagram given below, ∠𝐴𝐵𝐷 =

∠𝐶𝐷𝐵 = ∠𝑃𝑄𝐷 = 900. If AB: CD = 3:1,

the ratio of CD: PQ is

a) 1:0.69

b) 1:0.75

c) 1:0.72

d) None of these

Solution:

Let BQ = a, DQ = b

Triangles ABD and PQD are similar and CBD and

BPQ are similar

19. In the given figure, P is a point on AB,

such that AP: PB = 4:3, PQ is parallel to

AC and QD is parallel to CP. In triangle

ARC, ∠𝐴𝑅𝐶 = 900and in triangle PQS,

∠𝑃𝑆𝑄 = 900. The length of QS is 6 cm,

what is the ratio of AP:PD?

a) 10:3

b) 2:1

c) 7:3

d) None of these

Solution:

By basic proportionality theorem, BP/AP =

BQ/QC and BD/PD = BQ/QC.

Let AP = 4x and PB = 3x.

𝑃𝐷

𝐵𝐷=

𝑄𝐶

𝐵𝑄=

4

3

PD = 4 𝑃𝐵

7 =

12𝑥

7

AP : PD = 4x : 12𝑥

7 = 7:3

20. Find the sum of the medians of the

isosceles triangle, whose sides are

10,10 and 12 cm.



21. In a triangle ABC, AB = 10 cm, BC = 12

cm and AC = 14 cm, Find the length of

median AD, if G is the centroid, find the

length of GA?

Solution:

Use the appolonius theorem and we will

get, AB2 + AC2 = 2(AD2 + BD2)

Substitute the values to find AD = 5 cm.

The centroid in any triangle divides the

length of the median in the ratio 2:1, thus

GA = 2/3 * length of AD = 10/3 cm

22. In the equilateral triangle ABC, side BC

is trisected at D. Find the value of AD2?

a) 9/7 AB2

b) 7/9 AB2

c) 3/4 AB2

d) 4/5 AB2

Solution:

In an equilateral triangle ΔABC. The side BC is

trisected at D such that BD = (1/3) BC.

Draw the line AE such that it is perpendicular to

BC

In an equilateral triangle, median is same as

altitude (height). Hence AE divides the area of

triangle ABC into two equal parts, and BE = EC,

BE = EC = BC / 2

In a right angled triangle ADE



AD2 = AE2 + DE2 ---------(1)

In a right angled triangle ABE

AB2 = AE2 + BE2 ---------(2)

From equ (1) and (2) we obtain

⇒ AD2 – AB2 = DE2 – BE2

⇒ AD2 – AB2 = (BE – BD)2 – BE2

⇒ AD2 – AB2 = (BC/2 – BC/3)2- (BC/2)2

⇒ AD2 – AB2 = ((3BC – 2BC )/6)2 – (BC/2)2

⇒ AD2 – AB2 = BC2 /36 – BC2 /4 (In a equilateral

triangle ∆ABC, AB = BC = CA)

⇒ AD2 = AB2 + AB2 /36 – AB2 /4

⇒ AD2 = (36AB2 + AB2 – 9AB2)/36

⇒ AD2 = (28 AB2)/36

⇒ AD2 = (7AB)2/9

9AD2 = 7AB2

23. If A is the area, and S is the semi

perimeter of any triangle, then

a) A ≤ S2/ 3√3

b) A ≤ S2/2

c) A > S2/ √3

d) A ≥ S2/ 2√3

e) A ≤ S2/ 2√3

Solution:

We have A = √𝑆(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) where

a,b,c are sides of triangle

Using AM ≥ GM we have (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)

3 ≥ [(𝑠 − 𝑎)(𝑠 −

𝑏)(𝑠 − 𝑐)]1

3

∴ S - 𝑎+𝑏+𝑐

3 ≥ [(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 −

𝑐)]1

3

∴ S - 2𝑠

3 ≥ [(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)]

1

3

∴𝑠

3 ≥ (

𝐴2

𝑆)

1

3

∴ (𝑆

3)3 ≥

𝐴2

𝑆

∴ S4 ≥ 27 A2

Hence, A ≤ 𝑆2

3√3

24. In an equilateral triangle ABC of side 24

cm, a line parallel to the base and

passing through the centroid is drawn.

X is a point on this line, at a distance of

4 cm from the centroid, What is the

minimum possible area in sq. cm of the

triangle QXC, if XQ is drawn

perpendicular to BC?

a) 4 √3

b) 16 √3

c) 32 √3

d) 8 √3

Solution:

Let AP be the median of the triangle. Let G be

the centroid of the triangle.

∴ AG : GP = 2 : 1

As ∆ABC is an equilateral triangle.

AP = (√3

2) x 24 = 12 √3

∴ GP = 4 √3

Now, XQ = GP and XG = QP, as XQ || GP and X

lies on a line parallel to BC ∴ XQ = 4√3

Now, x can lie either to the right or left (X’) of

the centroid



As the heights of both resulting triangles are the

same the area of ∆XQC is less than that of

∆X1Q1C

in ∆XQC, QC = CP – PQ = 12 – 4 = 8 cm

Area of the ∆XOC = 1

2 x 8 x 4 √3 = 16 √3 sq.cm

25. Akash had a triangular plot as shown

below. He divided the plot into 4 areas

to build to house, a lawn, a parking area

and a swimming pool. The boundaries

of these areas, namely, QR, RU and TQ

were parallel to SV, PS and VP

respectively. He allotted area PQR for

his house, area SQT for his lawn, area

VUR for parking and area QTUR for

swimming pool.

Q divides PS in the ratio of 3:1, what is

the area of the swimming pool (sq. m) if

the area of the triangular plot PSV is

4800 sq.m ?

a) 1500

b) 900

c) 1600

d) 400

Solution:

As QR || SV, RU || PS and QT || PV, ∆PQR ~

∆PSV, ∆RUV ~ ∆PSV and ∆ QST ~ ∆PSV.

As Q divides PS in the ratio of 3 : 1, we have

∴𝑃𝑄

𝑄𝑆 =

3

1

∴ PQ = 3

4 x PS

∴ The area of partition PQR or the house = 9

16 x

Area of ∆PSV = 2700 sq.m.

Similarly, Lawn area = 1

16 x Area of ∆PSV = 300

Sq. m.

Similarly, Parking area = 1

16 x Area of ∆PSV = 300

Sq. m

∴ Remaining of swimming pool area = 4800 –

3300 = 1500 Sq. m.

26. Triangle ABC is a right angled triangle

intersecting the circle at E, B and D. AD

is the angle bisector of ∠𝐵𝐴𝐶. AB = 72

cm and AC = 75 cm, Also, BE and BD are

two equal chords of the circle. What is

the area of the given circle?

a) 1152 𝜋

9



b) 648 𝜋

9

c) 2592 𝜋

9

d) None of these

Solution:

Angle bisectors in a triangle have a

characteristic property of dividing the opposite

side in the ratio of the adjacent sides.

Let AD - with D on BC - be the bisector of ∠A in

ΔABC. If b = AC, c = AB, m = CD, and n = BD,

then

b/c = m/n.

As AD is the angle bisector, it divides BC in the

same ratio as AB to AC.

∴𝐵𝐷

𝐷𝐶 =

24

25

∴ BC = 24

49 x 21 =

72

7 cm

Also, BD = BE = 72

7 cm

DBE is an isosceles right angled triangle

∴ DE = 72√2

7

As ∠ABC is a right angle, clearly DE is the

diameter of the circle.

∴ The area given the circle = 𝜋(

72√2

7)2

4 =

2592𝜋

49

cm2

Hence, None of these

27. In a triangle ABC, the length of side BC

is 295. If the length of side AB is a

perfect square, then the length of side

AC is a power of 2, and the length of

side AC is twice the length of side AB,

what is the perimeter of the triangle?

a) 343

b) 487

c) 1063

d) None of these

Solution:

AC is a power of 2, so if 2x2 is a power

of 2, then x2 will be a power of 2 and

also x will be a power of 2. In any

triangle, sum of ay two sides must be

greater than the length of the third

side, so 3x2 > 295, hence x2 > 98.33, and

x > 10. For x to be a power of 2 and

greater than 10, let us take x = 16, then

we will get AB = 256, AC = 512, and

perimeter will be 1063.

28. A ladder of 7.6 m long is standing

against a wall and the difference

between the wall and the base of the

ladder is 6.4 m. If the top of the ladder



now slips by 1.2 m, then the foot of the

ladder shifts approximately by :

a) 0.4 m

b) 0.6 m

c) 0.8 m

d) 1.2 m

Solution:

29. Shyam, a fertilizer salesman, sells

directly to the farmers. He visits two

villages A and B. Shyam starts from A

and travels 50 m to the east, then 50 m

to the north east, at exactly 450 to his

earlier direction, and then another 50 m

east to reach village B. If the shortest

distance between villages A and B is in

the form of 𝑎√𝑏 + √𝑐 metres, find

a+b+c ?

Solution:

Construct the above scenario and you

will figure out that the total distance

travelled by shyam is

50 + 50 + 50

√2

Hence, smallest distance, say , between village

A and Village B is d =

√((100 + 25√2)2 + (25√2)2)

Hence, d2 = (100 + 25√2)2 + (25√2)2 = a2 (b +

√𝑐)

But, (100 + 25√2)2 + (25√2)2 = 2500 (5 + 2√2)

= 2500 (5 + √8)

Hence, a2 = 2500, b = 5 and c = 8

Hence, a + b + c = 50 + 5 + 8 = 63



30. In the diagram given below, CD = BF =

10 units, and ∠𝐶𝐸𝐷 = ∠𝐵𝐴𝐹 =

30 𝑑𝑒𝑔𝑟𝑒𝑒𝑠. What would be the area of

triangle AED?

a) 100 (√2 + 3)

b) 100 / (√3 + 4)

c) 50 / (√3 + 4)

d) 50 (√3 + 4)

Solution:

∠ECD = ∠BCF = 600

Also, ∠AFB = 600, ∠BFC = 300

∴ ∠AFC = 900

In a 300 – 600 – 900 triangle, sides are in the

ratio 1 : √3:2, I will take up this in next class , I

will take this up while teaching trigonometric

ratios

So, in ∆𝐸𝐷𝐶, ED = 10 √3 units ⇒ FC = 20

√3 units

and BC = 10

√3 units

In ∆AFC,FC = 20

√3 units ⇒ AC =

40

√3 units

∴ AD = (10 + 40

√3) units

Area(∆ADE) = 1

2 x 10 √3 x (10 +

40

√3)

= 50(√3 + 4) Sq. units

31. The centre of a circle inside a triangle is

at a distance of 625 cm. from each of

the vertices of the triangle. If the

diameter of the circle is 350 cm, and the

circle is touching only two sides of the

triangle, find the area of the triangle.

a) 240000

b) 387072

c) 480000

d) 506447

Solution:

OA ⊥ QP, OB ⊥ PR, OP = OQ = OR = 625 cm. In

∆OAQ, OA = 175 cm and OQ = 625 cm ⇒ AQ =

600 cm. Similarly, PA = PB = RB = 600 cm. ∆ PQR

is an isosceles triangle and PQ = PR = 1200 cm

So, PC ⊥ QR, In ∆PBO and ∆PCR

∠ OPB ≅ ∠RPC …..(common triangle)

∠ PBO ≅ ∠PCR …(Right angle)

∆PBO ~ ∆PCR …(AA test similarly)

∴ 𝑃𝐵

𝑃𝐶 =

𝐵𝑂

𝐶𝑅 =

𝑃𝑂

𝑃𝑅

∴ 600

𝑃𝐶 =

175

𝐶𝑅 =

625

1200

∴ PC = 1152 cm and CR = 336 cm, QR = 672 cm

A(∆PQR) = 1

2 x 672 x 1152 = 387072 cm2

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