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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?
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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:
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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
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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
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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?
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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)
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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
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= 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.
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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
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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
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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
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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
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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
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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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