Geometry -Unit Test 1 Similarity - · PDF fileGeometry -Unit Test 1 Similarity Answer Paper ... Unit Test 1 Page 8 of 18. From (1), (2) and (3) above, seg BD is the geometric mean

Geometry - Unit Test 1

Similarity

Answer Paper

Time: 2 Hrs..----------------------------------------------------------------------------------------------------------Max. Marks: 40

Note 1). All questions are compulsory. Note 2). Use of calculator is not allowed.

Q. 1. Solve the following: (Any 5 out of 6) (5 marks)

1. A line AB divides two sides of a triangle in the same ratio. If the slope of the third side of the triange is 1,

then what is the slope of the line AB?

Solution:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Since the slope of the parallel lines is equal, the slope of line AB is 1.

2. In ΔABC, find the ratio in which angle bisector of A divides the side BC if AB = 4 and AC = 2.

Solution:

In a triangle, the angle bisector divides the side opposite to the angle in the ratio of the remaining sides .

Now AB : AC = 4 : 2 = 2 : 1.

Hence angle bisector of A divides the side BC in the ratio 2 : 1

3. In ��ABC and ��PQR, ��P = ��A and ��Q = ��B. State if ��ABC ~ ��PQR. Justify.

Solution:

As per A-A test of similarity,

�ABC ~ �PQR

4. In 300

- 600

- 900

triangle, side opposite to 300

is 4. What is the length of the hypotenuse?

Solution:

In 300 - 600 - 900 triangle, side opposite to 300 is half of the hypotenuse.

Hence hypotenuse = 2 × side opposite to 300 = 2 × 4 = 8

5. In ΔABC, seg AD is the median and 2AD2

+ 2BD2

= 100 then find AB2

+ AC2.

Solution:

In ΔABC, seg AD is the median. Hence as per Appolonius theorem,

AB2 + AC2 = 2AD2 + 2BD2 = 100.

6. Find the hypotenuse of a right angled triangle if side making right angle are 3 and 4.

mySSC.in - Geometry - Unit Test 1mySSC.in - Geometry - Unit Test 1

Page 1 of 18

Solution:

We have,

Hypotenuse2 = sum of squares of sides making the right angle [.....Pythagoras theorem]

∴ Hypotenuse2 = 32 + 42 = 9 + 16 = 25

∴ Hypotenuse = 5

Q. 2. Solve the following : (Any 4 out of 6) (8 marks)

1.

In the figure, ��ABC �� APQ and A(��APQ) = 4A(��ABC).

Solution:

It is given that ,

1) �ABC � �APQ

2) A(�APQ) = 4 A(�ABC)

From (2),

...............(i)

From (1) and as per theorem on areas of similar triangles,


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2.

In the given figure, Point Q is on the side MP such that MQ = 2

and MP = 5.5. Ray NQ is the bisector of ��MNP of ��MNP. Find

MN : NP.

Solution:

It is given that,

(1) Point Q is on the side MP of �MNP

(2) MQ = 2, MP = 5.5

(3) Ray NQ is the bisector of �MNP

MN : NP = 4 : 7

3. The areas of two similar triangles are 81cm2 and 49 cm2 respectively. Find the ratio of their corresponding

heights. What is the ratio of their corresponding medians ?

Solution:


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The ratio of corresponding heights and medians is 9 : 7

4.

In ��ABC, AP is the median. If AP = 7, AB2

+ AC2

= 260 then find BC.

Solution:

It is given that,

1) seg AP is a median,

2) AP = 7 and

3) AB2

+ AC2

= 260.

From (1) and since B - P - C,

2BP = 2PC = AB ...............(i)

Since seg AP is the median,

AB2 + AC2 = 2AP2 + 2BP2 [...............Apollonius theorem ]

� 260 = 2(7)2

+ 2(BP)2[... From (2) and (3)]

� 260 = 2 × 49 + 2BP2

� 2BP2 = 260 - 98 = 162

Dividing each side by 2,

BP2 = 81

Taking square root of both the sides,

BP = 9


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Using (i),

BC = 2 × 9 = 18

BC = 18

5. Calculate the perimeter of the triangle if the sides making the right angle of a right-angled triangle are 3 cm

and 4 cm respectively.

Solution:

Let �ABC be the right angled triangle such that,

1) �B = 90°

2) AB = 3 cm

3) BC = 4 cm

As per Pythagoras theorem, for a right angled triangle,

(hypotenuse)2

= sum of the squares of the other two sides

� AC2 = AB2 + BC2

= (3)2 + (4)2 = 9 + 16 = 25

Taking square root of both the sides,

AC = 5 cm

Perimeter of �ABC = AB + BC + AC;

Using the values from (2), (3) and the value of AC above,

Perimeter = 3 + 4 + 5= 12 cm

Perimeter of the triangle is 12 cm.

6. Find the side of a square whose diagonal is 16��2 cm.

Solution:

Let ABCD be a square with diagonal 16�2 cm as shown in figure below.


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Then for �ABC,

1) �ABC = 90°

2) AC = 16�2 cm

Using Pythagoras theorem,

AC2 = AB2 + BC2

But AB = BC

� AC2

= 2(AB)2

Substituting using (2),

(16�2)2

= 2(AB)2

i.e. (AB)2

= 16 × 16

Taking square root of both sides,

AB = 16 cm

The side of the given square is 16 cm.


1. ��ABC has sides of length 5,6 and 7 units while ��PQR has perimeter of 360 units. If ��ABC is similar to��PQR

then find the sides of ��PQR.

Solution:


Page 6 of 18

The sides of �PQR are 100, 120 and 140 units.

2. ��PQR �� XYZ. PQ = 6, QR = 8, PR = 7, XY = 12 and ��Q = 75° Find YZ, XZ and ��Y.

Solution:

It is given that,

1) �PQR � � XYZ.

2) PQ = 6, QR = 8, PR = 7, XY = 12 and �Q = 75°

With this data, �PQR and �XYZ can be drawn as


Page 7 of 18

From (1) and using properties of similar triangles,

From (1) and using properties of similar triangles,

� �Q = �Y = 75° [...c.a.s.t. ]

YZ = 16

XZ = 14

�Y = 75°

3.

In the figure in ��ABC, ��ABC = 90°, seg BD �� seg AC, AD = 3, DC

= 15. Find the length BD and BC.

Solution:

From the figure,

1) AD = 3

2) DC = 15

3) �BDC = �ABC = 90°


Page 8 of 18

From (1), (2) and (3) above, seg BD is the geometric mean of seg AD and seg DC.

As per property of geometric mean,

(BD)2 = AD × DC

Substituting values using (1) and (2),

� BD2 = 3 × 15 = 45 = 9 × 5

Taking square root on both sides,

BD = 3�5

In �ABC, as per Pythagoras theorem,

BC2 = DC2 + BD2

Using (1) and the calculated value of BD,

BC2 = 152 + 3�52 = 225 + 45 = 270

Taking square root on both the sides,

BC = 3�30

BD = 3�5

BC = 3�30

4.

In the figure,

seg AB || seg DC.

Using the information given find the value of x.

Solution:

It is given that,

(1) seg AB || seg DC

(2) DO = 3, BO =x - 3, AO = 3x - 19 and CO= x - 5

Using (1) and since BD is tansversal,

�CDO � �OBA ...............[ alternate angles ]

�DCO � �OAB ...............[ alternate angles ]

� �ODC ~ �OBA ...............[ A - A test of similarity ]


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x = 8 or x = 9

5. A vertical stick 12 m log casts a shadow 8 m long on the ground. At the same time a tower casts the shadow

of length 40 m on the ground. Determine the height of the tower.

Solution:

Let TG be the tower, GQ be it's shadow and SL be the stick and LQ be its shadow as shown in figure below.

�TGQ = �SLQ= 900

Now, GQ = 40 m, SL = 12 m, and LQ = 8 m.

Since both tower and the stick are vertical,

TG || SL

�GTQ = �LSQ ...[ corresponding angles ]

and

�TGQ = �SLQ= 900

��In correspondence GTQ LSQ


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�GTQ � �LSQ [....A-A test of similarity ]

The height of the tower is 60 m


1. Prove that, if a line parallel to a side of a triangle intersects the others sides in two distinct points, then the

line divides those sides in proportion.

Solution:

Given : In ��PQR, line l ��side QR

Line l intersects side PQ and side PR in points M and N respectively.

P-M-Q and P-N-R.

To Prove :

Construction : Join seg QN and seg RM.

In �PMN and �QMN,

In �PMN and �RMN,

A( �QMN) = A( �RMN) ..........[triangles with common base MN and same height] ...(iii)


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Hence proved that if a line parallel to a side of a triangle intersects the others sides in two distinct

points, then the line divides those sides in proportion.

2.

In the given figure, DE || BC

(i) If DE = 4 cm, BC = 8 cm, A(��ADE) = 25 cm2, find A(��ABC).

(ii) If DE : BC = 3 : 5, then A(��ADE) : A( DBCE).

Solution:

It is given that, in �ABC,

seg DE || side BC


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i) A(�ABC) = 100 cm2

ii) A(�ADE) : A( DBCE) = 9 : 16

3. Prove that three times the square of a side of an equilateral triangle is equal to four times the square of an

altitude.

Solution:

Given:

(1) In an equilateral �ABC each side = a

(2) AH is the altitude of the triangle having length h

To Prove:

3a2

= 4h2

Proof:


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Hence proved that, three times the square of a side of an equilateral triangle is equal to four times the

square of an altitude.


1.

In the given figure, seg AD is the median of ��ABC and seg AM ��

seg BC. Prove that:

Solution:

Given :

1) AD is the median i.e. D is midpoint of side BC i.e. BD = DC i.e. 2BD = 2DC = BC

2) seg AM � side BC


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To Prove :

Proof :

From (1) , we can write that ,

MD + DC = MC [ ... M-D-C ] ...............(i)

From (2) and as per Pythagoras theorem for a right angled �AMC

AC2 = AM2 + MC2

Using (i),

AC2 = AM2 + (MD + DC)2

i.e. AC2 = AM2 + MD2 + 2 MD.DC + DC2 ...............(ii)

Similarly for a right angled �AMD,

AD2 = AM2 + MD2 ...............(iii)

Substituting (iii) in (ii),

AC2

= AD2

+ 2 MD.DC + DC2

Using (1),

From (2) and as per Pythagoras theorem for a right angled �AMB

AB2 = AM2 + BM 2

Since B - M - D,

AB2 = AM2 + (BD - MD)2

� AB2 = AM2 + BD2 - 2 BD.MD + MD2

�AB2

= AM2

+ MD2

- 2 BD.MD + BD2

� AB2

= AD2

- 2 BD.MD + BD 2

[.....AM2

+ MD2

= AD2

]

Using (1),


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Hence proved that,

.

2. ��ABC is a triangle where ��C = 900. Let BC = a, CA = b, AB = c and let 'p' be the length of the perpendicular

from C on AB. Prove that (i) cp = ab, (ii)

Solution:

Given:

1) In �ABC, �C = 90°

2) �ADC = 90°

3) The length of the perpendicular drawn from C on side AB is p

4) a, b and c are lengths of side BC, AC and AB respectively.

With this data �ABC can be drawn as shown in figure below.

Proof:

From (1) and as per Pythagoras theorem, for a right angled �ABC,

AB2

= BC2

+ AC2

Using (3) and (4),

c2 = a2 + b2 ...............(i)

Using (1) and (2) in �ABC,

AB is base and DC is height

And also BC is base and AC is height.

...............(ii)


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...............(iii)

Using (3), (4), (ii) and (iii),

...............(iv)

Substituting (i) in (iv),

3. In ��ABC, PQ is a line segment intersecting AB at P and AC at Q such that seg PQ || seg BC. If PQ divides

��ABC into two equal parts means equal in area, Find

Since PQ || BC,


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�APQ � �ABC [ ....corresponding angles ] ...............(1)

�PAQ � �BAC [ ....common angle ] ...............(2)

From (1) and (2),

�APQ � �ABC [... A-A test of similarity ]

[...theorem on areas of similar triangles ] ...............(3)

But 2A(�APQ) = A(�ABC) [...given ], hence from (3),


Page 18 of 18

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Geometry -Unit Test 1 Similarity - · PDF fileGeometry -Unit Test 1 Similarity Answer Paper ... Unit Test 1 Page 8 of 18. From (1), (2) and (3) above, seg BD is the geometric mean