Vedanta Excel in Mathematcs Book -10 Final (2078)

4. a) In the given 'ABC, P is any point on the median AD. Prove A
that ' APB = ' APC. P

B D C
Q C
A

b) In the adjoining ' ABC, P and Q are the
mid-points of the sides AB and AC respectively
and R be any point on BC. Prove that, P

area of ' PQR = 1 area of ' ABC. B
4
R

P S
R
c) In the figure, PQRS is a parallelogram. Q and S are M
joined to any point M on the diagonal PR. Prove that,
the area of ' PQM = the area of ' PSM.

Q

A

d) In the given figure, DE // BC. Prove that D E
(i) ' BOD = ' COE C
(ii) ' BAE = ' CAD O
B C

e) In the given figure, P is the mid-point of AB and Q A
is any point on the side BC.CR meets AB at R and
R
CR // PQ. Prove that P

area of ' BQR = 1 area of ' ABC. B Q
2

DC

f) The line drawn through the vertex C of the quadrilateral
ABCD parallel to the diagonal DB meets AB produced at E.
Prove that,
A B E
C
the area of quad. ABCD = the area of ' DAE.

A

g) In the given triangle ABC, two medians BE and CD are D E
intersecting at O. Prove that O

the area of 'BOC = the area of quad. ADOE

B

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Geometry – Area of Triangles and Quadrilateral

A D
E
h) In the adjoining figure, it is given that AD // BC and
BD // CE. Prove that, area of ' ABC = area of ' BDE.

i) In the given figure, BEST is a parallelogram. BC
Diagonal BS is produced to point L. Prove that L
'LST and 'LSE are equal in area.
TS
j) In the given figure, AD // BC. If the area of 'ABE and
'ACF are equal, prove that EF // AC. B E
A ED

F

B C
AD
k) In the given figure, AD // BE // GF and AB // DG // EF.
If the area of parallelograms ABCD and CEFG are equal, B CE
then prove that DE // BG.
G F
l) In the adjoining figure, PQRS and LQMN are P S
two parallelograms of equal in area. Prove that
LR // SN. L N

m) In the given triangle ABC, medians BN and CM are Q RM
intersected at O. Prove that A
the area of 'BOC = the area of quadrilateral AMON.
MN
n) In the given figure, M is the mid-point of AE, O
prove that area of 'ABE is equal to the area
of parallelogram ABCD. BC

E

D MC

AB

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Unit Geometry - Construction

14

14.1 Construction of a triangle whose area is equal to the area
of a given quadrilateral.

Example 1: Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 4.8 cm,
AC = 5.5 cm, CD = 2.6 cm, and AD = 4 cm. Construct a triangle equal in area to
the quadrilateral.

Steps 1 2.6 cm C
(i) Draw AB = 3.5 cm

(ii) From the point B, draw an arc with radius D 4.8 cm
BC = 4.8 cm and from A draw another arc with 4 cm
radius AC = 5.5 cm. These two arcs intersect each
other at C.

(iii) Join B, C and A, C.

(iv) From C, draw an arc with radius CD = 2.6 cm and A 3.5 cm B
from A, draw another arc with radius AD = 4 cm.
These two arcs intersect each other at D. Join C, D
and D, A. Thus, ABCD is the given quadrilateral.

Step 2 E D C
A B
(i) From the point D, draw DE parallel to CA to
meet BA produced at E.

(ii) Join C and E.
Now, ∆ BCE is the required triangle.
Here, ∆ BCE = ∆ ABC + ∆ ACE
= ∆ ABC + ∆ ACD
= Quad. ABCD

[∆ ACE and ∆ ACD are on the same base AC and between the same Parallels AC
and DE]

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Geometry – Construction

14.2 Construction of a parallelogram whose area is equal to the
area of a given triangle.

Example 2: Construct a triangle ABC in which a = 5.4 cm, b = 4.5 cm and c = 3.8 cm.
Construct a parallelogram equal in area to the triangle and having an angle
60°.

Step 1 A
(i) Draw a = BC = 5.4 cm

(ii) From B, draw an arc with radius c = BA = 3.8 cm and c = 3.8 cm b = 4.5 cm
from C, draw another arc with radius b = CA = 4.5 cm.
These two arcs intersect each other at A.

(iii) Join B, A and C, A. Thus, ABC is the given triangle. B a = 5.4 cm C

Step 2 A E FX
(i) Through the point A, draw AX parallel to BC.

(ii) With the help of the perpendicular bisector of
BC, mark the mid–point D on BC.

(iii) Draw ‘ EDC = 60° such that DE meet AX at E.

(iv) From the point E, draw an arc with radius B 60° C
EF = DC to cut AX at F. D

(v) Join F and C.
Now, EDCF is the required parallelogram.

Example 3: Construct a triangle ABC in which AB = 4.7 cm, BC = 5.8 cm and
CA = 3.5 cm. Construct a parallelogram with a side 6.5 cm and equal in
area to the triangle.

Step 1 4.7 cm A
(i) Construct the triangle ABC from the given 3.5 cm

measurements

Step 2 B 5.8 cm C
AE FX
(i) Follow the process (i) and (ii) of step 2 from
example 2.

(ii) From the point D, draw an arc with radius
DE = 6.5 cm to cut AX at E.

(iii) From the point E, draw an arc with radius 6.5 cm
EF = DC to cut AX at F.

(iv) Join F and C. D C
5.8 cm
B

Now, EDCF is the required parallelogram.

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Geometry – Construction

14.3 Construction of a triangle whose area is equal to the area of
a given parallelogram.

Example 4: Construct a parallelogram ABCD in which AB = 3.6 cm, BC = 5.2 cm
and ‘ ABC = 60°. Construct a triangle with one side 6.8 cm and equal
in area to the parallelogram.

Step 1 D X
(i) Draw AB = 3.6 cm C
(ii) At B, draw ‘ ABX = 60°

(iii) From B, draw an arc with radius
BC = 5.2 cm to cut BX at C.

(iv) From C, draw an arc with radius 5.2 cm
CD = AB and from A draw another arc
with radius AD = BC. These two arcs
intersect each other at D.

(v) Join, C, D and A, D. Thus, ABCD is the 60° B
given parallelogram.
A 3.6 cm

Step 2
(i) Produce AB to E such that AB = BE
(ii) From A, draw an arc with radius AF = 6.8 cm to cut DC (or DC produced, if necessary)

at F.
(iii) Join A, F and F, E. Now, AEF is the required triangle.

X
D CF

5.2 cm

60° B

A 3.6 cm E

14.4 Construction of a triangle whose area is equal to the area
of a given triangle.

Example 5: Construct a triangle ABC in which AB = 5.5 cm, ‘BAC = 60°and
AC = 5 cm . Construct another triangle DAB with the side AD = 7.8 cm and
the area equal to the area of ' ABC.

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Geometry – Construction Geometry – Construction

Step 1 D
(i) Construct the triangle ABC CX

from the given measurements. 5 cm

Step 2 A 5.5 cm B
(i) Through the point C, draw CX

parallel to AB.
(ii) From the point A, draw an arc

of radius AD = 7.8 cm to cut
CX at D.
(ii) Joing D, A and D, B.
Now, DAB is the required
triangle.

EXERCISE 14.1

General section
1. Construct triangle ABC by using the following measurements:

a) AB = 4.2 cm, BC = 3.6 cm, and CA = 5.5 cm b) AB = 5.4 cm., BC = 6.3 cm and CA = 4.5 cm

c) AB = 5.7 cm, ‘ A = 45°, and AC = 4.2 cm d) BC = 4.8 cm, ‘ B = 60° and CA = 6.8 cm

e) BC = 3.6 cm, ‘ B = 30°, and ‘ C = 45°.

2. Construct quadrilaterals ABCD by using the following measurements:
a) AB = 3.5 cm, BC = 4.2 cm, AC = 6.3 cm, CD = 2.5 cm, and AD = 5.2 cm

b) AB = 4.6 cm, BC = 3.9 cm, AC = 6.5 cm, CD = 3.7 cm, and AD = 4.8 cm

c) AB = 4.5 cm, BC = 4.2 cm, AC = 5.8 cm, BD = 6.4 cm, and AD = 5.3 cm

d) AB = 3.8 cm, ‘ B = 120°, BC = 4.6 cm, CD = 5.4 cm, and AD = 5 cm
e) AB = BC = 5.5 cm, CD = AD = 6.2 cm, and ‘ BAD = 75°.
3. Construct parallelograms ABCD by using the following measurements:
a) AB = 5.6 cm, BC = 4.5 cm, and ‘ ABC = 60°
b) AB = 4.3 cm, BC = 3.5 cm, and ‘ BAD = 120°
c) AB = 4.5 cm, BC = 5.5 cm, and AC = 6.7 cm
d) AB = 3.8 cm, ‘ ABD = 60°, and AD = 5.2 cm
e) AB = 5.5 cm, diagonal AC = 7.6 cm, and diagonal BD = 5.6 cm.

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Geometry – Construction Geometry – Construction

Creative section

4. a) Construct a quadrilateral PQRS in which PQ = QR = 5 cm, RS = SP = 4 cm,
and ‘P = 60°. Then, construct a triangle whose area is equal to the area of the
quadrilateral.

b) Construct a quadrilateral PQRS having PQ = QR = 5.9 cm, RS = PS = 6.1 cm and
‘QPS = 75°. Then, construct ' PST which is equal in area to the given quadrilateral.

c) Construct a quadrilateral MNOP in which MN = 4 cm, NO = 5 cm, OP = 5.5 cm,
PM = 3.5 cm, and ‘PMN = 30°. Construct a triangle MPQ whose area is equal to
the quadrilateral MNOP.

d) Construct a quadrilateral ABCD having AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm,
AD = 4 cm, and ‘ABC = 60°. Construct ' ADE whose area is equal to the
quadrilateral ABCD.

e) Construct a quadrilateral ABCD in which AB = 4 cm. BC = 5 cm, CD = 5.5 cm,
DA = 4.5 cm, and diagonal AC = 6 cm. Also construct a triangle ADE equal in area
to the quadrilateral.

f) Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 5.1 cm, CD = 4.9 cm,
AD = 6.1 cm, and the diagonal BD = 5.7 cm. Also construct a triangle equal in area
to the quadrilateral ABCD.

g) Construct a quadrilateral PQRS in which QR = 5.3 cm, RS = 5 cm PS = 5.7 cm,
PQ = 6.2 cm, and PR = 5.6 cm. Then, construct a triangle RQT equal in area to the
quadrilateral PQRS.

h) Construct a 'WXB equal in area to the quadrilateral WXYZ having WX = XY = 5.5
cm, YZ = ZW = 4.5 cm, and ³WXY = 75°.

5. a) Construct a triangle in which a = 7 cm, b = 6 cm and c = 5 cm and also construct a
parallelogram whose area is equal to the area of triangle ABC and one of the angles
being 60°.

b) Construct a triangle ABC in which a = 7.8 cm, b = 7.2 cm and C = 6.3 cm. Then
construct a parallelogram DBEF equal in area to ' ABC and ‘ DBC = 75°.

c) Construct a triangle PQR in which PQ = 7.5 cm, QR = 6.8 cm, and RP = 6 cm.
Then construct a parallelogram TQSU equal in area to ' PQR having a side
TQ = 6.4 cm.

d) Construct a triangle ABC from the following data. Also construct a parallelogram
having one side 7.5 cm and equal in area to the triangle.
AB = 7.1 cm, ‘ BAC = 60° and AC = 5.7 cm

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Geometry – Construction

e) Construct a triangle ABC in which AB = 5 cm, ‘ ABC = 75°and ‘ ACB = 60°.
Then, construct a rectangle equal in area to the triangle ABC.

f) Construct a triangle ABC having sides AB = 4 cm, BC = 6.8 cm, and CA = 6.5 cm.
Then, construct a rectangle equal in area of ' ABC.

6. a) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and
‘BAD = 45°. Construct a triangle APQ having one angle 60° and equal in area of
the parallelogram.

b) Construct a parallelogram ABCD having AB = 5 cm, BC = 4 cm, and ‘B = 120°.
Also construct a triangle with the area equal to the parallelogram.

c) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and
‘ BAD = 60°. Then construct a triangle AEF equal in area to the parallelogram and
the side AE = 7.5 cm.

d) Construct a parallelogram PQRS in which PQ = 5 cm, diagonal PR = 6 cm, and
diagonal QS = 8cm. Construct a triangle PSA whose area is equal to the area of the
parallelogram.

7. a) Construct a triangle ABC having sides a = 6.4 cm, b = 6cm, and c = 5.6 cm. Also
construct another triangle equal in area to the 'ABC and having a side 7 cm.

b) Construct a triangle PQR in which PQ = 5.2 cm, QR = 6 cm and ‘Q = 60°.
Construct another triangle SPQ which is equal in area to the 'PQR and side
PS = 8.5 cm.

c) Construct a triangle XYZ in which XY = 6.3 cm, ‘X = 30°, and ‘Y = 45°. Construct
another triangle WXY equal in area to ' XYZ and a side WY = 7.5 cm.

8. a) Construct a parallelogram ABCD in which AB = 3.5 cm, AC = 5.9 cm, and
BC = 4.2 cm. Construct another parallelogram ABEF where BE = 6.5 cm.

b) Construct a parallelogram PQRS in which PQ = 4 cm, ‘P = 45°, and PS = 4.5 cm.
Construct another parallelogram PQXY where ‘PQX = 30°.

9. a) Construct a rectangle with length 7.1 cm and breadth 6.1 cm. Also, construct a
triangle having one angle 60° and equal to the area of the rectangle.

b) Construct a rectangle ABCD with length 5.5 cm and breadth 4.5 cm. Also, construct
a triangle EBF having one angle 45° and equal to the area of the rectangle.

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Unit Trigonometry

16

16.1 Trigonometric ratios of an acute angle of a right-angled

triangle - review C

In the given right-angled triangle ABC, right angled at hypotenuse (h) perpendicular (p)
B. ‘BAC = T is an acute angled and taken as angle of
reference.
Here, side AC is opposite to the right angled and it is
hypotenuse (h).

Side BC is opposite to the angle of reference (T) and it is T B

perpendicular (p). A base (b)

Side BC is adjacent to the angle of reference (T) and it is base (b).

Now, we can make six different ratios of the various pairs of the sides of a right-
angled triangle. These ratios are given below:

(i) perpendicular = p which is called sine of T or sinT .
hypotenuse h

(ii) base = b which is called cosine of T or cosT.
hypotenuse h

(iii) perpendicular = p Which is called tangent of T or tanT.
base b

(iv) hypotenuse = h Which is called cosecant of T or cosecT
perpendicular p

(v) hypotenuse = h which is called secant of T or secT.
base b

(vi) base = b which is called cotangent of T or cotT
perpendicular p

Here, sinT, cosT, and tanT are the main ratios and cosecT, secT, and cotT are inverse
ratios of the main ratios respectively.

16.2 Values of trigonometric ratios of some standard angles

The angles like 0°, 30°, 45°, 60°, and 90° are commonly known as the standard angles.
The values of trigonometric ratios of these standard angles can also be obtained
geometrically without using the table. The table given below shows the value of
trigonometric ratios of these standard angles

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Trigonometry

Trigonometric ratios Angles
sin
cos 0° 30° 45° 60° 90°
tan 0 1
1 1 13 0
0 2 22 f

3 1 1
2 2 2

1 1 3
3

16.3 Height and distance A
B
One of the important applications of trigonometry

is to find the heights of objects like, tree, building, Line of sight Height
tower, etc. and distance between objects. In such
cases, we have to solve right angled triangles.

For example, in the figure, AB is the height of a

tree, BC is the distance between the tree and the C T

observer at C. Distance

Here, if T and BC are known, we can calculate the height AB or if T and AB are
known, we can calculate the distance between the tree and the observer by using a
trigonometric ratio.

16.4 Angle of elevation and angle of depression Angle of eleLivnateioofnsight

In the adjoining figure, an observer is looking at the top Horizontal
of a tower. Here, OA is called the line of sight and OB
is the horizontal line. ‘BOA is the angle between the
upward line of sight and the horizontal line. ‘BOA is
called the angle of elevation.

Thus, the angle made by upward line of sight with
horizontal line is known as angle of elevation.

In this figure, ‘BOA is the angle between the OB
downward line of sight and the horizontal line. Angle of depression
‘BOA is called the angle of depression.

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Trigonometry

Thus, the angle made by downward line of sight with horizontal line is known as

angle of depression. C A
In the adjoining figure, the horizontal lines AC and BO are Angle of depression

parallel to each other. The angle of depression CAO and the

angle of elevation BOA are alternate angles between parallel Angle of elevation B

lines. O

? ‘ CAO = ‘ BOA

i.e., Angle of depression= Angle of elevation.

The instrument which is used to measure the angle of elevation or the angle of
depression is called Theodolite.

Worked-out examples

Example 1: A man observes the top of a tree 30 3 m high and finds the elevation of
30°. Find the distance between the man and the foot of the tree.

Solutions:

Let AB be the height of the tree and BC be the distance A
between the foot of the tree and the man. 30 3 m
Here, AB = 30 3 m B

Angle of elevation, ‘ BCA = 30°. C 30°

In rt. ‘ ed ' ABC, tan 30° = p = AB = 30 3
b BC BC
1 30 3
or, 3 = BC

or, BC = 30 3 × 3 = 30 × 3 = 90 m

Hence, the required distance is 90 m.

Example 2: An electric pole 15 m high is supported by a wire fixing its one end on the

ground at some distance from the pole. If the wire joining the top of the pole

is inclined to the ground at an angle of 60q, find the length of the wire.

Solution:

Let AB be the height of the pole and AC be the length of wire. A

Here, AB = 15 m.

Angle of elevation BCA = 60q 15 m

In rt. ‘ed ' ABC, B 60° C

sin60q = p = AB
b AC

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Trigonometry

or, 3 = 15
2 AC

or, AC 3 = 30 m

AC = 30 × 3 = 10 3= 10 × 1.732 = 17.32 m
3 3

Hence, the required length of the wire is 17.32 m.

Example 3: The diameter of a circular pond is 90 m and a pole is fixed at the centre of
Solutions: the pond. The height of the pole is 48 m and the pond is 3 m deep. Find the
angle of elevation of the top of the pole from a point in the circumference
of the pond.

Let BC be the height of the pole above the surface of water. C
AB be the radius of the circular pond.
Here, diameter of the pond = 90 m

Then, radius of the pond = 90 m = 45 m
2

Height of the pond above the surface of water, BC = 48 m – 3 m 45 m

= 45 m A T B
45 m 3m

In right-angled triangle ABC,

tan T = p = BC = 45 =1
b AB 45

or, tan T = tan45°

? T = 45°
Hence, the required angle of elevation is 45°.

Example 4: A man 1.5 m tall standing at a distance of 50 m from a tree observes the
angle of elevation of the top of the tree to be 45°. Find the height of the tree.

Solution:

Let CE be the height of the tree and AB be the height of the man. BC be the distance between

the foot of the tree and the man. E
Here, AB = 1.5 m

BC = 50 m A 45° D
Angle of elevation DAE = 45q 1.5 m 50 m C
In rectangle ABCD,
B

DC = AB = 1.5 m and AD = BC = 50 m

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Trigonometry

In rt. ‘ed ' ADE,

tan45q = p = ED
b AD

or, 1 = ED
50

or, ED = 50 m

Now, CE = ED + DC = 50 m + 1.5 m = 51.5 m

So, the required height of the tree is 51.5 m.

Example 5: A girl, 1.2 m tall, is flying a kite. When the length of the string of the kite is

180 m, it makes an angle of 30° with the horizontal line. At what height is

the kite from the ground?

Solution:

Let CK be the height of the kite from the ground, AB be the height of the girl and AK be the

length of the string. K

Here, AB = 1.2 m 180 m
AK = 180 m

Angle of elevation DAK = 30q. A 30° D
1.2 m C
In rectangle ABCD, AB = DC = 1.2 m.
B
In rt. ‘ ed ' ADK,

sin30q = p = KD
h AK

or, 1 = KD
2 180

or, KD = 90 m

Now, CK = KD + DC = 90 m + 1.2 m = 91.2 m

So, the kite is at a height of 91.2 m from the ground.

Example 6: A woman is 1.6 m tall and the length of her shadow in the sun is 1.6 3 m.
Find the altitude of the sun.

Solution:
Let AB be the height of the woman and BC be the length of her shadow. Let T be the altitude of
the sun.

Here, AB = 1.6 m A

BC = 1.6 3 m

In rt. ‘ ed ' ABC,
p AB
tanT = b = BC 1.6 m
B
or, tanT = 1.6 T =?
1.6 3 C

1.6 3 m

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Trigonometry

or, tanT = 1
3

or, tanT = tan30q
? T = 30q
So, the required altitude of the sum is 30q.

Example 7: A boy 3 m tall is 72 m away from a tower 25 3 m high. Find the angle of
elevation of the top of the tower from his eyes.

Solutions:

Let AB be the height of the boy and CD be the height of the tower. Let BD be the distance
between the boy and the tower.

Here, BD = AE = 72 m C

AB = 3 m and CD = 25 3 25 3 m
E
? CE = CD – ED = 25 3 – AB = 25 3 – 3 = 24 3 m D

Now, in rt. ‘ ed ' AEC,

tan A = p = CE A
b AE 3m

24 3 B 72 m
72
or, tan A =

tan A = 3 = 3 3 = 1
3 3× 3

A = tan–1 1 = 30°
3

So, the required angle of elevation is 30°.

Example 8: The top of a tree which is broken, by the wind makes an angle of 60° with the
Solutions: ground at a distance 3 3 m from the foot of the tree. Find the height of the tree
before it was broken.

A

Let AC be the height of the tree before it was broken, BD be the broken part
of the tree and CD be the distance between the foot of the tree and the point
on the ground at which the top of the tree touched.

Here, CD = 3 3 m B

The angle of elevation, ‘ CDB = 60°

Now, in rt. ‘ed ' BCD, 60°
3m
tan 60° = p = BC = BC C 3 D
b CD 33

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