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Ncrt math

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Summary of 28 pages for the course Mathematic at (Ncrt math book)

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  • March 19, 2023
  • 28
  • 2022/2023
  • Book review
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  • Secondary school
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Class IX Chapter 14 – Quadrilaterals Maths
______________________________________________________________________________

Exercise – 14.1

1. Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth
angles.
Sol:
Given
Three angles are 110,50 and 40
Let fourth angle be 𝑥
We have,
Sum of all angles of a quadrilaterals  360
110  50  40  x  360
 x  360  200
 x  160
Required fourth angle  160 .

2. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure
of each angles of the quadrilateral.
Sol:
Let the angles of the quadrilateral be
A  x, B  2 x, C  4 x and D  5x then,
A  B  C  D  360
 x  2x  4x  5x  360
 12x  360
360
x
12
 x  30
 A  x  30
B  2 x  60
C  4 x  30  4   120
D  5 x  5  30   150


3. In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that
1
∠COD = 2 (∠𝐴 + ∠𝐵).
Sol:
In DOC
1  COD  2  180 [Angle sum property of a triangle]
 COD  180  1  2
 COD  180  1  2

,Class IX Chapter 14 – Quadrilaterals Maths
______________________________________________________________________________

1 1 
 COD  180   C  D 
2 2 
[ OC and OD are bisectors of C and D represents]
1
 COD  180   C  D   ..... 1
2
In quadrilateral ABCD
A  B  C  D  360
C  D  360  A  B ......  2  [Angle sum property of quadrilateral]
Substituting (ii) in (i)
 COD  180   360   A  B  
1
2
1
 COD  180  180   A  B 
2
1
 COD   A  B 
2

4. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Sol:
Let the common ratio between the angle is ‘𝑥’ so the angles will be 3x,5 x,9 x and 13x
respectively
Since the sum of all interior angles of a quadrilateral is 360
3x  5x  9x  13x  360
 30x  360
 x  12
Hence, the angles are
3x  312  36
5x  5 12  60
9 x  9 12  108
13x  1312  156

, Class IX Chapter 14 – Quadrilaterals Maths
______________________________________________________________________________
Exercise – 14.2

1. Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each
angle of the parallelogram.
Sol:
We know that
Opposite sides of a parallelogram are equal
 3x  2  50  x
 3x  x  50  2
 4 x  52
 x  13
  3 x  2     3  13  2   37
 50  x     50  13   37
Adjacent angles of a parallelogram are supplementary
 x  37  180
 x  180  37  143
Hence, four angles are : 37,143,37,143

2. If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the
parallelogram.
Sol:
Let the measure of the angle be x
2x
 The measure of the angle adjacent is
3
We know that the adjacent angle of a parallelogram is supplementary
2x
Hence x   180
3
2x  3x  540
 5x  540
 x  108
Adjacent angles are supplementary
 x  108  180
 x  180 108  72
 x  72
Hence, four angles are : 180, 72,108, 72

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