Class 9: Maths Chapter 16 solutions. Complete Class 9 Maths Chapter 16 Notes.
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Selina Class 9 ICSE Solutions Mathematics : Chapter 16- Area Theorems [Proof and Use]
Selina 9th Maths Chapter 16, Class 9 Maths Chapter 16 solutions
Exercise 16(A)
Solution 1:
Solution 2:
Since from the figure, we get CD//FE therefore FC must parallel to DE. Therefore it is proved that the quadrilateral CDEF is a parallelogram.
Area of parallelogram on same base and between same parallel lines is always equal and area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between same parallel lines.
So Area of CDEF= Area of ABDC + Area of ABEF
Hence Proved
Solution 3:
Solution 4:
Given ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.
proof:
since triangles with same base and between same set of parallel lines have equal areas
area (CPD)=area(BCD)…… (1)
again, diagonals of the parallelogram bisects area in two equal parts
area (BCD)=(1/2) area of parallelogram ABCD…… (2)
from (1) and (2)
area(CPD)=1/2 area(ABCD)…… (3)
similarly area (AQD)=area(ABD)=1/2 area(ABCD)…… (4)
from (3) and (4)
area(CPD)=area(AQD),
hence proved.
(ii)
We know that area of triangles on the same base and between same parallel lines are equal
So Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC
Hence Proved
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Solution 15:
Solution 16:
We know that area of triangles on the same base and between same parallel lines are equal.
Consider ABED quadrilateral; AD||BE
With common base, BE and between AD and BE parallel lines, we have
Area of ΔABE = Area of ΔBDE
Similarly, in BEFC quadrilateral, BE||CF
With common base BC and between BE and CF parallel lines, we have
Area of ΔBEC = Area of ΔBEF
Adding both equations, we have
Area of ΔABE + Area of ΔBEC = Area of ΔBEF + Area of ΔBDE
=> Area of AEC = Area of DBF
Hence Proved
Solution 17:
Given: ABCD is a parallelogram.
We know that
Area of ΔABC = Area of ΔACD
Consider ΔABX,
Area of ΔABX = Area of ΔABC + Area of ΔACX
We also know that area of triangles on the same base and between same parallel lines are equal.
Area of ΔACX = Area of ΔCXD
From above equations, we can conclude that
Area of ΔABX = Area of ΔABC + Area of ΔACX = Area of ΔACD+ Area of ΔCXD = Area of ACXD Quadrilateral
Hence Proved
Solution 18:
Join B and R and P and R.
We know that the area of the parallelogram is equal to twice the area of the triangle, if the triangle and the parallelogram are on the same base and between the parallels
Consider ABCD parallelogram:
Since the parallelogram ABCD and the triangle ABR lie on AB and between the parallels AB and DC, we have
Exercise 16(B)
Solution 1:
Solution 2:
Solution 3:
Solution 4:
We have to join PD and BD.
Solution 5:
Solution 6:
Ratio of area of triangles with same vertex and bases along the same line is equal to ratio of their respective bases. So, we have
Solution 7:
Solution 8:
Exercise 16(C)
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Selina Class 9 ICSE Solutions Mathematics : Chapter 16- Area Theorems [Proof and Use]
Chapterwise Selina Publishers ICSE Solutions for Class 9 Mathematics :
- Chapter 1 – Rational and Irrational Numbers
- Chapter 2- Compound Interest (Without using formula)
- Chapter 3 – Compound Interest (Using Formula)
- Chapter 4- Expansions (Including Substitution)
- Chapter 5- Factorisation
- Chapter 6 – Simultaneous (Linear) Equations (Including Problems)
- Chapter 7- Indices (Exponents)
- Chapter 8- Logarithms
- Chapter 9- Triangles [Congruency in Triangles]
- Chapter 10- Isosceles Triangles
- Chapter 11- Inequalities
- Chapter 12- Mid-point and Its Converse [ Including Intercept Theorem]
- Chapter 13- Pythagoras Theorem [Proof and Simple Applications with Converse]
- Chapter 14- Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
- Chapter 15- Construction of Polygons
- Chapter 16- Area Theorems [Proof and Use]
- Chapter 17- Circle
- Chapter 18- Statistics
- Chapter 19- Mean and Median
- Chapter 20 – Area and Perimeter of Plane Figures
- Chapter 21- Solids [Surface Area and Volume of 3-D Solids]
- Chapter 22- Trigonometrical Ratios
- Chapter 23- Trigonometrical Ratios of Standard Angles
- Chapter 24- Solution of Right Triangles
- Chapter 25- Complementary Angles
- Chapter 26- Co-ordinate Geometry
- Chapter 27- Graphical Solution (Solution of Simultaneous Linear Equations, Graphically)
- Chapter 28- Distance Formula
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