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CBSE Class 9 Mathematics Worksheet
Unit 4-Geometry - Areas of Parallelograms and Triangles
ABCD is a paralleogram and line segment AX and CY bisects angles A and C respectively where X is a point on AB. To prove ?
A parallelogram with equal sides is
Rhombus
B)Square
C)Rectangle
D)Kite
ABCD is a trapezium with AB DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
A rhombus has an angle sixty degrees. If the side length is 10cm find its area.
40 X sq cm
50 X sq cm
30 X
sq cm
100 X
sq cm
AD is a median of triangle ABC and E is the midpoint of AD. BE produced meets AC in F. Prove that AF = 1/3 AC.
Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at O. Prove that ar (AOD) = ar (BOC).
. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar() = ar(
)
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
The longest chord in a circle is 20cm. Find its area if a pizza divided between 4 members
in a family how much one person gets if the division is in the ratio 1:2:3:4 area wise
Smallest share is 10 sq. cm.
Smallest share is 15
sq. cm.
Biggest share is 20 sq. cm.
Can not say
a
B)b
C)c
D)d
A circle of diameter 20cm two chords are drawn of length 16cm on either side and parallel to the diameter. . Find the shortest distance between them?
16cm
B)12cm
C)18cm
D)20cm
Do all cyclic quadrilaterals inscribed within a half circle need to be isosceles trapezia, if they are trapezia? Take help by drawing a diagram and observing the result.
ABCD is a quadrilateral in which L, M, N and O are the mid points of the sides AB, BC, CD and DA respectively. Show that (i) and ON = 1/2 AC (ii) ON = LM (iii) LMNO is a parallelogram
Can a cyclic quadrilateral have four angles in the ratio
All of them are valid cyclic quadrilaterals.
a
B)v
C)c
D)d
Three regular polygons are inscribed in a circle. Can you find out how many diagonals they have in total?
21
B)20
C)23
D)24
a
B)b
C)c
D)d
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Area of figure shown above is in square units is
100
B)99
C)90
D)97
a
B)b
C)c
D)d
CBSE Class 9 Mathematics Worksheet
Unit 4-Geometry - Areas of Parallelograms and Triangles
Answers
proved.
Solution:
Given, ABCD is a parallelogram (Opposite angles of the parallelogram are equal)
=> ------------------(1) [ AX is bisector of ∠ A and CY is bisector of ∠ C ] .
Now, and CY is the transversal.
...(2) [Alternate interior angles].
From (1) and (2), we get Transversal AB intersects AX and CY at A and Y respectively such that
i.e., corresponding angles formed are equal.
hence, proved.
Given,
ABCD is a trapezium with AB || DC.
XY || AC
Construction,
CX is joined.
To Prove,
ar(ADX) = ar(ACY)
Proof:
ar(â?³ADX) = ar(â?³AXC) — (i) (On the same base AX and between the same parallels AB and CD)
also,
ar(â?³ AXC)=ar(â?³ ACY) — (ii) (On the same base AC and between the same parallels XY and AC.)
From (i) and (ii),
ar(â?³ADX)=ar(â?³ACY)
Construction: Draw DG parallel to BF.
In triangle ADG, EF parallel to base and E is the midpoint of AD. Therefore AF = FG.(Midpoint theorem) ------------(i)
In triangle BCF, DG is parallel to BF and D is mid pint of BC FG = GC (Midpoint theorem) -----------(ii)
From (i) and (ii) Now, AF = FG = GC = 1/3 AC.
â?³DAC and â?³DBC lie on the same base DC and between the same parallels AB and CD.
∴ ar(â?³DAC) = ar(â?³DBC)
⇒ ar(â?³DAC) − ar(â?³DOC) = ar(â?³DBC) − ar(â?³DOC)
⇒ ar(â?³AOD) = ar(â?³BOC)
Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔAC)
Proof:
EY || BC (XY || BC) — (i)
also,
BEâ?¥ CY (BE || AC) — (ii)
From (i) and (ii),
BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)
Similarly,
BXFC is a parallelogram.
Parallelograms on the same base BC and between the same parallels EF and BC.
⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) — (iii)
Also,
â?³AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(â?³AEB) = 1/2ar(BEYC) — (iv)
Similarly,
â?³ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(â?³ ACF) = 1/2ar(BXFC) — (v)
From (iii), (iv) and (v),
ar(â?³AEB) = ar(â?³ACF)
Solution:
and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.
ar (
) = ½ ar (ABCD) ... (1) Similarly,
and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
ar (
) = ½ ar (ABCD) ... (2) From equation (1) and (2), we obtain ar (
) = ar (
)
Solution:
In Δ ACD, N and O are the mid points of the sides CD and DA respectively
=> and ON = 1/2 AC ----------------(i)
Similarly, In , L and M are the mid points of the sides AB and BC respectively
=> and LM = 1/2 AC ------------------(ii)
From (i) and (ii) ON = LM and Thus, LMNO is a parallelogram.
O is the mid point of AC and BD. (diagonals of bisect each other)
In ΔABC, BO is the median.
∴ ar(AOB) = ar(BOC) — (i)
also,
In ΔBCD, CO is the median.
∴ ar(BOC) = ar(COD) — (ii)
In ΔACD, OD is the median.
∴ ar(AOD) = ar(COD) — (iii)
In ΔABD, AO is the median.
∴ ar(AOD) = ar(AOB) — (iv)
From equations (i), (ii), (iii) and (iv),
ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)
So, the diagonals of a parallelogram divide it into four triangles of equal area.
