NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles
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NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles Page/Excercise 9.1
(i) Yes. Trapezium ABCD and triangle PCD are having a common base CD and these are lying between the same parallel lines AB and CD. (ii) No. The parallelogram PQRS and trapezium MNRS are having a common base RS but their vertices (i.e. opposite to common base) P, Q of parallelogram and M, N of trapezium are not lying on a same line. (iii) Yes. The Parallelogram PQRS and triangle TQR are having a common base QR and they are lying between the same parallel lines PS and QR. (iv) No. We see that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC but these are not having any common base. (v) Yes. We may observe that parallelogram ABCD and parallelogram APQD have a common base AD and also these are lying between same parallel lines AD and BQ. (vi) No. We may observe that parallelogram PBCS and PQRS are laying on same base PS, but these are not between the same parallel lines.
NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles Page/Excercise 9.2
In parallelogram ABCD, CD = AB = 16 cm [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base corresponding attitude
Area of parallelogram ABCD = CD AE = AD CF
16 cm 8 cm = AD 10 cm AD = cm = 12.8 cm. Thus, the length of AD is 12.8 cm.
Construction: Join HF.
In parallelogram ABCD
AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)
AB = CD [Opposite sides of a parallelogram are equal] AH = BF and AH || BF Therefore, ABFH is a parallelogram.
Since HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF. area (HEF) = area (ABFH) ...(1) Similarly, we can prove area (HGF) = area (HDCF) ...(2) On adding equations (1) and (2), we have
Here BQC and parallelogram ABCD lie on same base BC and these are between same parallel lines AD and BC.
Similarly, APB and parallelogram ABCD lie on the same base AB and between same parallel lines AB and DC
From equation (1) and (2), we have
area (BQC) = area (APB)
(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.
In parallelogram ABCD we find that
AB || EF (By construction) ...(1)
ABCD is a parallelogram From equations (1) and (2), we have
AB || EF and AE || BF
So, quadrilateral ABFE is a parallelogram
Now, we may observe that APB and parallelogram AB || FE are lying on the same base AB and between the same parallel lines AB and EF. Similarly, for PCD and parallelogram EFCD
area (PCD) = area (EFCD) ... (4)
Adding equations (3) and (4), we have
(ii)Draw a line segment MN, passing through point P and parallel to line segment AD.
In parallelogram ABCD we may observe that
MN || AD (By construction) ...(6)
ABCD is a parallelogram From equations (6) and (7), we have
MN || AD and AM || DN
So, quadrilateral AMND is a parallelogram Now, APD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN. Similarly, for PCB and parallelogram MNCB
area (PCB) = area (MNCB) ... (9)
Adding equations (8) and (9), we have
On comparing equations (5) and (10), we have
Area (APD) + area (PBC) = area (APB) + area (PCD)
(i) Parallelogram PQRS and ABRS lie on the same base SR
and also these are in between same parallel lines SR and PB. (ii) Now consider AXS and parallelogram ABRS
As these lie on the same base and are between same parallel lines AS and BR
From figure it is clear that point A divides the field into three parts. These parts are triangular in shape - PSA, PAQ and QRA
Area of PSA + Area of PAQ + Area of QRA = area of PQRS ... (1)
We know that if a parallelogram and triangle are on the same base and between the same parallels, the area of triangle is half the area of the parallelogram. From equations (1) and (2), we have
Area (PSA) + area (QRA) = area (PQRS) ... (3) Clearly, farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular part PSA and QRA and pulses in triangular parts PAQ.
NCERT Solution for Class 9 Mathematics Chapter 9 - Areas of Parallelograms and Triangles Page/Excercise 9.3
AD is median of ABC. So, it will divide ABC into two triangles of equal areas.
AD is median of ABC. So, it will divide ABC into two triangles of equal areas.
(i) In ABC
E and F are mid points of side AC and AB respectively
So, EF || BC and EF = BC (mid point theorem)
But BD = BC (D is mid point of BC)
So, BD = EF
Now the line segments BF and DE are joining two parallel lines EF and BD of same length.
So, line segments BF and DE will also be parallel to each other and also these will be equal in length.
Now as each pair of opposite sides are equal in length and parallel to each other.
Therefore BDEF is a parallelogram (ii) Using the result obtained as above we may say that quadrilaterals BDEF, DCEF, AFDE are parallelograms.
We know that diagonal of a parallelogram divides it into two triangles of equal area. Now,
Area (AFE) + area (BDF) + area (CDE) + area (DEF) = area (ABC) (iii) Area (parallelogram BDEF) = area (DEF) + area (BDE) Area (parallelogram BDEF) = area (DEF) + area (DEF) Area (parallelogram BDEF) = 2 area (DEF) Area (parallelogram BDEF) = 2 area (ABC) Area (parallelogram BDEF) = area (ABC)
Let us draw DN AC and BM AC
(i) In DON and BOM
DNO = BMO (By construction)
DON = BOM (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule
DON BOM On adding equation (2) and (3), we have
Area (DON) + area (DNC) = area (BOM) + area (BMA)
So, area (DOC) = area (AOB) (ii) We have
Area (DOC) = area (AOB) Area (DOC) + area (OCB) = area (AOB) + area (OCB) (Adding area (OCB) to both sides)
Area (DCB) = area (ACB) (iii) Area (DCB) = area (ACB)
Now if two triangles are having same base and equal areas, these will be between same parallels For quadrilateral ABCD, we have one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel
(DA || CB). Therefore, ABCD is parallelogram
Since, BCE and BCD are lying on a common base BC and also having equal areas, so, BCE and BCD will lie between the same parallel lines.
XY || BC EY || BC
BE || AC BE || CY
So, EBCY is a parallelogram.
It is given that
XY || BC XF || BC
FC || AB FC || XB So, BCFX is a parallelogram.
Now parallelogram EBCY and parallelogram BCFX are on same base BC and between same parallels BC and EF Consider parallelogram EBCY and AEB
These are on same base BE and are between same parallels BE and AC Also parallelogram BCFX and ACF are on same base CF and between same parallels CF and AB From equations (1), (2), and (3), we have
Area (ABE) = area (ACF)
Here, DAC and DBC lie on same base DC and between same parallels AB and CD
(i) ACB and ACF lie on the same base AC and are between
the same parallels AC and BF (ii) Area (ACB) = area (ACF)
Let quadrilateral ABCD be original shape of field. Join diagonal BD and draw a line parallel to BD through point A.
Let it meet the extended side CD of ABCD at point E. Join BE and AD. Let them intersect each other at O. Now portion AOB can be cut from the original field so that new shape of field will be BCE.
Now we have to prove that the area of AOB (portion that was cut so as to construct Health Centre) is equal to the area of the DEO (portion added to the field so as to make the area of new field so formed equal to the area of original field) DEB and DAB lie on same base BD and are between same parallels BD and AE.
Therefore, ABCD is a trapezium
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