NCERT Solutions Class 9 Mathematics Chapter 7 Areas Of Parallelograms And Triangles Download In Pdf

Chapter 9 Areas of Parallelograms and Triangles Download in Pdf

**Question1.** Which of the following figures lie on the same base and between the same parallels.
In such a case, write the common base and the two parallels.

EXERCISE 9.2

**Question 1.** In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC
and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and
CF = 10 cm, find AD.

**Question 2.** If E,F,G and H are respectively the mid-points of
the sides of a parallelogram ABCD, show that
ar (EFGH) =
1 ar (ABCD)
2
.

**Question 3.** 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).

**Question 4.** In Fig. 9.16, P is a point in the interior of a
parallelogram ABCD. Show that:

(i) ar (APB) + ar (PCD) =
1 ar (ABCD)
2

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint : Through P, draw a line parallel to AB.]

**Question 5.** . In Fig. 9.17, PQRS and ABRS are parallelograms
and X is any point on side BR. Show that

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AX S) =
1 ar (PQRS)

**Question 6.** A farmer was having a field in the form of a parallelogram PQRS. She took any point A
on RS and joined it to points P and Q. In how many parts the fields is divided? What
are the shapes of these parts? The farmer wants to sow wheat and pulses in equal
portions of the field separately. How should she do it?

**Question 1.** In Fig.9.23, E is any point on median AD of a
Δ ABC. Show that ar (ABE) = ar (ACE).

**Question 2.** In a triangle ABC, E is the mid-point of median
AD. Show that ar (BED) =
1 ar(ABC)
4
.

**Question 2.** Show that the diagonals of a parallelogram divide
it into four triangles of equal area.

**Question 4.** In Fig. 9.24, ABC and ABD are two triangles on
the same base AB. If line- segment CD is bisected
by AB at O, show that ar(ABC) = ar (ABD).

**Question 5.** D, E and F are respectively the mid-points of the sides BC, CA and AB of a Δ ABC.
Show that

(i) BDEF is a parallelogram.

(ii) ar (DEF) =
1
4
ar (ABC)

(iii) ar (BDEF) =
1
2
ar (ABC)

**Question 6.** In Fig. 9.25, diagonals AC and BD of quadrilateral
ABCD intersect at O such that OB = OD.
If AB = CD, then show that:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]

**Question 7.** D and E are points on sides AB and AC respectively of Δ ABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.

**Question 8.** 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 (ABE) = ar (ACF)

**Question 9.** The side AB of a parallelogram ABCD is produced
to any point P. A line through A and parallel to CP
meets CB produced at Q and then parallelogram
PBQR is completed (see Fig. 9.26). Show that
ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ)
and ar (APQ).]

**Question 10.** Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.
Prove that ar (AOD) = ar (BOC).

**Question 11.** In Fig. 9.27, ABCDE is a pentagon. A line through
B parallel to AC meets DC produced at F. Show
that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

**Question 12.** A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat
of the village decided to take over some portion of his plot from one of the corners to
construct a Health Centre. Itwaari agrees to the above proposal with the condition
that he should be given equal amount of land in lieu of his land adjoining his plot so
as to form a triangular plot. Explain how this proposal will be implemented.

**Question 13.** 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).
[Hint : Join CX.]

**Question 14.** In Fig.9.28, AP || BQ || CR. Prove that
ar (AQC) = ar (PBR).

**Question 15.** . Diagonals AC and BD of a quadrilateral
ABCD intersect at O in such a way that
ar (AOD) = ar (BOC). Prove that ABCD is a
trapezium.

**Question 16.** In Fig.9.29, ar (DRC) = ar (DPC) and
ar (BDP) = ar (ARC). Show that both the
quadrilaterals ABCD and DCPR are
trapeziums.

**Question 1.** Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal
areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

**Question 2.** In Fig. 9.30, D and E are two points on BC
such that BD = DE = EC. Show that
ar (ABD) = ar (ADE) = ar (AEC).
Can you now answer the question that you have
left in the ‘Introduction’ of this chapter, whether
the field of Budhia has been actually divided
into three parts of equal area? triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n
equal parts and joining the points of division so obtained to the opposite vertex of
BC, you can divide ΔABC into n triangles of equal areas.]

**Question 3.** In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that
ar (ADE) = ar (BCF).

**Question 4.** In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that
AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
[Hint : Join AC.]

**Question 5.** In Fig.9.33, ABC and BDE are two equilateral
triangles such that D is the mid-point of BC. If AE
intersects BC at F, show that:

(i) ar (BDE) =
1
4
ar (ABC)

(ii) ar (BDE) =
1
2
ar (BAE)

(iii) ar (ABC) = 2 ar (BEC)

(iv) ar (BFE) = ar (AFD)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) =
1
8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

**Question 6.** Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]

**Question 7.** P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R
is the mid-point of AP, show that
:

(i) ar (PRQ) =
1
2
ar (ARC)

(ii) ar (RQC) =
3
8 ar (ABC)

(iii) ar (PBQ) = ar (ARC)

**Question 8.** In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are
squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC
at Y. Show that:

(i) Δ MBC Δ ABD

(ii) ar (BYXD) = 2 ar (MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB Δ ACE

(v) ar (CYXE) = 2 ar (FCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Note : Result

(vii) is the famous Theorem of Pythagoras. You shall learn a simpler
proof of this theorem in Class X.

Please send your queries to ncerthelp@gmail.com you can aslo visit our facebook page to get quick help. Link of our facebook page is given in sidebar

- Chapter 3 Coordinate Geometry
- Chapter 4 Linear Equations In Two Variables
- Chapter 7 Triangles
- Chapter 8 Quadrilaterals
- Chapter 9 Areas of Parallelograms and Triangles
- Chapter 10 Circles
- Chapter 11 Constructions
- Chapter 12 Heron’s Formula
- Chapter 13 Surface Areas and Volumes
- Chapter 14 Statistics
- Chapter 15 Probability

- NCERT Solutions for Class 9 Science Maths Hindi English Math
- NCERT Solutions for Class 10 Maths Science English Hindi SST
- Class 11 Maths Ncert Solutions Biology Chemistry English Physics
- Class 12 Maths Ncert Solutions Chemistry Biology Physics pdf

- Class 1 Model Test Papers Download in pdf
- Class 5 Model Test Papers Download in pdf
- Class 6 Model Test Papers Download in pdf
- Class 7 Model Test Papers Download in pdf
- Class 8 Model Test Papers Download in pdf
- Class 9 Model Test Papers Download in pdf
- Class 10 Model Test Papers Download in pdf
- Class 11 Model Test Papers Download in pdf
- Class 12 Model Test Papers Download in pdf

Copyright @ ncerthelp.com A free educational website for CBSE, ICSE and UP board.