NCERT Grade 9-Areas of Parallelograms and Triangles-Questions

NCERT Solutions for Class 9 Maths

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Areas of Parallelograms and Triangles

Exercise 1

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

Exercise 2

2.   In figure, ABCD is a parallelogram. AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

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

4.    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).

5.    In figure, P is a point in the interior of a parallelogram ABCD. Show that:

(i) ar (APB) + ar (PCD) =  ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)


6.    In figure, PQRS and ABRS are parallelograms and X is any point on the side BR. Show that:

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

(ii) ar (AXS) = ar (PQRS)

7.    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 field 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?

Exercise 3

8.    In figure, E is any point on median AD of a Δ ABC. Show that ar (Δ ABE) = ar (Δ ACE).

9.    In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) =  ar (ABC).

10.  Show that the diagonals of a parallelogram divide it into four triangles of equal area.

11.   In figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at 0, show that ar (ABC) = ar (ABD).

12.  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) =  ar (ABC)

(iii) ar (BDEF) =  ar (ABC)

13.  In figure, diagonals AC and BD of quadrilateral ABCD intersect at 0 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.

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

15.  XY is a line parallel to side BC of triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).

16.  The side AB of 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. Show that ar (ABCD) = ar (PBQR).

17.  Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at 0. Prove that ar(AOD) = ar (BOC).

18.  In figure, 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)

19.  A villager Itwaari has a plot of land of the shape of quadrilateral. The Gram Panchyat of two villages decided to take over some portion of his plot from one of the corners to construct a health centre. Itwaari agrees to the above personal 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.

20.  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).

21.  In figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

22.  Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar (AOD) = ar(BOC). Prove that ABCD is a trapezium.

23.  In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Exercise 4


24.  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.

25.  In figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Can you know 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?

26.  In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

27.  In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).


28.  In figure, ABC and BDF are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:

(i) ar (BDE) = ar (ABC)

(ii) ar (BDE) =  ar (BAE)

(iii) ar (ABC) = 2 ar (BEC)

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

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) =  ar (AFC)

29.  Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar (APB) x ar (CPD) = ar (APD) x ar (BPC)

30.  P and Q are respectively the mid-points of sides AB and BC or a triangle ABC and R is the mid­point of AP, show that:

31.  In figure, 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).