# NCERT Solutions | Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

NCERT Class 9 Maths Solutions PDF: In this post, we have discussed the solution of the Maths class 9 book which is followed in all CBSE schools. Solutions are given below with proper Explanation and utmost care has been taken to ensure that the solutions are correct. Answers provided will not only help in completing all the assignments but also help students in clearing their concepts. Students can download the solutions by printing the chapters by using the command Ctrl+P in google chrome and saving it in PDF format. All the best !! Please support us by sharing this website with your school friends.

## Exercise 9.3 Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

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

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

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

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).

5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ ABC. Show that
(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC) (iii) ar (BDEF) = 1/2ar (ABC)

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.]

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

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)

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).]

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

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)

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.

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.]

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

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.

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