ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that
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.] 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.] 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.] [hint: get chx.] Question 13: ABCD is a trapezium in which side AB is parallel to side DC. A line is drawn parallel to diagonal AC, and it intersects side AB at point X and side BC at point Y. Prove that: The area of triangle A-D-X is equal to the area of triangle A-C-Y. (That is: ar(ADX) = ar(ACY)) Hint: Join point C to point X. You can contact with us - / thestudyadda / thestudy01 Telegram- https://t.me/TheStudyAdda Video Link Chapter 9 Exercise 9.1- • Chapter 9 Areas of Parallelograms and Tria... Video Link Chapter 9 Exercise 9.2- • Chapter 9 Areas of Parallelograms and Tria... Video Link Chapter 9 Exercise 9.3- • Chapter 9 Areas of Parallelograms and Tria... Chapterwise Playlist of Class 9 Math NCERT - • CLASS 9th NCERT MATHS Chapterwise Solution... QuestionWise Playlist of Class 9th NCERT Math - • Class 9th NCERT Maths | All NCERT Mathemat... Class 9th NCERT Chapter 9 Areas of Parallelogram & Triangles Exercise 9.3 Questions 1. In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar(ABE) = ar(ACE). 2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 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 ΔABC.Show that:-(i) BDEF is a parallelogram. (ii) ar(DEF) = 1/4 ar(ABC)(iii) ar(BDEF) = 1/2 ar(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 ΔABC 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 meets XY at E and F respectively, show thatar(ΔABE) = ar(ΔAC) 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 in 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. 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. 1. In figure 9.23, ai is a point on the median edge of a cube. ar(abbreviation) = ar(access). 2. In a triangle, ai is the mid-point of the median edge. ar(abbreviation) = 1/4 ar(abbreviation). 3. The diagonals of a parallel quadrilateral divide it into four triangles of equal area. 4. In figure 9.24, now and now are two triangles situated on the same base. If the line chd is divided equally on o by now, then ar(abch) = ar(abd). 5. d, ai and f are the mid-points of the same side of the triangle, now and then. Example:- (i) bdaif is a parallel quadrilateral. (ii) ar(daif) = 1/4 ar(abch) (iii) ar (bdaif) = 1/2 ar(abch) 6. In Figure 9.25, the diagonals of the quadrilateral abch are divided by the other side at point o in such a way that ob = od. If ab = chd, then see: (i) ar (doch) = ar (aob) (ii) ar (dchb) = ar (achb) (iii) da || chb or abchd are parallel quadrilaterals. [Hint: Draw a lamb on top of d and b.]
