Question

NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 -Maths 9th

Answer 1

1. In the figure, E is any point on median AD of a DABC.  Show that ar (ABE) = ar (ACE). We have a ΔABC such that AD is a median. ∴ ar (ΔABD) = ar (ΔADC)                ...(1) [∵ A median divides the triangle into two triangles of equal areas.] Similarly, in DBEC, we have ar (ΔBED) = ar (ΔDEC)                …(2) Subtracting (2) from (1), we have ar (ΔABD) - ar (ΔBED) = ar (ΔADC) - ar (ΔDEC) ⇒ ar (ΔABE) = ar (ΔACE). 2. In a triangle ABC, E is the mid-point of median AD.  Show that ar (BED) =  ar (ABC).  We have a ΔABC and its median AD. Since, a median divides the triangle into two triangles equal in area. ∴ ar (ΔABD) = (1/2) ar (ΔABC)            …(1) Let us join B and E. Now, in ΔABD, BE is a median. [∵ E is the mid-point of AD] ∴ ar (ΔBED) = (1/2)ar (ΔABD) From (1) and (2), we have ar (ΔBED) = (1/2)[(1/2)]ar ( ABC) ] ⇒ ar (ΔBED) = (1/4)ar (ΔABC) 3. Show that the diagonals of a parallelogram divide it into four triangles of equal area. We have a parallelogram ABCD such that its diagonals intersect at O. ∵ Diagonals of a parallelogram bisect each other. ∴ AO = OC and BO = OD Let us draw CE ⊥ BD. Now, ar (ΔBOC) = (1/2) BO x CE and ar (ΔDOC) = (1/2) OD x CE Since, BO = OD ∴ ar (ΔBOC) = ar (ΔDOC)     …(1) Similarly, ar (ΔAOD) = ar (ΔDOC)   …(2) and ar (ΔAOB) = ar (BOC)        …(3) From (1), (2) and (3), we have ar (ΔAOB) = ar (ΔBOC) = ar (ΔBOC) = ar (ΔAOD) Thus, the diagonals of a parallelogram divide it into four triangles of equal area. 4. In figure, 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). We have ΔABC and ΔABD are on the same base AB. ∵ CD is bisected at O             [Given] ∴ AO = BO Now, in ΔACD, AO is a median. ∴  ar (ΔOAC) = ar (ΔOAD)                  …(1) Again, in DBCD, BO is a median. ∴  ar (ΔOBC) = ar (ΔODC)                     …(2) Adding (1) and (2), we have ar (ΔOAC) + ar (ΔOBC) = ar (ΔOAD) + ar (ΔODC) ⇒ ar (ΔABC) = ar (ΔABD) 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)  We have: ΔABC such that the mid-points of BC, CA and AB are respectively D, E and F. (i) To prove that BDEF is a parallelogram. In ΔABC, E and F are the mid-points of AC and AB. ∴ EF || BC ⇒ EF || BD Also EF =(1/2) BC ⇒ EF = BD           [∵ D is the mid-point of BC.] ∵ BDEF is a quadrilateral whose one pair of opposite sides is parallel and of equal lengths. ∴ BDEF is a parallelogram. (ii) To prove that ar (DEF) = (1/4)ar (ABC). We have proved that BDEF is a parallelogram. Similarly, CDEF is a parallelogram and ΔEAF is a parallelogram. Now, parallelogram BDEF and parallelogram CDEF are on the same base EF and between the same parallels BC and EF. ∴ ar (BDEF) = ar (CDEF) ⇒(1/2) ar (BDEF) = (1/2)ar (DEF) ⇒ ar (ΔBDF) = ar (ΔCDF) …(1) [∵ Diagonal of a parallelogram divides it into two triangles of equal area] Similarly, ar (ΔCDE) = ar (ΔDEF)         …(2) ar (ΔAEF) = ar (ΔDEF)           …(3) From (1), (2) and (3), we have ar (ΔAEF) = ar (ΔFBD) = ar (ΔDEF) = ar (ΔCDE) Thus, ar (ΔABC) = ar (ΔAEF) + ar (ΔFBD) + ar (ΔDEF) + ar (ΔCDE) = 4 ar (ΔDEF) ⇒ ar (ΔDEF) =(1/4) ar (ΔABC) (iii) To prove that ar (BDEF) = (1/2)ar (ABC). We have ar (BDEF) = ar (ΔBDF) + ar (ΔDEF) = ar (ΔDEF) + ar (ΔDEF)           [∵ ar (ΔDEF) = ar (ΔBDF)] = 2 ar (ΔDEF) = 2 [ (1/4)ar ( ABC)] = 2 x (1/4)ar (ΔABC) = (1/2)ar (ΔABC) Thus, ar (BDEF) =(1/2)ar (ΔABC) 6. In the figure, diagonals AC and BD of a 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.  Solution: We have a quadrilateral ABCD whose diagonals AC and BD intersect at O. We also have that OB = OD AB = CD Let us draw DE ⊥ AC and BF ⊥ AC (i) To prove that ar (ΔDOC) = ar (ΔAOB) In ΔDEO and ΔBFO, we have DO = BO             [Given] –DOE = –BOF [Vertically opposite angles] –DEO = –BFO              [Each = 90º] ∴ ΔDEO ≌ ΔBFO               [ASA criteria] ⇒ DE = BF [c.p.c.t] and ar (ΔDEO) = ar (ΔBFO) …(1) [∵ Congruent triangle are equal in areas] Now, in ΔDEC and ΔBFA, we have –DEC = –BFA [Each = 90º] DE = BF [Proved] DC = BA [Given] ∴ ΔDEC ≌ ΔBFA                [RHS criteria] ⇒ ar (ΔDEC) = ar (ΔBFA)                     …(2) Adding (1) and (2), we have ar (ΔDEO) + ar (ΔDEC) = ar (ΔBFO) + ar (ΔBFA) ⇒ ar (ΔDOC) = ar (ΔAOB) (ii) To prove that ar (DCB) = ar (ACB) ∵ ar (ΔDOC) = ar (ΔAOB)            [Proved] ∴ Adding ar (ΔBOC), on both sides, we have ar (ΔDOC) + ar (ΔBOC) = ar (ΔAOB) + ar (ΔBOC) ⇒ ar (ΔDCB) = ar (DACB) (iii) To prove that DA || CB ∵ ΔDCB and ΔACB are on the same base CB and having equal areas. ∴ They lie between the same parallels CB and DA. ⇒ CB || DA (or ABCD is a parallelogram) 7. D and E are points on sides AB and AC respectively of ΔABC such that ar (DBC) = ar (EBC). Prove that DE || BC.  We have ΔABC and points D and E are such that ar (DBC) = ar (EBC) Since, ΔDBC and ΔEBC are on the same base DE and having same area. ∴ They lie between the same parallels DE and BC. ⇒ 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). We have a DABC such that  XY || BC, BE || AC and CF || AB. Since, XY || BC and BE || CY ∴ BCYE is a parallelogram. Now, the parallelogram BCYE and ΔABE are on the same base BE and between the same parallels BE and AC. ∴ ar (ΔABE) = (1/2)ar (BCYE)            …(1) Again, CF || AB                   [Given] XY || BC                        [Given] ⇒ CF || AB and XF || BC ∴ BCFX is a parallelogram. Now, ΔACF and parallelogram BCFX are on the same base CF and between the same parallels AB and FC. ∴ ar (ΔACF) = (1/2)ar (BCFX)               …(2) Also parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels. ∴ ar (BCFX) = ar (BCYE)                 …(3) From (1), (2) and (3) we get ar (ΔABE) = ar(ΔACF) 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 figure). Show that ar (ABCD) = ar (PBQR). Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ). Solution: We have a parallelogram ABCD. AB is produced to P. CB is produced to Q and PBQR is completed. Let us join AC and PQ. ∵ ABCD is a parallelogram [Given] and AC is its diagonal. ar (ΔABC) = (1/2)ar (parallelogram ABCD)               …(1) ∵ BQRP is a parallelogram [Given] and QP is its diagonal. ∴   ar (ΔBPQ) = (1/2)ar (parallelogram BQRP)              …(2) Since ΔACQ and ΔAPQ are on the same base AQ and between the same parallels AQ and CP. ∴  ar (ΔACQ) = ar (ΔAPQ)                                [∵ Triangles on the same base and between the same parallels are equal.] ⇒ ar (ΔACQ)  - ar (ΔABQ) = ar (DAPQ) - ar (DABQ)               [Subtracting ar (ΔABQ) from both sides] ⇒ ar (ΔABC) = ar (ΔBPQ)                       …(3) From (1), (2) and (3), we get (1/2)ar (parallelogram ABCD) = (1/2)ar (parallelogram BQRP) ⇒ 2 [ (1/2)ar (parallelogram ABCD)] = 2 [(1/2)ar (parallelogram BQRP)] ⇒ ar (parallelogram ABCD) = ar (parallelogram BQRP) 10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC). We have a trapezium ABCD having AB || CD and its diagonals AD and BC are joined. Since, triangles on the same base and between the same parallels have equal areas. ∵ ΔABD and ΔABC are on the same base AB and between the same parallels AB and DC. ∴ ar (ΔABD) = ar (ΔABC) Subtracting ar (ΔAOB) from both sides, we get [ar (ΔABD) – ar (ΔAOB)] = [ar (ΔABC) – ar (ΔAOB)] ⇒ ar (ΔAOD) = ar (ΔBOC)