Advertisements
Advertisements
Question
In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:
(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)
Advertisements
Solution
(1) Clearly Triangles LMB and LMC are on the same base LM and between the same
parallels LM and BC.
∴ ar (ΔLMB) = ar (ΔLMC) ......(1)
(2) We observe that triangles LBC and MBC area on the same base BC and between the
same parallels LM and BC
∴ arc ΔLBC = ar (MBC) ..........(2)
(3) We have
ar (ΔLMB) = ar (ΔLMC) [from (1)]
⇒ ar ( ΔALM) + ar (ΔLMB) = ar (ΔALM) + ar (LMC)
⇒ ar (ΔABM) = ar (ΔACL)
(4) We have
ar(ΔCBC) = ar (ΔMBC) ∴ [from (1)]
⇒ ar (ΔLBC) = ar (ΔBOC) = a (ΔMBC) - ar (BOC)
⇒ ar (ΔLOB) = ar (ΔMOC)
APPEARS IN
RELATED QUESTIONS
In the given 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).
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)
In the given 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)
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
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.
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.]
In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
