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प्रश्न
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)
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उत्तर
It is given that
XY || BC ⇒ EY || BC
BE || AC ⇒ BE || CY
Therefore, EBCY is a parallelogram.
It is given that
XY || BC ⇒ XF || BC
FC || AB ⇒ FC || XB
Therefore, BCFX is a parallelogram.
Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.
∴ Area (EBCY) = Area (BCFX) ... (1)
Consider parallelogram EBCY and ΔAEB
These lie on the same base BE and are between the same parallels BE and AC.
∴ Area (ΔABE) = 1/2Area (EBCY) ... (2)
Also, parallelogram BCFX and ΔACF are on the same base CF and between the same parallels CF and AB.
∴ Area (ΔACF) = 1/2Area (BCFX) ... (3)
From equations (1), (2), and (3), we obtain
Area (ΔABE) = Area (ΔACF)
संबंधित प्रश्न
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
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).
In the given figure, 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.]
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)
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
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).
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 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).
In the following figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).
