Advertisements
Advertisements
प्रश्न
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Advertisements
उत्तर
It is given that
Area (ΔDRC) = Area (ΔDPC)
As ΔDRC and ΔDPC lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines.
∴ DC || RP
Therefore, DCPR is a trapezium.
It is also given that
Area (ΔBDP) = Area (ΔARC)
⇒ Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)
⇒ Area (ΔBDC) = Area (ΔADC)
Since ΔBDC and ΔADC are on the same base CD and have equal areas, they must lie between the same parallel lines.
∴ AB || CD
Therefore, ABCD is a trapezium.
APPEARS IN
संबंधित प्रश्न
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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).
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
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).
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
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).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
