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प्रश्न
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).
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उत्तर
Consider ΔACD.
Line-segment CD is bisected by AB at O. Therefore, AO is the median of
ΔACD.
∴ Area (ΔACO) = Area (ΔADO) ... (1)
Considering ΔBCD, BO is the median.
∴ Area (ΔBCO) = Area (ΔBDO) ... (2)
Adding equations (1) and (2), we obtain
Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO)
⇒ Area (ΔABC) = Area (ΔABD)
संबंधित प्रश्न
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.]
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)
A villager Itwaari has a plot of land of 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.
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
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)
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
