Advertisements
Advertisements
प्रश्न
In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
Advertisements
उत्तर
Given: In the following figure, CD || AE and CY || BA
To prove: ar (ΔCBX) = ar (ΔAXY) .
Proof: We know that, triangles on the same base and between the same parallels are equal in areas.
Here, ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA.
ar (ΔABY) = ar (ΔABC)
⇒ ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX)
⇒ ar (AXY) = ar (CBX) ...[Eliminating ar (ABX) from both sides]
APPEARS IN
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.
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.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
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 (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)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
