Advertisements
Advertisements
Question
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove
that: (1) ar (ΔADO) = ar (ΔCDO) (2) ar (ΔABP) = ar (ΔCBP)
Advertisements
Solution

Given that ABCD is a parallelogram
To prove: (1) ar (ΔADO) = ar (ΔCDO)
(2) ar ( ΔABP) = ar (ΔCBP)
Then area (ΔADO) = area (ΔCDO)
Then; area (ΔBAO) area (ΔBCO) ......(1)
In a ΔPAC, Since PO is a median
Then, area (ΔPAO) = area (ΔPCO) ......(2)
⇒ Area (ΔABP) = Area of ΔCBP
APPEARS IN
RELATED QUESTIONS
If ABCD is a parallelogram, then prove that
𝑎𝑟 (Δ𝐴𝐵𝐷) = 𝑎𝑟 (Δ𝐵𝐶𝐷) = 𝑎𝑟 (Δ𝐴𝐵𝐶) = 𝑎𝑟 (Δ𝐴𝐶𝐷) = `1/2` 𝑎𝑟 (||𝑔𝑚 𝐴𝐵𝐶𝐷) .
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)
In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is
bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)

In the given figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ΔEFG.

ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AErespectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF.
In a ΔABC if D and E are mid-points of BC and AD respectively such that ar (ΔAEC) = 4cm2, then ar (ΔBEC) =
ABCD is a trapezium in which AB || DC. If ar (ΔABD) = 24 cm2 and AB = 8 cm, then height of ΔABC is
A floor is 40 m long and 15 m broad. It is covered with tiles, each measuring 60 cm by 50 cm. Find the number of tiles required to cover the floor.
The length and breadth of a rectangular piece of land are in the ratio 5 : 3. If the total cost of fencing it at the rate of ₹24 per meter is ₹9600, find its :
(i) length and breadth
(ii) area
(iii) cost of levelling at the rate of ₹60 per m2.
What is the area of a closed shape?
