Advertisements
Advertisements
Question
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O
intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).
Advertisements
Solution
In triangles and , POA QOC we have
∠AOP = ∠COQ [vertically opposite angles]
OA = OC [ Diagonals of a parallelogram bisect each other]
∠PAC = ∠QCA [ AB || DC ; alternative angles ]
So, by ASA congruence criterion, we have
ΔPOA ≅ QOC
Area (ΔPOA) = area (ΔQOC) .
APPEARS IN
RELATED QUESTIONS
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)
If P is any point in the interior of a parallelogram ABCD, then prove that area of the
triangle APB is less than half the area of parallelogram.
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6: 5: 4. Find the lengths of its sides.
Let the sides be x cm, y cm, and (37 - x - y) cm. Also, let the lengths of altitudes be 6a cm, 5a cm, and 4a cm.
Find the area of a rectangle whose length = 3.6 m breadth = 90 cm
Find the area of a square, whose side is: 7.2 cm.
An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
A magazine charges Rs 300 per 10 sq cm area for advertising. A company decided to order a half page advertisment. If each page of the magazine is 15 cm × 24 cm, what amount will the company has to pay for it?
Find the area of the following figure by counting squares:
Which statement correctly describes how area is determined?
