Advertisements
Advertisements
प्रश्न
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)
Advertisements
उत्तर

Construction: Draw BQ ⊥ AC and DR ⊥ AC
Proof:
L.H.S
= ar (Δ APB ) × ar (ΔCP)
= `1/2[ ( AP xx BQ )] xx (1/2 xx PC xx DR)`
=`( 1/2 xx PC xx BQ )xx (1/2 xx AP xx DR)`
= ar (Δ BPC ) × ar (APR)
= RHS
∴ LHS = RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
D is the mid-point of side BC of ΔABC and E is the mid-point of BD. if O is the mid-point
of AE, prove that ar (ΔBOE) = `1/8` ar (Δ ABC).
A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
Find the area of a rectangle whose length = 15 cm breadth = 6.4 cm
Find the area of a rectangle whose length = 3.6 m breadth = 90 cm
Is the area of the blue shape more than the area of the yellow shape? Why?


What is the area of the rectangle? ________ square cm
Measure the length of the floor of your classroom in meters. Also, measure the width.
- So how many children can sit in one square meter?
The King was very happy with carpenters Cheggu and Anar. They had made a very big and beautiful bed for him. So as gifts the king wanted to give some land to Cheggu, and some gold to Anar. Cheggu was happy. He took 100 meters of wire and tried to make different rectangles.
He made a 10 m × 40 m rectangle. Its area was 400 square meters. So he next made a 30 m × 20 m rectangle.
- What other rectangles can he make with 100 meters of wire? Discuss which of these rectangles will have the biggest area.
Each line gives a story. You have to choose the question which makes the best story problem. The first one is already marked.
- 352 children from a school went on a camping trip. Each tent had a group of 4 children.
a) How many children did each tent have? b) How many tents do they need? c) How many children in all are in the school?
Is the area of both your footprints the same?
