Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Draw DN ⊥ AB and PM ⊥ AB.
Now,
`Area (ΙΙ^(gm) ABCD) = AB xx DN , ar (ΔAPB ) = 1/2 (AB xx PM)`
Now , PM < DN
⇒ `AB xx PM < AB xx DN`
⇒ ` 1/2 (AB xx PM) < 1/2 (AB xx DN)`
⇒ `Area ( ΔAPB ) <1/2 ar ( Parragram ABCD)`
APPEARS IN
संबंधित प्रश्न
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is ______.
The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m2, find
(i) its perimeter
(ii) cost of fencing it at the rate of ₹40 per meter.
Find the area of a rectangle whose length = 3.6 m breadth = 90 cm
Find the area of a square, whose side is: 4.1 cm.
Length of a rectangle is 30 m and its breadth is 20 m. Find the increase in its area if its length is increased by 10 m and its breadth is doubled.
The table given below contains some measures of the rectangle. Find the unknown values.
| Length | Breadth | Perimeter | Area |
| 13 cm | ? | 54 cm | ? |
This stamp has an area of 4 square cm. Guess how many such stamps will cover this big rectangle.
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?
Whose footprint is larger - yours or your friend’s?
Is the area of both your footprints the same?
