Advertisements
Advertisements
प्रश्न
A plot of land of area 20km2 is represented on the map with a scale factor of 1:200000. Find: The ground area in km2 that is represented by 2cm2 on the map.
उत्तर
1cm on the map = 200,000cm on the land (as the scale is 1:200000)
1cm2 on the map = (200000)2 on the land
1km = 100000cm
⇒ 1km2 = 100000 x 100000cm2
`"distance(map)"/"distance(land)"` = Scale
`(2)/("distance(land)" xx (100000)) = (1)/((200000)^2`
Hence 2cm on map
= `(2 xx 200000 xx 200000)/(100000 xx 100000)`
= 8km2.
APPEARS IN
संबंधित प्रश्न
Prove that the angle bisector of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
- Prove that ΔACD is similar to ΔBCA.
- Find BC and CD.
- Find area of ΔACD : area of ΔABC.
P and Q are points on the sides AB and AC respectively of a ΔABC. If AP = 2cm, PB = 4cm, AQ = 3cm and QC = 6cm, show that BC = 3PQ.
On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and ∠ABC = 90°.
Calculate : the actual lengths of AB and BC in km.
In the following figure, point D divides AB in the ratio 3 : 5. Find :
- `(AE)/(EC)`
- `(AD)/(AB)`
- `(AE)/(AC)`
Also, if: - DE = 2.4 cm, find the length of BC.
- BC = 4.8 cm, find the length of DE.
In the following figure, point D divides AB in the ratio 3 : 5. Find : `(AE)/(AC)`
In ΔABC, D and E are the mid-point on AB and AC such that DE || BC.
If AD = 8cm, AB = 12cm and AE = 12cm, find CE.
ΔABC is right angled at A. AD is drawn perpendicular to BC. If AB = 8cm and AC = 6cm, calculate BD.
A map is drawn to scale of 1:20000. Find: The distance on the map representing 4km
Construct a triangle similar to a given triangle PQR with its sides equal to `2/3` of the corresponding sides of the triangle PQR (scale factor `2/3 < 1`)
