Advertisements
Advertisements
Question
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.
Solution
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
RELATED QUESTIONS
In the figure given below, Ray PT is bisector of ∠QPR. If PQ = 5.6 cm, QT = 4 cm and TR = 5 cm, find the value of x .
In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form:
The scale of a map is 1 : 200000. A plot of land of area 20km2 is to be represented on the map. Find:
The area in km2 that can be represented by 1 cm2
If ΔABC ~ ΔDEF, then writes the corresponding congruent angles and also write the ratio of corresponding sides.
Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are `3/5` of the corresponding sides of the first triangle.
In ΔABC, D and E are the mid-point on AB and AC such that DE || BC.
If AD = 4x - 3, AE = 8x - 7, BD = 3x - 1 and CE = 5x - 3,Find x.
The sides PQ and PR of the ΔPQR are produced to S and T respectively. ST is drawn parallel to QR and PQ: PS = 3:4. If PT = 9.6 cm, find PR. If 'p' be the length of the perpendicular from P to QR, find the length of the perpendicular from P to ST in terms of 'p'.
Find the scale factor in each of the following and state the type of size transformation:
Image length = 6cm, Actual length = 4cm.
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, ∠ABC = 90°. Calculate:
(i) The actual length of AB in km.
(ii) The area of Plot in sq. km.
If in triangles PQR and XYZ, `"PQ"/"XY" = "QR"/"ZX"` then they will be similar if