मराठी

Two Straight Roads Ab and Cd Cross Each Other at Pat an Angle of 75• .

Advertisements
Advertisements

प्रश्न

Two straight roads AB and CD cross each other at Pat an angle of 75°  . X is a stone on the road AB, 800m from P towards B. BY taking an appropriate scale draw a figure to locate the position of a pole, which is equidistant from P and X, and is also equidistant from the roads. 

बेरीज
Advertisements

उत्तर

Steps of construction: 

(i) Draw two lines AB and CD crossing at an angle of 75 °

(ii) Draw an angle bisector for  ∠ BPD 

(iii) Draw perpendicular from X on angle bisector meeting at 0. 

(iv) From point Y, PX = PY, draw a perpendicular on angle bisector meeting at 0. 

(v) 0 is the point which is equidistant from P, X and both the roads. 

cos θ = `"hypotenuse"/"base"`

cos `75/2 = "PO"/"PX"`

cos (37.5) = `"PO"/800`

0.980243 = `"PO"/800`

PO = 784.19 m

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 15: Loci - Exercise 16.1

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:

  1. equidistant from BA and BC.
  2. 4 cm from M.
  3. 4 cm from N.
    Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.

Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained. 


Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. prove your construction.
(ii) Construct the locus of points, inside the circle that are equidistant from AB and AC. 


Plot the points A(2, 9), B(–1, 3) and C(6, 3) on graph paper. On the same graph paper draw the locus of point A so that the area of ΔABC remains the same as A moves. 


In  Δ PQR, s is a point on PR such that ∠ PQS = ∠  RQS . Prove thats is equidistant from PQ and QR. 


In the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC. 


Describe completely the locus of a point in the following case:

Point in a plane equidistant from a given line. 


Construct a triangle BPC given BC = 5 cm, BP = 4 cm and .

i) complete the rectangle ABCD such that:
a) P is equidistant from AB and BCV
b) P is equidistant from C and D.
ii) Measure and record the length of AB.


Without using set squares or a protractor, construct:

  1. Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
  2. Draw the locus of a point which moves so that it is always 2.5 cm from B.
  3. Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
  4. Mark the point of intersection of the loci with the letter P and measure PC.

How will you find a point equidistant from three given points A, B, C which are not in the same straight line?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×