मराठी

Describe the Locus for Questions 1 to 13 Given Below: 1. the Locus of a Point at a Distant 3 Cm from a Fixed Point. - Mathematics

Advertisements
Advertisements

प्रश्न

Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point. 

Advertisements

उत्तर

  

The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle. 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC. 


Use ruler and compasses only for this question.

  1. Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
  2. Construct the locus of points inside the triangle which are equidistant from BA and BC.
  3. Construct the locus of points inside the triangle which are equidistant from B and C.
  4. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.

Prove that: 


F is equidistant from A and B.


The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.

Prove that: 


F is equidistant from AB and AC.


Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B. 


Describe the locus of the moving end of the minute hand of a clock. 


Describe the locus of points at distances less than or equal to 2.5 cm from a given point. 


In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP. 


In Fig. AB = AC, BD and CE are the bisectors of ∠ABC and ∠ACB respectively such that BD and CE intersect each other at O. AO produced meets BC at F. Prove that AF is the right bisector of BC.


Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×