हिंदी

Describe the locus of vertices of all isosceles triangles having a common base.

Advertisements
Advertisements

प्रश्न

Describe the locus of vertices of all isosceles triangles having a common base.

Draw and describe the locus in the following case:

The locus of vertices of all isosceles triangles having a common base.

आकृति
अति संक्षिप्त उत्तर
Advertisements

उत्तर

The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Locus - Exercise 14 [पृष्ठ ३०२]

APPEARS IN

नूतन Mathematics [English] Class 10 ICSE
अध्याय 14 Locus
Exercise 14 | Q 2. (ii) | पृष्ठ ३०२
सेलिना Concise Mathematics [English] Class 10 ICSE
अध्याय 16 Loci (Locus and Its Constructions)
Exercise 16 (B) | Q 10. | पृष्ठ २४१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distantce from the given line is always equal to 3 units 


State the locus of a point in a rhombus ABCD, which is equidistant

  1. from AB and AD;
  2. from the vertices A and C.

Construct a rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Measure ∠ ABC. Find the point R on AD such that RB = RC. Measure the length of AR. 


Construct a rhombus ABCD whose diagonals AC and BD are 8 cm and 6 cm respectively. Find by construction a point P equidistant from AB and AD and also from C and D. 


Construct a Δ XYZ in which XY= 4 cm, YZ = 5 cm and ∠ Y = 1200. Locate a point T such that ∠ YXT is a right angle and Tis equidistant from Y and Z. Measure TZ. 


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. 


Draw and describe the locus in the following case:

The locus of a point in rhombus ABCD which is equidistant from AB and AD.


Construct a Δ ABC, with AB = 6 cm, AC = BC = 9 cm; find a point 4 cm from A and equidistant from B and C.


Using ruler and compasses construct:
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of point equidistant from A and C.
(iii) a circle touching AB at A and passing through C.


Using only a ruler and compass construct ∠ABC = 120°, where AB = BC = 5 cm.
(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
(ii) In the above figure, join AD, DC, AE and EC. Describe the figures:
(a) AECB, (b) ABD, (c) ABE.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×