English

Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.

Advertisements
Advertisements

Question

Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude. 

Diagram
Advertisements

Solution

Steps of construction:

 
Draw a line XY parallel to the base BC from the vertex A.


This line is the locus of vertex A. All the triangles which have the base BC and length of altitude equal to AD.

shaalaa.com
  Is there an error in this question or solution?
Chapter 16: Loci (Locus and Its Constructions) - Exercise 16 (B) [Page 241]

APPEARS IN

Selina Concise Mathematics [English] Class 10 ICSE
Chapter 16 Loci (Locus and Its Constructions)
Exercise 16 (B) | Q 16. | Page 241

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C. 


In each of the given figures; PA = PB and QA = QB. 

i.
ii.

Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.


Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR. 


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. 


Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B. 


In the figure given below, find a point P on CD equidistant from points A and B. 


Describe the locus of points at a distance 2 cm from a fixed line. 


Describe the locus of the centres of all circles passing through two fixed points. 


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.


Use ruler and compasses for the following question taking a scale of 10 m = 1 cm. A park in a city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×