Advertisements
Advertisements
प्रश्न
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°
Hence:
1) Construct the locus of points equidistant from BA and BC
2) Construct the locus of points equidistant from B and C.
3) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Advertisements
उत्तर
1) Draw a line segment AB of length 5.5 cm.
2) Make an angle m∠BAX = 105° using a protractor
3) Draw an arc AC with radius AC = 6 cm on AX with centre at A.
4) Join BC.
Thus ΔABC is the required triangle.
a) Draw BR, the bisector of ∠ABC, which is the locus of points equidistant from BA and BC.
b) Draw MN, the perpendicular bisector of BC, which is the locus of points equidistant from B and C
c) The angle bisector of ∠ABC and the perpendicular bisector of BC meet at point P. Thus, P satisfies the above two loci.
Length of PC = 4.8 cm
APPEARS IN
संबंधित प्रश्न
Describe the locus of vertices of all isosceles triangles having a common base.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
- Construct a ΔABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
- Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
- Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
- Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
- Measure and record the length of CQ.
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
- Plot the points A(1, 1), B(5, 3) and C(2, 7).
- Construct the locus of points equidistant from A and B.
- Construct the locus of points equidistant from AB and AC.
- Locate the point P such that PA = PB and P is equidistant from AB and AC.
- Measure and record the length PA in cm.
A and B are fixed points while Pis a moving point, moving in a way that it is always equidistant from A and B. What is the locus of the path traced out by the pcint P?
In given figure, ABCD is a kite. AB = AD and BC =CD. Prove that the diagona AC is the perpendirular bisector of the diagonal BD.
In Δ PQR, bisectors of ∠ PQR and ∠ PRQ meet at I. Prove that I is equidistant from the three sides of the triangle , and PI bisects ∠ QPR .
Using a ruler and compass only:
(i) Construct a triangle ABC with BC = 6 cm, ∠ABC = 120° and AB = 3.5 cm.
(ii) In the above figure, draw a circle with BC as diameter. Find a point 'P' on the circumference of the circle which is equidistant from Ab and BC.
Measure ∠BCP.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
Ruler and compass only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
(ii) Construct the locus of point inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
