Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
(i) Steps of Construction:
1. Draw a line segment BC = 8 cm.
2. Make ∠CBX = 60°
3. Set off BA = 5 cm, along BX.
4. Join CA.
Then, ΔABC is the required triangle.
(ii) We know that the locus of point equidistant from two intersecting straight lines consist of a pair of straight lines that bisect the angles between the given straight lines.
Therefore in this case is the angle bisector of angle B, It is shown in the adjoining figure.
(iii) We know that the locus of a point equidistant from two fixed points is the right bisector of the straight line joining the two fixed points.
Therefore, in this case the right bisector of side BC of ΔABC. It is shown in the given figure.
(iv) The point P, is the point in intersecting of angle bisector of ∠ABC and the right bisector of BC.
It is shown in the following figure.
(v) On measuring we find the length of PB = 3 cm.
APPEARS IN
संबंधित 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.
On a graph paper, draw the lines x = 3 and y = –5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
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.
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.
Without using set squares or protractor, construct a quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6 cm, BC= 3.6 cm and CD=5 cm. Locate the point P on BD which is equidistant from BC and CD.
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.
Draw and describe the locus in the following case:
The locus of points inside a circle and equidistant from two fixed points on the circle.
Draw and describe the locus in the following case:
The locus of a point in the rhombus ABCD which is equidistant from the point A 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.
Ruler and compasses 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 a ΔABC, in which BC = 6 cm, AB = 9 cm and ∠ABC = 60°.
(ii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ΔABC.
(iii) Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
(iv) Measure and record the length of CQ.
