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प्रश्न
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.
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उत्तर
Steps of construction:
- Draw a line segment BC = 6 cm.
- At B, draw a ray BX making an angle 60 degree and cut off BA = 9 cm.
- Join AC. ABC is the required triangle.
- Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
- Through A, draw a line m || BC.
- The perpendicular bisector of BC and the parallel line m intersect each other at Q.
- Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.
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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.
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. prove your construction.
(ii) Construct the locus of points, inside the circle that are equidistant from AB and AC.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
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.
AB and CD are two intersecting lines. Find a point equidistant from AB and CD, and also at a distance of 1.8 cm from another given line EF.
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.
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.
Without using set squares or protractor.
(i) Construct a ΔABC, given BC = 4 cm, angle B = 75° and CA = 6 cm.
(ii) Find the point P such that PB = PC and P is equidistant from the side BC and BA. Measure AP.
