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प्रश्न
Use ruler and compasses only for the following questions:
Construct triangle BCP, when CB = 5 cm, BP = 4 cm, ∠PBC = 45°.
Complete the rectangle ABCD such that :
(i) P is equidistant from AB and BC and
(ii) P is equidistant from C and D. Measure and write down the length of AB.
उत्तर
Given: BC = 5 cm, BP = 4 cm and ∠PBC = 45°
Steps of construction :
1. Constant ΔBCP with BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
2. Draw perpendicular BE and CF and B and C respectively.
3. Draw perpendicular from on CF meeting CF in K.
4. Cut CD from CF, such that CK = KD.
5. Cut BA from BE, such that BA = CD.
6. Join AD.
Hence, ABCD is the required rectangle and AB = 5·7 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.
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.
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
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.
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 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.
In Δ ABC, B and Care fixed points. Find the locus of point A which moves such that the area of Δ ABC remains the same.
Without using set squares or protractor construct:
(i) Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
(ii) Draw the locus of a point which moves so that it is always 2.5 cm from B.
(iii) Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
(iv) Mark the point of intersection of the loci with the letter P and measure PC.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
