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प्रश्न
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.
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उत्तर
(i) Draw BC = 4 cm. Draw BA at B such that ∠ABC = 75°. Cut CA = 6 cm. Then ΔABC is the required Δ.
(ii) Draw single bisector of ∠B. Draw ⊥ bisector of BC. Their point of intersection (P) is the requisite point.
AP = 3·9 cm.
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संबंधित प्रश्न
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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.
