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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.
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.
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.
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