हिंदी

Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that: P is equidistant from B and C. P is equidistant from AB and BC.

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प्रश्न

Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:

  1. P is equidistant from B and C.
  2. P is equidistant from AB and BC.
  3. Measure and record the length of PB.
ज्यामितीय चित्र
विस्तार में उत्तर
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उत्तर


Steps of construction: 

    • Draw a line segment AB = 7 cm.
    • Draw angle ∠ABC = 60° with the help of a compass.
    • Cut off BC = 8 cm.
    • Join A and C.
    • The triangle ABC so formed is the required triangle.
  1. Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
  2. Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
    The point that fulfills the condition required in i. and ii. is the intersection point of the bisector of line BC and the angular bisector of ∠ABC.
    P is the required point, which is equidistant from AB and AC as well as from B and C.
    On measuring the length of line segment PB, it is equal to 4.5 cm.
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अध्याय 14: Locus - Exercise 14 [पृष्ठ ३०३]

APPEARS IN

नूतन Mathematics [English] Class 10 ICSE
अध्याय 14 Locus
Exercise 14 | Q 15. | पृष्ठ ३०३
सेलिना Concise Mathematics [English] Class 10 ICSE
अध्याय 16 Loci (Locus and Its Constructions)
Exercise 16 (A) | Q 18. | पृष्ठ २३८

वीडियो ट्यूटोरियलVIEW ALL [1]

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