costract a triangle ABC having angleABC=60degree, BC=7.5cm & (AB-AC)=1.5cm

Question

from class 10 math wbbse board

Answer 1

1. To construct the required △ABC, here we use two facts on triangles: one is that, in a equilateral triangle, all three sides and three included angles are equal to each other, and the other is that, in a isoscelese triangle, the altitude bisects its base perpendicularly and also bisects the opposite vertex held by two sides of equal length.
2. Mark a point B. From B, draw a ray rightwards, then an arc with radius of given length, BC=7.5cm, crossing the ray at C.  We are to construct vertex A above BC.  From B and C, draw 2 arcs with the same radius BC, crossing each other above BC at D.  From B thru D, draw a ray upwareds.  △DBC is an equilateral triangle.  ∠DBC=60°  Point A will be on the ray.
3. From B, draw an arc with radius of given length, AB－AC=1.5cm, crossing BD at E.  Conect C to E.  Here we draw a perpendicular bisector of CE.  From C and E, draw 2 arcs with the same radius CE, crossing each othere above and below CE.  Draw a line thru the 2 crossings, intersecting the ray from B at A.  Connect A to C.  △AEC is an isoscelese triangle.
4. CK: From A, draw an arc with radius AE.  The arc crosses AC at C. AE=AC, so AB－AC=AB－AE=BE=1.5cm.  CKD.  △ABC is the required triangle with ∠ABC=60°, BC=7.5cm and (AB－AC)=1.5cm.  (AB=approx.12.0 cm, AC=approx. 10.5cm).