An 80 mm long line PQ has the end Q lying both in the HP and VP. The line is inclined at 300 to HP and 450 to the VP. Draw its projections.
Answers
Given Area of circle=154cm²
⇒πr²=154cm²⇒r² = 154×7/22=49cm
ABC is an equilateral △, h is the altitude of
△ABC
0 is the incenter of △ABC, and this is the point of intersection of the angular bisectors. Hence, these bisectors are also the altitude and medians whose point of intersection divides the medians in the ratio 2:1
∴∠ADB=90° & OD=1/3 AD & OD is radius of circle. Then,
r= h/3⇒h=3r=3×7=21cm
Let each side of an equilateral triangle be 'a', then altitude of an equilateral triangle is (√3/2) times its side. So that
h=3/2 a⇒a=2h/3 = 2×21/3 = 2×21√3/3=14√3cm
Perimeter of triangle ABC=3a=3×14√3cm=42√3 cm=42×1.73
=72.66cm²
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