20. Construct an equilateral triangle if its altitude is 6 cm.
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Draw any line l.
Take any point M on it and draw a line p perpendicular to l.
With M as centre, cut off MC = 6 cm.
4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.
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(i) Draw a line XY.
(ii) Construct perpendicular PD at any point D on the line XY.
(iii) From point D, cut-off line segment AD = 6 cm.
(iv) Construct ∠BAD = ∠CAD = 30°. Then ABC is the required triangle.
Justification
As ∠A = ∠BAD + ∠CAD = 30 °+ 30° = 60° and AD perpendicular BC therefore, △ABC is an equilateral triangle with altitude AD = 6 cm.
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