Q. 65.0( 1 Vote )

Find the area of isosceles triangle with base 16cm and vertical angle 60° 40’

Answer :


Draw CD perpendicular to AB.


D is the midpoint of AB and ACB = 60° 40’


Then ACD = = 30° 20’ = BCD


In right angled triangle ACD,


tan 30° 20’ =


tan 30° 20’ =


From tangent table, tan 30° 18’ = 0.5844 and Mean difference of 2’ = 0.0008.


tan 30° 20’ = 0.5844 + 0.0008 = 0.5852


CD = 8 ÷ 0.5852 = 13.672 cm


We know that Area of right angled triangle = 1/2 bh.


Area of ΔACD = 1/2 × 8 × 13.672 = 54.688 cm2


Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 54.688 cm2.


Area of ΔABC = Area of (ΔACD + ΔBCD)


= 54.688 + 54.688


= 109.376 cm2


The area of given isosceles triangle = 109.376 cm2.


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