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Area of isoscles triangle = √s(s-a)(s-b)(s-c)

S = ( a+ b + c) / 2 =( 5+5 + 8 )/2 = 9

Area = √9 (4) (4)(1) = 3 * 2 * 2 = 12

Answer - b

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Since ABC is an isosceles triangle, drawing a perpendicular line will bisect side BC and give us the height.

We now have two right angle triangles. To find the length of the height:
(AB)^2 = (height)^2 + 1/2(BC)^2
25 -16= H
3 = H

Area of a triangle = 1/2BH = 1/2* 8* 3 =12

OA: B
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Solution



Given

    • ABC is a triangle with AB = AC = 5 and BC = 8

To Find

    • The area of the triangle ABC.

Approach and Working Out

    • As given in the figure AD is perpendicular to BC.
      o BD = DC = 8/2 = 4
    • ABD is a right-angled triangle with base = 4 and hypotenuse = 5.
      o The height will be 3 (Pythagorean triplet).
    • Area of ABC = ½ × 3 × 8 = 12

Correct Answer: Option B
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We can calculate area of isoscles triangle with two similar sides 'A' and base 'B' as follows:
__________
Area= B/4 (√4*A^2 - B^2)
____________
Area= 8/4 (√4*5^2 - 8^2 )
_________
Area= 2 ( √4*25 - 64 )
______
Area = 2 (√100-64)
__
Area = 2(√36)

Area= 2*6

Area= 12.

Answer - b

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