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ABC is an ososceles triangle right …

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ABC is an ososceles triangle right angled at B two equilateral triangle BDC and AEC are constructed with side BC and AC prove that area of teiangle BCD = 1/2 area of triangle ACE

    • Posted by Priyanshu Sinha 6 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 6 years, 6 months ago

    Let AB = x, BC = x

    Area of an equilateral triangle of side a = {tex}\frac { \sqrt { 3 } } { 4 }{/tex}a2
    Area of {tex}\triangle{/tex}BCD = {tex}\frac { \sqrt { 3 } } { 4 }{/tex}x2 ..(i)
    Now side AC = {tex}\sqrt { x ^ { 2 } + x ^ { 2 } } = \sqrt { 2 x ^ { 2 } } = x \sqrt { 2 }{/tex} (by pythagoras theorem)
    Area of {tex}\triangle{/tex}ACE = {tex}\frac { \sqrt { 3 } } { 4 } ( \sqrt { 2 } x ) ^ { 2 }{/tex} = 2
    {tex}\frac { \sqrt { 3 } } { 4 }{/tex}x2 = 2. Area of BCD [from (i)]
    {tex}\therefore{/tex} ar{tex}\triangle{/tex}BCD = {tex}\frac{1}{2}{/tex}area of {tex}\triangle{/tex}ACE.

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