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how to do ncert textbook maths ex 12.2 question 2 ?????

    • Posted by Stella Antony 4 years, 11 months ago

    CBSE > Class 09 > Mathematics
    • 3 answers

    Mehak Choudhary 4 years, 11 months ago

    Triangle ABC ac sq = ab sq +=bc sq (5) sq = 3 sq + 4sq Therefore, triangle ABC is a right angled triangle right angle at point B . AREA of triangle ABC = HALF AB = BC ( half × 3× 4) =6cmsq For triangle ADC perimeter=2s =AC+CD+DA=(5+4+5) CM 14cm S =7cm By herons formula area of triangle =√ s(s - a ) ( s - b ) ( s - c ) Area of triangle ADC =√ 7 ( 7-5) (7-5) (7-4) cm sq = √ 7 (2)(2)(3) cm sq =2 √ 21 cm sq = ( 2× 4. 583 ) cmsq =9.166cm sq Area of ABCD = area of triangle ABC + area of triangle ACD

    1Thank You

    Aryan Kothari 4 years, 11 months ago

    Ar(abc) = 1÷2 bh (because irs a pythagoreas triplet so, ar(abc)= 6cm*2 Ar(adc) (by herons formula) S= (5+5+4)÷2=7 Area = √84 Area=2√21 cm*2 Area abcd = 6cm*2 +2√21 cm2

    2Thank You

    Vinayak Killedar 4 years, 11 months ago

    I dont know

    2Thank You

    ANSWER
    http://mycbseguide.com/examin8/

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