Question

9. AABC is an isosceles triangle with AB = AC=10 cm, BC = 12 cm, AD 1 BC
and AD = 8 cm. Find the area of AABC. Also, find the length of CE.​

Answer 1

Answer:

Area of triangle ABC=24√4cm^2

Area of triangle ABC=24√4cm^2EC=4.8√4

Step-by-step explanation:

Area of triangle ABC:-

s=(a+b+c)/2 and let a=AB=10, b=AB=10, c=BC=12

s=(10+10+12)/2

=32/2

s=16

now, area of a triangle is:-

√s(s-a)(s-b)(s-c)

=√16(16-10)(16-10)(16-12)

=√16*6*6*4

=√4*4*6*6*4

Area of triangle ABC=24√4cm^2........... eq.1

we also know that area of a triangle is also:-

1/2*base*hight and the base is AB and hight is EC

Now,

Area of triangle ABC=1/2*10*EC

Area of triangle ABC=5EC ............ eq.2

from eq.1 and 2 we get:-

5EC=24√4

EC=24√4/5

EC=4.8√4

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Answer 2

Answer:

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