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∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm, Find the area of ∆ABC.

Q. 8.    \Delta ABC is isosceles with AB=AC=7.5\; cm and BC=9\; cm (Fig 11.26). The height AD fromA to BC, is 6\; cm, Find the area of

            \Delta ABC.  What will be the height from C to AB i.e., CE ?

            

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R Riya

We know that 

Area of triangle = \frac{1}{2} \times base \times height

Now,

When base(BC) = 9 cm and height(AD) = 6 cm 

Then, the area is equal to

\Rightarrow \frac{1}{2} \times 9 \times 6 = 27 \ cm^2

Now, 

When base(AB) = 7.5 cm and height(CE) = h , area remain same 

Therefore,

\Rightarrow 27= \frac{1}{2} \times 7.5 \times CE

\Rightarrow CE = \frac{54}{7.5}= 7.2 \ cm 

Therefore, value of CE is  7.2cm  and area is equal to 27 \ cm^2

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