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Q. 8. $\Delta ABC$ 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 $\Delta ABC.$ What will be the height from $C$ to $AB$ i.e., $CE$ ?

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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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Riya

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