Question

Triangle abc is isosceles with ab=ac=7.5 cm and bc=9 cm. the height ad from a to bc is 6cm.find the area of triangle abc.what will be the height from c to ab ie., ce

Answer 1

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Answer 2

Answer:Area of Δ ABC = $27 cm^2$

             CE = 7.2 cm

Step-by-step explanation:

Given Δ ABC is an isosceles triangle.

AB= AC= 7.5 cm ; BC= 9 cm

Therefore $\text { area of triangle } = \dfrac{1}{2} \times base \times height$

So, $\text { area of } \triangle ABC } = \dfrac{1}{2} \times BC \times AD= \dfrac{1}{2} \times 9\times 6 = 27 cm^2$

Now

$\text { area of } \triangle ABC } = \dfrac{1}{2} \times AB \times CE\\\\\Rightarrow \dfrac{1}{2} \times 7.5\times CE = 27 cm^2\\\\\Rightarrow CE= \dfrac{27\times 2}{7.5} = 7.2 cm$

Hence, area of Δ ABC is $27 cm^2$ and CE is 7.2 cm