Solve this:

1. Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm. 
2. Find the area ofa triangle whose sides are 13 cm. 14 cm and 15 cm.
3. In a $△$ABC, AB = 30 cm,  BC=26 and AC = 28 cm. Find the area of $△$ABC and its altitude on AC. 
 

 

Dear Student,

Please find the below required solution of Q-1.

Let a = 18 cm, b = 24 cm  and c =30 cm
By Hero's formula area of a triangle =$\sqrt{s(s-a)(s-b)(s-c)} where s =\frac{1}{2}(a+b+c)$
Now for the given triangle , $s=\frac{1}{2}(18 cm+24 cm+30 cm) =\frac{1}{2}\times 72=36 cm$

$$\begin{gathered} \therefore Area of the given triangle =\sqrt{36(36 -18)(36-24)(36-30)} \\ =\sqrt{36\times 18 \times 12\times 6} \\ =\sqrt{46656 }c{m}^2=216 c{m}^2 \end{gathered}$$

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