Question

what is the area of δabc? add text

Explanation:

Step1: Find the length of the base of $\triangle ABC$

Since $\triangle ABC\cong\triangle DEF$ (by the markings of equal sides and angles), and in $\triangle DEF$, using the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $c = 10$ (hypotenuse) and $a = 6$ (one - leg), we find the other leg. Let the other leg be $b$. Then $b=\sqrt{10^{2}-6^{2}}=\sqrt{100 - 36}=\sqrt{64}=8$.

Step2: Calculate the area of $\triangle ABC$

The area of a right - triangle is given by $A=\frac{1}{2}\times base\times height$. For $\triangle ABC$, base $= 6$ and height $= 8$. So $A=\frac{1}{2}\times6\times8$.

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Answer:

The area of $\triangle ABC$ is $24\space mm^{2}$.