Question

In $\triangle ABC,$ $AB=BC=29,$ and $AC=42.$ What is the area of $\triangle ABC?$

$100$

$420$

$500$

$609$

$701$

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Drop the altitude from $B$ to $AC$ , meeting $AC$ at $X$ . Since $AB=BC$ , point $X$ is the midpoint of $AC$ , so $AX=21$ .

In right triangle $ABX$ , the altitude is $\begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20. \end{aligned}$

The area of $\triangle ABC$ is $\dfrac12\cdot42\cdot20=420$ .

Thus, B is the correct answer.