Consider a square ABCD of side 12 and let M, | vedclass

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(B) Let the square $ABCD$ be placed in the coordinate plane with $A(0, 12), B(12, 12), C(12, 0), D(0, 0)$.$M$ is the midpoint of $AB$,so $M = (6, 12)$

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) Let the square $ABCD$ be placed in the coordinate plane with $A(0, 12), B(12, 12), C(12, 0), D(0, 0)$.$M$ is the midpoint of $AB$,so $M = (6, 12)$.$N$ is the midpoint of $CD$,so $N = (6, 0)$.$P$ lies on $MN$,so $P = (6, y)$ for some $y \in [0, 12]$.$AP = r = \sqrt{(6-0)^2 + (y-12)^2} = \sqrt{36 + (12-y)^2}$.$PC = s = \sqrt{(6-12)^2 + (y-0)^2} = \sqrt{36 + y^2}$.Consider the triangle with sides $r, s, 12$. Note that $BC = 12$. The point $P$ has a horizontal distance of $6$ from the side $BC$ (which lies on the line $x=12$).Thus,the area of $	riangle PBC$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes 	ext{distance of } P 	ext{ from } BC = \frac{1}{2} 	imes 12 	imes 6 = 36$.

Consider a square $ABCD$ of side $12$ and let $M, N$ be the midpoints of $AB, CD$ respectively. Take a point $P$ on $MN$ and let $AP=r, PC=s$. Then,the area of the triangle whose sides are $r, s, 12$ is

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