, 2016

Solution 1

In $\triangle ABC$, $AD$ is a median and so $D$ is the midpoint of $BC$. Since $BC=12$ and $D$ is the midpoint of $BC$, then $CD=\frac{12}{2}=6$. In $\triangle ACD$, base $CD$ has length $6$, and corresponding height $AB$ has length $4$. (Since $\angle ABC=90^{\circ}$, $AB$ is the height of $\triangle ACD$ even though $AB$ is outside $\triangle ACD$.) Thus, $\triangle ACD$ has area $\frac12(6)(4)=12$. Solution 2 In $\triangle ABC$, $AD$ is a median and so $D$ is the midpoint of $BC$. Since $BC=12$ and $D$ is the midpoint of $BC$, then $CD=DB=6$. [[IMAGE0]] In $\triangle ABD$, $AB=4$, $DB=6$, and $\angle ABD=90^{\circ}$, and so $\triangle ABD$ has area $\frac12(6)(4)=12$. Similarly, $\triangle ABC$ has area $\frac12(12)(4)=24$, and so the area of $\triangle ACD$ is the area of $\triangle ABC$ minus the area of $\triangle ABD$, or $24-12=12$. Solution 3 In $\triangle ABC$, $AB=4$, $BC=12$, and $\angle ABC=90^{\circ}$, and so $\triangle ABC$ has area $\frac12(12)(4)=24$. A median of $\triangle ABC$ divides the triangle into two equal areas. Why? In $\triangle ABC$, $AD$ is a median and so $D$ is the midpoint of $BC$. Therefore, $\triangle ACD$ and $\triangle ABD$ have equal bases ($CD=BD$). Further, the height of $\triangle ABD$ is equal to the height of $\triangle ACD$ (both are $AB$). Thus, $\triangle ACD$ and $\triangle ABD$ have equal bases and equal heights. Since the area of each triangle equals one-half times the base times the height, then $\triangle ABD$ and $\triangle ACD$ have equal areas and so median $AD$ divides $\triangle ABC$ into equal areas. Since $\triangle ABC$ has area 24, then $\triangle ACD$ has area $\frac{24}{2}=12$. Solution 1 In $\triangle FSG$, $FS=18$, $SG=24$, and $\angle FSG=90^{\circ}$. Thus, by the Pythagorean Theorem, $FG=\sqrt{18^2+24^2}=\sqrt{324+576}=\sqrt{900}=30$ (since $FG>0$). Since $S$ is on $FH$ so that $FS=18$ and $SH=32$, then $FH=FS+SH=18+32=50$. In $\triangle FGH$, $FH=50$, $FG=30$, and $\angle FGH=90^{\circ}$. Thus, by the Pythagorean Theorem, $GH=\sqrt{50^2-30^2}=\sqrt{2500-900}=\sqrt{1600}=40$ (since $GH>0$). In $\triangle FGH$, $FT$ is a median and so $T$ is the midpoint of $GH$. In $\triangle FHT$, base $HT=\frac{40}{2}=20$, and height $FG=30$. (Since $\angle FGH=90^{\circ}$, $FG$ is the height of $\triangle FHT$ even though $FG$ is outside $\triangle FHT$.) [[IMAGE1]] Thus, $\triangle FHT$ has area $\frac12(20)(30)=300$. Solution 2 Since $S$ is on $FH$ so that $FS=18$ and $SH=32$, then $FH=FS+SH=18+32=50$. In $\triangle FGH$, base $FH=50$, and height $SG=24$ (since $SG$ is perpendicular to $FH$, $SG$ is a height of $\triangle FGH$). Thus, $\triangle FGH$ has area $\frac12(50)(24)=600$. The median of a triangle divides the area of the triangle in half. (Solution 3 to (a) shows an example of why a median divides a triangle’s area in half.) Since $FT$ is a median of $\triangle FGH$, then the area of $\triangle FHT=\frac{600}{2}=300$. We use the notation $|\triangle KLM|$ to represent the area of $\triangle KLM$, $|KPMQ|$ to represent the area of $KPMQ$, and so on. In $\triangle KLM$, $KP$ is a median and so $2|\triangle KPM|=|\triangle KLM|$. (Solution 3 to (a) shows an example of why a median divides a triangle’s area in half.) In $\triangle KMN$, $KQ$ is a median and so $2|\triangle KMQ|=|\triangle KMN|$. [[IMAGE2]] Therefore, $$|KLMN| = |\triangle KLM| + |\triangle KMN| = 2|\triangle KPM| + 2|\triangle KMQ|$$ and $$|KPMQ| = |\triangle KPM| + |\triangle KMQ|$$ which tells us that $|KLMN| = 2|KPMQ|$. Since $|KPMQ| = 63$, then $|KLMN| = 2|KPMQ| = 2(63) = 126$. Now $|KLMN| = |\triangle KRL| + |\triangle LRM| + |\triangle KRN| + |\triangle NRM|$. Each of these four triangles is right-angled. Since $KR = x$ and $LR = 6$, then $|\triangle KRL| = \frac{1}{2}x(6) = 3x$. Since $LR = 6$ and $RM = 2x+2$, then $|\triangle LRM| = \frac{1}{2}(6)(2x+2) = 6x+6$. Since $KR = x$ and $RN = 12$, then $|\triangle KRN| = \frac{1}{2}x(12) = 6x$. Since $RN = 12$ and $RM = 2x+2$, then $|\triangle NRM| = \frac{1}{2}(12)(2x+2) = 12x+12$. Therefore, $126 = 3x+(6x+6)+6x+(12x+12)$ or $126 = 27x + 18$ or $27x = 108$ and so $x = 4$.