","comment":{"@type":"Comment","text":" Use the formula for a centroid to calculate the point of concurrency. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( \\dfrac{14}{3},\\dfrac{14}{3} \\right)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( \\dfrac{10}{3},\\dfrac{10}{3} \\right)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( \\dfrac{10}{3},\\dfrac{14}{3} \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( \\dfrac{14}{3},\\dfrac{10}{3} \\right)$$","position":1,"answerExplanation":{"@type":"Comment","text":"$$C=\\left( \\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \\right) \\\\ \\Rightarrow C=\\left( \\dfrac{0+4+10}{3},\\dfrac{0+10+0}{3} \\right) \\\\ $$ Therefore, $$C=\\left( \\dfrac{14}{3},\\dfrac{10}{3} \\right)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Points of concurrency Quiz 1","text":" An equilateral triangle with length $$15$$ units is being constructed. A student wishes to find the point of concurrency. Given that orthocenter is the point of concurrency by dropping altitudes from each side, determine the orthocenter of the triangle. $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Use the formula, $O=\\left( \\dfrac{{{x}_{1}}\\tan A+{{x}_{2}}\\tan B+{{x}_{3}}\\tan C}{\\tan A+\\tan B+\\tan C},\\dfrac{{{y}_{1}}\\tan A+{{y}_{2}}\\tan B+{{y}_{3}}\\tan C}{\\tan A+\\tan B+\\tan C} \\right)$ for calculating the orthocentre of the triangle. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$O=\\left( \\dfrac{13}{3},4.5 \\right)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$O=\\left( \\dfrac{13}{3},7.5 \\right)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$O=\\left( 4.5,\\dfrac{13}{3} \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$O=\\left( 7.5,\\dfrac{13}{3} \\right)$$","position":1,"answerExplanation":{"@type":"Comment","text":"Altitude of an equilateral triangle: $$\\dfrac{s\\sqrt{3}}{2}$$ $$ \\\\ $$ Therefore, altitude of the equilateral triangle dropped from any side is $$H=\\dfrac{15\\sqrt{3}}{2}=12.99$$ $$ \\\\ $$ The altitude bisects the side of the equilateral triangle, therefore, $$ \\\\ $$ Vertex of point $$C=(7.5,12.99)$$ $$ \\\\ $$ Using the formula for the orthocenter, $$\\\\ O=\\left( \\dfrac{{{x}_{1}}\\tan A+{{x}_{2}}\\tan B+{{x}_{3}}\\tan C}{\\tan A+\\tan B+\\tan C},\\dfrac{{{y}_{1}}\\tan A+{{y}_{2}}\\tan B+{{y}_{3}}\\tan C}{\\tan A+\\tan B+\\tan C} \\right) \\\\ \\Rightarrow O=\\left( \\dfrac{0\\tan 60+7.5\\tan 60+15\\tan 60}{\\tan 60+\\tan 60+\\tan 60},\\dfrac{0\\tan 60+12.99\\tan 60+0\\tan 60}{\\tan 60+\\tan 60+\\tan 60} \\right) \\\\ \\Rightarrow O=\\left( \\dfrac{7.5\\tan 60+15\\tan 60}{3\\tan 60},\\dfrac{12.99\\tan 60}{3\\tan 60} \\right) \\\\ \\Rightarrow O=\\left( \\dfrac{7.5\\sqrt{3}+15\\sqrt{3}}{3\\sqrt{3}},\\dfrac{12.99\\sqrt{3}}{3\\sqrt{3}} \\right)=\\left( 7.5,\\dfrac{13}{3} \\right) $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Points of concurrency Quiz 1","text":" A triangle with the points $$\\left( 0,0 \\right),\\left( 3,4 \\right)$$ and $$\\left( 6,0 \\right)$$ is constructed on a cartesian plane. Calculate the incenter of the given triangle.$$ \\\\ $$
","comment":{"@type":"Comment","text":" Use the formula, $IC=\\left( \\dfrac{a{{x}_{a}}+b{{x}_{b}}+c{{x}_{c}}}{a+b+c},\\dfrac{a{{y}_{a}}+b{{y}_{b}}+c{{y}_{c}}}{a+b+c} \\right)$ for calculating the incentre of the triangle."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$IC=\\left( 1.5,1.5 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$IC=\\left( 3,3 \\right)$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$IC=\\left( 1.5,3 \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$IC=\\left( 3,1.5 \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":"$$\\overline{AD}=\\sqrt{{{\\left( 4-0 \\right)}^{2}}+{{\\left( 3-0 \\right)}^{2}}}=5 \\\\ \\overline{AB}=\\sqrt{{{\\left( 6-0 \\right)}^{2}}+{{\\left( 0-0 \\right)}^{2}}}=6 \\\\ \\overline{BD}=\\sqrt{{{\\left( 4-0 \\right)}^{2}}+{{\\left( 3-6 \\right)}^{2}}}=5 \\\\ IC=\\left( \\dfrac{5{{x}_{a}}+6{{x}_{b}}+5{{x}_{c}}}{5+6+5},\\dfrac{5{{y}_{a}}+6{{y}_{b}}+5{{y}_{c}}}{5+6+5} \\right) \\\\ \\Rightarrow IC=\\left( \\dfrac{5\\cdot 0+6\\cdot 3+5\\cdot 6}{16},\\dfrac{5\\cdot 0+6\\cdot 4+5\\cdot 0}{16} \\right) \\\\ \\Rightarrow IC=\\left( 3,1.5 \\right)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Points of concurrency Quiz 1","text":" A triangle with points $$\\left( 0,0 \\right),\\left( 10,0 \\right)$$ and $$\\left( 5,5 \\right)$$ with an angle of $$90^\\circ$$ has a circumcenter lying on the triangle. Given that the triangle is constructed on a cartesian plane, calculate the point of the circumcenter.$$ \\\\ $$
","comment":{"@type":"Comment","text":" Use the formula of a circumcenter to calculate the point of concurrency. You can use properties such as $\\sin (90)=1$ and $\\sin (180)=0$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 15,10 \\right)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 10,15 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 0,5 \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 5,0 \\right)$$","position":2,"answerExplanation":{"@type":"Comment","text":"$$C=\\left( \\dfrac{{{x}_{1}}\\sin 2A+{{x}_{2}}\\sin 2B+{{x}_{3}}\\sin 2C}{\\sin 2A+\\sin 2B+\\sin 2C},\\dfrac{{{y}_{1}}\\sin 2A+{{y}_{2}}\\sin 2B+{{y}_{3}}\\sin 2C}{\\sin 2A+\\sin 2B+\\sin 2C} \\right) \\\\ \\Rightarrow C=\\left( \\dfrac{0\\sin \\left( 2\\times 45 \\right)+10\\sin \\left( 2\\times 45 \\right)+5\\cdot \\sin \\left( 2\\times 90 \\right)}{\\sin \\left( 2\\times 45 \\right)+\\sin \\left( 2\\times 45 \\right)+\\sin \\left( 2\\times 90 \\right)},\\dfrac{0\\sin \\left( 2\\cdot 45 \\right)+0\\sin \\left( 2\\times 45 \\right)+5\\sin \\left( 2\\times 90 \\right)}{\\sin \\left( 2\\times 45 \\right)+\\sin \\left( 2\\times 45 \\right)+\\sin \\left( 2\\times 90 \\right)} \\right) \\\\ \\Rightarrow C=\\left( \\dfrac{0+10+0}{1+1+0},\\dfrac{0+0+0}{1+1+0} \\right) \\\\ \\Rightarrow C=\\left( 5,0 \\right)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Points of concurrency Quiz 1","text":" A triangle has an angle of $$90^\\circ$$ in the $$1\\text{st}$$ quadrant at point $$(0,0)$$. Given that the triangle follows the $$1:\\sqrt{2}:1$$ special triangle rule, the length of its shortest side is $$6$$ units, calculate the circumcenter of the triangle. ","comment":{"@type":"Comment","text":" Use the formula of a circumcenter to calculate the point of concurrency. You can use properties such as $\\sin (90)=1$ and $\\sin (180)=0$ to help with ease of calculation. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$(2,2)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$(4,2)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$(4,4)$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$(2,4)$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$90^\\circ$$ at point $$(0,0)$$ in the $$1\\text{st}$$ quadrant $$ \\\\ $$ Length of shortest side $$= 6$$ units, therefore, $$2\\text{nd}$$ point is $$(6,0)$$ $$ \\\\ $$ The triangle follows the $$1:\\sqrt{2}:1$$ special triangle rule, therefore, length of 2nd longest side is $$6$$ units $$ \\\\ $$ This implies that the $$3\\text{rd}$$ point is on $$(0,6)$$ $$ \\\\ C=\\left( \\dfrac{{{x}_{1}}\\sin 2A+{{x}_{2}}\\sin 2B+{{x}_{3}}\\sin 2C}{\\sin 2A+\\sin 2B+\\sin 2C},\\dfrac{{{y}_{1}}\\sin 2A+{{y}_{2}}\\sin 2B+{{y}_{3}}\\sin 2C}{\\sin 2A+\\sin 2B+\\sin 2C} \\right) \\\\ \\Rightarrow C=\\left( \\dfrac{0\\cdot \\sin \\left( 2\\cdot 90 \\right)+6\\cdot \\sin \\left( 2\\cdot 45 \\right)+0\\cdot \\sin \\left( 2\\cdot 45 \\right)}{\\sin \\left( 2\\cdot 90 \\right)+\\sin \\left( 2\\cdot 45 \\right)+\\sin \\left( 2\\cdot 45 \\right)},\\dfrac{0\\cdot \\sin \\left( 2\\cdot 90 \\right)+0\\cdot \\sin \\left( 2\\cdot 45 \\right)+6\\cdot \\sin \\left( 2\\cdot 45 \\right)}{\\sin \\left( 2\\cdot 90 \\right)+\\sin \\left( 2\\cdot 45 \\right)+\\sin \\left( 2\\cdot 45 \\right)} \\right) \\\\ \\Rightarrow C=\\left( \\dfrac{0+6+0}{0+1+1},\\dfrac{0+0+6}{0+1+1} \\right) \\\\ \\Rightarrow C=\\left( 3,3 \\right) \\\\ $$
","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}