$$ \\\\ $$ Apply Pythagoras Theorem, $$ \\\\ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} \\\\ \\Rightarrow AC=\\sqrt{{{4}^{2}}+{{8}^{2}}}=\\sqrt{80} \\\\ \\sqrt{5}$$ times the AC becomes, $$ \\\\ \\sqrt{5}AC=\\sqrt{5}\\times \\sqrt{80}=20cm$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Pythagorean theorem proof Quiz 1","text":" What is the area of the rectangle given below. $$ \\\\ $$","comment":{"@type":"Comment","text":" Apply Pythagoras theorem to find the value of base, then apply the formula, area of rectangle $=2\\times \\text{area of }\\Delta \\text{ABC}$"},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2\\sqrt{23}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$6\\sqrt{29}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":"$$4\\sqrt{29}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$8\\sqrt{23}$$","position":1,"answerExplanation":{"@type":"Comment","text":" Apply Pythagoras Theorem on $$\\Delta ABC$$, $$ \\\\ A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}} \\\\ \\Rightarrow {{(6\\sqrt{3})}^{2}}={{4}^{2}}+A{{C}^{2}} \\\\ \\Rightarrow AC=\\sqrt{108-16}=\\sqrt{92} \\\\ $$ Area of rectangle, $$ =2\\times \\text{area of }\\Delta \\text{ABC} \\\\ =2\\times \\dfrac{1}{2}\\times AC\\times BC \\\\ =\\sqrt{92}\\times 4 \\\\ =8\\sqrt{23} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Pythagorean theorem proof Quiz 1","text":" Find $$\\dfrac{BC}{DF} \\\\ $$
","comment":{"@type":"Comment","text":" Find the values of BC and DF using the Pythagoras Theorem and find their ratios."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{11}{13}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{21}{11}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{19}{37}}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\sqrt{\\dfrac{13}{41}}$$","position":3,"answerExplanation":{"@type":"Comment","text":" Apply Pythagoras Theorem on $$\\Delta ABC$$, $$ \\\\ B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}} \\\\ \\Rightarrow BC=\\sqrt{{{6}^{2}}+{{4}^{2}}} \\\\ \\Rightarrow BC=\\sqrt{52} \\\\ $$ Apply Pythagoras Theorem on $$\\Delta DEF$$, $$ \\\\ D{{F}^{2}}=D{{E}^{2}}+E{{F}^{2}} \\\\ \\Rightarrow DF=\\sqrt{{{8}^{2}}+{{10}^{2}}} \\\\ \\Rightarrow DF=\\sqrt{164} \\\\ \\therefore \\dfrac{BC}{DF}=\\dfrac{\\sqrt{52}}{\\sqrt{164}}=\\sqrt{\\dfrac{13}{41}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Pythagorean theorem proof Quiz 1","text":" A man is climbing up a ladder of $$13\\text{ m}$$ which is $$\\text{5 m}$$ away from the base of post A. Another man is climbing up a $$\\text{10 m}$$ long ladder which is $$\\text{6 m}$$ away from the base of a post B. Which is longer and by how much?","comment":{"@type":"Comment","text":" Draw corresponding diagrams and then apply Pythagoras theorem."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{post A,6 m}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{post B,8 m}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\text{post B,6 m}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{post A,4 m}$$","position":0,"answerExplanation":{"@type":"Comment","text":" Let PQ and ST be 2 ladders away from the post at A and B respectively. As per the given criteria, we get $$ \\\\ $$
$$ \\\\ $$ Apply Pythagoras Theorem on $$\\Delta PQA$$, $$ \\\\ P{{A}^{2}}=P{{Q}^{2}}-Q{{A}^{2}} \\\\ \\Rightarrow PA=\\sqrt{{{13}^{2}}-{{5}^{2}}} \\\\ \\Rightarrow PA=12m \\\\ $$ Apply Pythagoras Theorem on $$\\Delta STB$$, $$ \\\\ S{{B}^{2}}=S{{T}^{2}}-T{{B}^{2}} \\\\ \\Rightarrow SB=\\sqrt{{{10}^{2}}-{{6}^{2}}} \\\\ \\Rightarrow SB=8m \\\\ $$ Therefore, post A is longer by $$(12-8)=4m$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Pythagorean theorem proof Quiz 1","text":" Find BC $$ \\\\ $$
","comment":{"@type":"Comment","text":" Apply Pythagoras theorem to find the value of BD. Then again Pythagoras theorem to find the value of BC."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{46}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{53}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\sqrt{91}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{85}$$","position":2,"answerExplanation":{"@type":"Comment","text":" Apply Pythagoras Theorem on $$\\Delta ADC$$, $$ \\\\ A{{D}^{2}}=A{{C}^{2}}-D{{C}^{2}} \\\\ \\Rightarrow AD=\\sqrt{{{20}^{2}}-{{8}^{2}}} \\\\ \\Rightarrow AD=4\\sqrt{21} \\\\ $$ From given figure, $$ \\\\ AB=AD+DB \\\\ \\Rightarrow DB=5\\sqrt{21}-4\\sqrt{21}=\\sqrt{21} \\\\ $$ Now apply Pythagoras Theorem on ∆BDC, $$ \\\\ B{{C}^{2}}=D{{B}^{2}}+D{{C}^{2}} \\\\ \\Rightarrow BC=\\sqrt{{{\\left( \\sqrt{21} \\right)}^{2}}+{{8}^{2}}} \\\\ \\Rightarrow BC=\\sqrt{85}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}