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GeometryQuestion and Answers: Page 82

Question Number 68305    Answers: 1   Comments: 3

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Question Number 68289    Answers: 1   Comments: 0

A circle is divided into two equal parts By An arc with center on the circle. Determine (a) The length of the arc (b)The ratio in which the arc divides the diameter meeting the center of the arc.

$$\mathrm{A}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{parts} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{By} \\ $$$$\:\mathrm{An}\:\mathrm{arc}\:\mathrm{with}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathcal{D}{etermine} \\ $$$$\:\:\left({a}\right)\:{The}\:{length}\:{of}\:{the}\:{arc} \\ $$$$\:\:\left({b}\right){The}\:{ratio}\:{in}\:{which}\:{the}\:{arc} \\ $$$$\:\:\:\:\:\:\:\:{divides}\:{the}\:{diameter}\: \\ $$$$\:\:\:\:\:\:\:\:{meeting}\:{the}\:{center}\:{of}\:{the}\:{arc}. \\ $$

Question Number 68183    Answers: 0   Comments: 1

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Question Number 68132    Answers: 0   Comments: 3

Question Number 68110    Answers: 0   Comments: 2

Question Number 68092    Answers: 0   Comments: 4

Question Number 68041    Answers: 0   Comments: 1

Question Number 67977    Answers: 1   Comments: 1

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Question Number 67969    Answers: 0   Comments: 0

Two triangles △_1 and △_2 are given,such that length of sides of triangle 1,are equail to length of medians of triangle 2. 1.find the ratio of areas of triangles. 2.given that small side of △_1 , be equail to:(√2) and one angle be:90^• . find at least one angle of △_2 . 3.solve part#2,if replace: △_2 with: △_1 . 4.solve part#2,if great side of:△_1 ,be equail to :(√2).

$$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{triangles}}\:\bigtriangleup_{\mathrm{1}} \:\boldsymbol{\mathrm{and}}\:\bigtriangleup_{\mathrm{2}} \:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}},\boldsymbol{\mathrm{such}}\: \\ $$$$\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{1},\boldsymbol{\mathrm{are}}\: \\ $$$$\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{medians}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{2}. \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{areas}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{triangles}}. \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{of}}\:\bigtriangleup_{\mathrm{1}} ,\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{be}}:\mathrm{90}^{\bullet} . \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{least}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{of}}\:\bigtriangleup_{\mathrm{2}} . \\ $$$$\mathrm{3}.\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{part}}#\mathrm{2},\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{replace}}:\:\bigtriangleup_{\mathrm{2}} \boldsymbol{\mathrm{with}}:\:\bigtriangleup_{\mathrm{1}} . \\ $$$$\mathrm{4}.\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{part}}#\mathrm{2},\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{great}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{of}}:\bigtriangleup_{\mathrm{1}} ,\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{equail}}\: \\ $$$$\boldsymbol{\mathrm{to}}\::\sqrt{\mathrm{2}}. \\ $$

Question Number 67948    Answers: 0   Comments: 0

Question Number 67860    Answers: 1   Comments: 3

Question Number 67852    Answers: 1   Comments: 4

Question Number 67849    Answers: 1   Comments: 7

Question Number 67807    Answers: 1   Comments: 2

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