جواب تمرین صفحه 148 درس 9 ریاضی هشتم (دایره)

الف

\(\begin{array}{l}\widehat e = \mathop {AB}\limits^\frown \, = 2 \times A\widehat C\,B = 2 \times {57^ \circ } = {114^ \circ }\\\widehat d = \frac{{\mathop {AB}\limits^\frown }}{2} = \frac{{{{114}^ \circ }}}{2} = {57^ \circ }\end{array}\)

ب

\(\begin{array}{l}\mathop j\limits^\frown \, = 2 \times {42^ \circ } = {84^ \circ }\\\mathop k\limits^\frown \, = 2 \times {33^ \circ } = {66^ \circ }\\\mathop h\limits^\frown \, = \frac{{\mathop j\limits^\frown \,}}{2} = \frac{{{{84}^ \circ }}}{2} = {42^ \circ }\\\mathop i\limits^\frown \, = \frac{{\mathop k\limits^\frown \,}}{2} = \frac{{{{66}^ \circ }}}{2} = {33^ \circ }\end{array}\)

الف برابر \({72^ \circ }\) است.

ب 5 کمان مساوی

الف شکل دارای 8 خط تقارن است.

ب هرکمان برابر: \(\frac{{{{360}^ \circ }}}{8} = {45^ \circ }\)

\(y = z = \frac{{2 \times andaze\,kaman}}{2} = \frac{{2 \times {{45}^ \circ }}}{2} = {45^ \circ }\)

\(x = {180^ \circ } - (y + z) = {180^ \circ } - {90^ \circ } = {90^ \circ }\)

زوایای B و D، زوایای رو به روی هم هستند؛ بنابراین:

\(\begin{array}{l}\widehat B + \widehat D = \frac{{\mathop {ABC}\limits^\frown \,}}{2} + \frac{{\mathop {CDA}\limits^\frown }}{2} = \frac{{\mathop {ABC}\limits^\frown + \mathop {CDA}\limits^\frown }}{2}\\\begin{array}{*{20}{c}}{}&{}\end{array}\;\, = \frac{{\mathop {ABCDA}\limits^\frown }}{2} = \frac{{{{360}^ \circ }}}{2} = {180^ \circ }\end{array}\)

الف

\(\widehat {{O_1}} = x\)

ب 

\(\mathop {ADB}\limits^\frown = {360^ \circ } - \mathop {ACB}\limits^\frown = {360^ \circ } - x\)

ج

\(\widehat C = \frac{{\mathop {ADB}\limits^\frown }}{2} = \frac{{{{360}^ \circ } - x}}{2} = {180^ \circ } - \frac{x}{2}\)

الف

\(\overline {TX} = \overline {TR} + \overline {RX} = 60\;cm\)

ب متساوی الاضلاع؛ زیرا :

\(\left\{ {\begin{array}{*{20}{c}}{\overline {RZ} = \overline {RT} \begin{array}{*{20}{c}}{}&{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\\{\widehat R = 2 \times {{30}^ \circ } = {{60}^ \circ }}\end{array}} \right. \Rightarrow \widehat T = \widehat Z = {60^ \circ }\)

ج

\(\overline {ZT} = \overline {RT} = 30\;cm\)

هـ

\(\begin{array}{l}Z{X^2} = T{X^2} - Z{T^2}\\\quad \;\;{\kern 1pt} {\kern 1pt} = {60^2} - {30^2} = 2700\\ \Rightarrow ZX = \sqrt {2700} = 30\sqrt 3 \end{array}\)