Given: $PQ = 5\ cm$, $m\angle\ PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$.To do:  To construct $\triangle PQR$.Solution: Here given, $PQ = 5\ cm$, $m\angle PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$Let us find the value of $\angle QPR=?$$\angle PQR + \angle QRP + \angle QPR = 180^{\circ}$   [angle ... Read More

Given: $m\angle Q = 90^{\circ}, \ QR = 8cm$ and $PR = 10\ cm$.To do: To construct the right-angled $\triangle PQR$Steps of construction:Draw a line segment $QR$ of length $8\ cm$. At the point $Q$, let us draw a perpendicular $QX$ such that $QX\perp QR$. Assuming $R$ as a center, ... Read More

Given: A right-angled triangle whose hypotenuse is $6\ cm$ long and one of the legs is $4\ cm$ long.To do: To construct a right-angled triangle whose hypotenuse is $6\ cm$ long and one of the legs is $4\ cm$ long.Steps of construction:Let us draw a line segment $EF$ of length $4\ ... Read More

Given: Pairs of rational numbers:$(i)$. $-\frac{7}{21}$ and $\frac{3}{9}$$(ii)$. $-\frac{16}{20}$ and $\frac{20}{-25}$$(iii)$. $\frac{-2}{-3}$ and  $\frac{2}{3}$$(iv)$. $\frac{-3}{5}$ and $\frac{-12}{20}$$(v)$. $\frac{8}{-5}$ and $\frac{-24}{15}$$(vi)$. $\frac{1}{3}$ and $\frac{-1}{9}$$(viii)$ $\frac{-5}{-9}$ and $\frac{5}{-9}$To do: To find pairs that represent the same rational number.Solution: $(i)$. $-\frac{7}{21}$ and $\frac{3}{9}$Given pairs are: $(i)$. $-\frac{7}{21}$ and $\frac{3}{9}$On reducing the given fractions to the ... Read More

Given: Rational numbers-$(i)$. $\frac{-8}{6}$$(ii)$. $\frac{25}{45}$ $(iii)$. $\frac{-44}{72}$$(iv)$. $\frac{-8}{10}$To do:We have to write the given rational numbers in the simplest form.Solution:$(i)$. $\frac{-8}{6}$$=\frac{2\times(-4)}{2\times3}$$=\frac{-4}{3}$ $(ii)$. $\frac{25}{45}$$=\frac{5\times5}{5\times9}$$=\frac{5}{9}$$(iii)$. $\frac{-44}{72}$$=\frac{4\times(-11)}{4\times18}$$=\frac{-11}{18}$ $(iv)$. $\frac{-8}{10}$$=\frac{2\times(-4)}{2\times5}$$=\frac{-4}{5}$Read More

Given: $(i)$. $-\frac{5}{7}\square\ \frac{2}{3}$$(ii)$. $-\frac{4}{5}\square\ -\frac{5}{7}$$(iii)$. $-\frac{7}{8}\square\ \frac{14}{-16}$$(iv)$. $-\frac{8}{5}\square\ -\frac{7}{4}$$(v)$. $\frac{1}{-3}\square\ \frac{-1}{4}$$(vi)$. $\frac{5}{-11}\ \square\ \frac{-5}{11}$$(vii)$. $0\ \square\ \frac{-7}{6}$To do: To fill in the boxes with the correct symbol out of >, $-\frac{7}{4}$$(v)$. $\frac{1}{-3}\square -\frac{1}{4}$Here LCM of $3$ and $4$ is $12$.So, $\frac{1}{-3}=\frac{(1\times 4)}{(-3\times 4)}$$=-\frac{4}{12}$And $-\frac{1}{4}=\frac{(-1\times 3)}{(4\times 3)}$$=-\frac{3}{12}$$-\frac{4}{12}$    $\frac{-7}{6}$Read More

Given:$(i)$. $\frac{2}{3}, \ \frac{5}{2}$$(ii)$. $-\frac{5}{6}, \ -\frac{4}{3}$$(iii)$. $-\frac{3}{4}, \ \frac{2}{-3}$$(iv)$. $-\frac{1}{4}, \ \frac{1}{4}$$(v)$. $-3\frac{2}{7\ }, \ -3\frac{4}{5}$To do: To find the greater rational number in each of the given pairs. Solution: $(i)$. $\frac{2}{3}, \ \frac{5}{2}$Taking the LCM of the denominators $3$ and $2$ of both the rational numbers, we get $6$.So, $\frac{2}{3}=\frac{2}{3}\times\frac{2}{2}$$=\frac{4}{6}$And ... Read More

Given: $(i)$. $\frac{-3}{5}, \ \frac{-2}{5}, \ \frac{-1}{5}$$(ii)$. $\frac{-1}{3}, \ \frac{-2}{9}, \ \frac{-4}{3}$$(iii)$. $\frac{-3}{7}, \ \frac{-3}{2}, \ \frac{-3}{4}$To do: To write the given rational numbers in ascending order.Solution:$(i)$. $-\frac{3}{5}, \ -\frac{2}{5}, \ -\frac{1}{5}$Here, denominators are the same for each rational number. We just need to compare the numerators of the given rational ... Read More

Given: $(i)$. $(-4)\div\frac{2}{3}$$(ii)$. $-\frac{3}{5}\div2$$(iii)$. $-\frac{4}{5}\div(-3)$$(iv)$. $-\frac{1}{8}\div\frac{3}{4}$$(v)$. $-\frac{2}{13}\div\frac{1}{7}$$(vi)$. $-\frac{7}{12}\div(-\frac{2}{13})$$(vii)$. $\frac{3}{13}\div(-\frac{4}{65})$To do: To find the value of the given expressions.Solution:$(i)$. $(-4)\div\frac{2}{3}$$=-4\times\frac{3}{2} = -6$ $(ii)$. $-\frac{3}{5}\div2$$=-\frac{3}{5}\times\frac{1}{2}$$=-\frac{3}{10}$$​(iii)$. $-\frac{4}{5}\div(-3)$$=-\frac{4}{5}\times \frac{1}{-3}$$=\frac{4}{15}$$(iv) $. $-\frac{1}{8}\div\frac{3}{4}$$=-\frac{1}{8}\times\frac{4}{3}$$=-\frac{1}{6}$$(v)$. $-\frac{2}{13}\div\frac{1}{7}$$=-\frac{2}{13}\times\frac{7}{1}$$=-\frac{14}{13}$$=-\frac{14}{13}$$(vi)$. $-\frac{7}{12}\div(-\frac{2}{13})$$=-\frac{7}{12}\times \frac{13}{-2}$$=\frac{91}{24}$$=3\frac{19}{24}$$(vii) $. $\frac{3}{13}\div(-\frac{4}{65})$$=\frac{3}{13}\times(-\frac{65}{4})$$=\frac{15}{-4}$$=-3\frac{3}{4}$Read More

Given: $(i)$. $\frac{9}{2}\times(-\frac{7}{4})$$(ii)$. $\frac{3}{10}\times(-9)$$(iii)$. $-\frac{6}{5}\times\frac{9}{11}$$(iv)$. $\frac{3}{7}\times(-\frac{2}{5})$ $(v)$. $\frac{3}{11}\times\ \frac{2}{5}$$(vi)$. $\frac{3}{-5}\times(-\frac{5}{3})$To do: To find the product of the given expression.Solution:$(i)$. $\frac{9}{2}\times(-\frac{7}{4})$$=\frac{9\times(-7)}{2\times4}$$=-\frac{63}{8}$$=-7\frac{7}{8}$​$(ii)$. $\frac{3}{10}\times(-9)$$=-\frac{27}{10}$$=-2\frac{7}{10}$$(iii)$. $-\frac{6}{5}\times\frac{9}{11}$$=\frac{(-6)\times9}{5\times11}$$=-\frac{54}{55}$​$(iv)$. $\frac{3}{7}\times(-\frac{2}{5})$$=\frac{3\times(-2)}{7\times5}$ $=-\frac{6}{35}$$(v)$. $\frac{3}{11}\times\ \frac{2}{5}$ $=\frac{3\times2}{11\times5}$  $=\frac{6}{55}$​$(vi)$. $\frac{3}{-5}\times(-\frac{5}{3})$$=\frac{3\times(-5)}{(-5)\times3}$$=\frac{-15}{-15}$$=1$Read More