$(i)$ Coloured part $=\frac{1}{4}$$\therefore$ Percent of coloured part $=\frac{1}{4}\times100$ % $=25$ % $(ii)$ Coloured part $=\frac{3}{5}$$\therefore$ Percent of coloured part $=\frac{3}{5}\times100$ %$=60$ % $(iii)$ Coloured part $=\frac{3}{8}$$\therefore$ Percent of coloured part$=\frac{3}{8}\times100$ %$=\frac{3}{2}\times25$ % $=37.5$ %

No, because two triangles with equal corresponding angles need not be congruent. In such correspondence, one of them may be an enlarged copy of the other. For example: Here are two triangles $∆ABC$, with $\angle A=30^{\circ}, \ \angle B=40^{\circ}$ and $\angle C=110^{\circ}$And $∆PQR$, $\angle P=30^{\circ}, \ \angle Q=40^{\circ}$ and $\angle R=110^{\circ}$But $∆ABC$ ... Read More

Given: Given two triangles are congruent. And their corresponding parts are given.To do: To write $∆RAT≅?$Solution:In the given figure, corresponding triangles and congruent.$A\Leftrightarrow O$,$R\Leftrightarrow W$,$T\Leftrightarrow N$,Therefore we can write, $\Delta RAT\cong\Delta WON$    [SAS congruence]

In $\Delta BAT$  & $\Delta ABC$, given triangle are congruent;$B\Leftrightarrow B$$A\Leftrightarrow A$$T\Leftrightarrow C$                                           [By SSS congruence]In $\Delta QRS$ and $\Delta QPT$,$PQ=RS$     [given]$∠R=∠P$     [given]$QR=PT$     [given]Thus, $\Delta QRS ≅ \Delta TPQ$                         [By SAS rule]

To do: To draw two triangles of equal areas such that$(i)$. the triangles are congruent.$(ii)$. the triangles are not congruent.To compare the perimeters of two triangles if they are congruent and not congruent.Solution:$(i)$. If two triangles are congruent, then all the corresponding parts of the triangles will be equal.Let us ... Read More

Given: Two triangles $∆ABC$ and $∆FED$.To do: To explain that $∆ABC ≅ ∆FED$.Solution:Given $\angle A\cong\angle F$$BC=ED$$\angle B=\angle E$In $\triangle ABC$ and $\triangle FED$$\angle B=\angle E=90^{\circ}$$\angle A=\angle F$$BC=ED$Therefore, $\triangle ABC\cong\triangle FED$          [RHS congruence rule]Read More

Given: ∆ABC and ∆PQR are congruent in the given figure.To do: To name one additional pair of corresponding parts.Solution:In $\triangle ABC$ and $\triangle PQR$$\angle B=\angle Q=90^{\circ}$$\angle C=\angle R$      [Given]$\overline{BC}=\overline{QR}$     [Given]Therefore, $\triangle ABC\cong\triangle PQR$    [ASA congruence]Read More

StepsReasons$(i).\ PM = QM$$(i)$ Given in the question$(ii).\ \angle PMA = \angle QMA$$(ii)$  Given in the question$(iii).\ AM = AM$$(iii)$ Common side$(iv).\ ∆AMP ≅ ∆AMQ$$(iv)$ SAS congruence rule

$(a)$ Two triangles are said to be congruent by the SSS criterion if and only if, all the three sides of one triangle are equal to all the three sides of another triangle.Hence, if $∆ ART ≅ ∆ PEN$ by SSS criterion, then$(i).\ AR=PE$$(ii).\ RT=EN$ and$(iii).\ AT=PN$$(b)$ Two triangles are ... Read More

Given: $∆DEF ≅ ∆BCA$.To do: To write the part(s) of $∆BCA$ that correspond to$(i).\ \angle E$ $(ii).\ \overline{EF}$ $(iii).\ \angle F$ $(iv).\ \overline{DF}$Solution:Given that $∆ DEF ≅ ∆ BCA$Since we know that corresponding parts of congruent triangles are equal,   hence we can say that the corresponding sides and angles of both congruent triangles are ... Read More