Criteria for Similarity of Triangles Questions in Gujarati

(D) સંગતતા $ABC \leftrightarrow EFD$ સાથે $\Delta ABC \sim \Delta DEF$ આપેલ હોવાથી,અનુરૂપ બાજુઓનો ગુણોત્તર સમાન થાય.
તેથી,$\frac{AB}{EF} = \frac{BC}{FD} = \frac{CA}{DE} = k$.
$AB : BC : CA = 4 : 3 : 5$ આપેલ છે,તેથી ધારો કે $AB = 4x, BC = 3x, CA = 5x$.
સંગતતા $ABC \leftrightarrow EFD$ પરથી,$EF = BC = 3x$,$FD = CA = 5x$,અને $DE = AB = 4x$ મળે.
$\Delta DEF$ ની પરિમિતિ $= EF + FD + DE = 3x + 5x + 4x = 12x$.
પરિમિતિ $36$ આપેલ હોવાથી,$12x = 36$,જેનો અર્થ છે કે $x = 3$.
તેથી,બાજુઓ $EF = 3(3) = 9$,$FD = 5(3) = 15$,અને $DE = 4(3) = 12$ થાય.
આમ,$\Delta DEF$ ની બાજુઓ $12, 9, 15$ છે.

(C) આપેલ છે કે $\Delta ABC \sim \Delta PQR$,તેથી તેમની અનુરૂપ બાજુઓનો ગુણોત્તર તેમની પરિમિતિના ગુણોત્તર જેટલો હોય છે.
ધારો કે $\Delta ABC$ ની પરિમિતિ $P_1 = 48$ અને $\Delta PQR$ ની પરિમિતિ $P_2 = 60$ છે.
અનુરૂપ બાજુઓનો ગુણોત્તર $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{P_1}{P_2} = \frac{48}{60} = \frac{4}{5}$ થાય.
આમ,$\frac{AB}{PQ} = \frac{4}{5}$.
બીજા ગુણોત્તર માટે,સમાન ગુણોત્તરના ગુણધર્મનો ઉપયોગ કરતા: જો $\frac{a}{b} = \frac{c}{d} = k$ હોય,તો $\frac{a+c}{b+d} = k$ થાય.
તેથી,$\frac{AB + BC}{PQ + QR} = \frac{AB}{PQ} = \frac{4}{5}$.
બંને ગુણોત્તર $\frac{4}{5}$ છે.

(N/A) બે ત્રિકોણો સમરૂપ ત્યારે કહેવાય જો:
$1$. તેમના અનુરૂપ ખૂણાઓ સમાન હોય.
$2$. તેમની અનુરૂપ બાજુઓ સમાન ગુણોત્તરમાં (પ્રમાણમાં) હોય.
ધારો કે બે સમબાજુ ત્રિકોણો $\triangle ABC$ અને $\triangle PQR$ છે.
સમબાજુ ત્રિકોણમાં,બધા ખૂણાઓ $60^{\circ}$ હોય છે અને બધી બાજુઓ સમાન હોય છે.
તેથી,$\triangle ABC$ માટે: $\angle A = \angle B = \angle C = 60^{\circ}$ અને $AB = BC = CA = a$.
$\triangle PQR$ માટે: $\angle P = \angle Q = \angle R = 60^{\circ}$ અને $PQ = QR = RP = b$.
પગલું $1$: ખૂણાઓની સરખામણી કરતા:
$\angle A = \angle P = 60^{\circ}$,$\angle B = \angle Q = 60^{\circ}$,અને $\angle C = \angle R = 60^{\circ}$.
આમ,અનુરૂપ ખૂણાઓ સમાન છે.
પગલું $2$: બાજુઓની સરખામણી કરતા:
$\frac{AB}{PQ} = \frac{a}{b}$,$\frac{BC}{QR} = \frac{a}{b}$,અને $\frac{CA}{RP} = \frac{a}{b}$.
આમ,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{a}{b}$.
જેથી,અનુરૂપ ખૂણાઓ સમાન હોવાથી અને અનુરૂપ બાજુઓ સમાન ગુણોત્તરમાં હોવાથી,સમરૂપતાની વ્યાખ્યા મુજબ,$\triangle ABC \sim \triangle PQR$.
આમ,તમામ સમબાજુ ત્રિકોણો સમરૂપ છે.

Given: $ABCD$ is a parallelogram. $P$ is on $BC$,and $DP$ produced meets $AB$ produced at $L$.
$(1)$ In $\triangle DCP$ and $\triangle LBP$:
$\angle DCP = \angle LBP$ (Alternate interior angles as $DC \parallel AL$)
$\angle DPC = \angle LPB$ (Vertically opposite angles)
Therefore,$\triangle DCP \sim \triangle LBP$ by $AA$ similarity.
Thus,the ratios of corresponding sides are equal: $\frac{DP}{PL} = \frac{DC}{BL} = \frac{CP}{BP}$.
Hence,$\frac{DP}{PL} = \frac{DC}{BL}$ is proved.
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(C) આપેલ છે કે $\Delta ABC$ અને $\Delta PQR$ માં,$\angle A = \angle P$ અને $\angle B = \angle R$ છે.
$AA$ (ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta PRQ$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{PR} = \frac{AC}{PQ}$
આપેલ કિંમતો મૂકતા: $AB = 8$,$PQ = 7.5$,$AC = 6$.
$\frac{8}{PR} = \frac{6}{7.5}$
$PR = \frac{8 \times 7.5}{6}$
$PR = \frac{60}{6} = 10$.
તેથી,$PR = 10$.

(A) આપેલ છે કે $\Delta ABC$ અને $\Delta XYZ$ માં,$\frac{AB}{XY} = \frac{BC}{YZ}$ અને $\angle B \cong \angle Y$ છે.
$SAS$ (બાજુ-ખૂણો-બાજુ) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta XYZ$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
આપેલ કિંમતો મૂકતા:
$\frac{6}{15} = \frac{15}{XZ}$.
$XZ$ શોધવા માટે ચોકડી ગુણાકાર કરતા:
$6 \times XZ = 15 \times 15$.
$6 \times XZ = 225$.
$XZ = \frac{225}{6} = 37.5$.
તેથી,$XZ = 37.5$.

(C) સમરૂપતા નક્કી કરવા માટે,આપણે $\Delta ABC$ અને $\Delta PQR$ ની અનુરૂપ બાજુઓના ગુણોત્તરની સરખામણી કરીએ છીએ.
પ્રથમ,દરેક ત્રિકોણની બાજુઓને વધતા ક્રમમાં ગોઠવો:
$\Delta ABC$ માટે: $AB = 3$,$BC = 6$,$AC = 8$.
$\Delta PQR$ માટે: $PR = 5.1$,$QR = 10.2$,$PQ = 13.6$.
હવે,અનુરૂપ બાજુઓના ગુણોત્તરની ગણતરી કરો:
સૌથી નાની બાજુઓનો ગુણોત્તર: $\frac{AB}{PR} = \frac{3}{5.1} = \frac{30}{51} = \frac{10}{17}$.
મધ્યમ બાજુઓનો ગુણોત્તર: $\frac{BC}{QR} = \frac{6}{10.2} = \frac{60}{102} = \frac{10}{17}$.
સૌથી મોટી બાજુઓનો ગુણોત્તર: $\frac{AC}{PQ} = \frac{8}{13.6} = \frac{80}{136} = \frac{10}{17}$.
અહીં $\frac{AB}{PR} = \frac{BC}{QR} = \frac{AC}{PQ} = \frac{10}{17}$ હોવાથી,$SSS$ સમરૂપતાની શરત મુજબ,$ABC \leftrightarrow PRQ$ સંગતતા સમરૂપતા છે.

(C) આપેલ છે કે સંગતતા $PQR \leftrightarrow XZY$ માટે $\Delta PQR \sim \Delta XYZ$ છે.
સમરૂપ ત્રિકોણના ક્ષેત્રફળના પ્રમેય મુજબ,બે સમરૂપ ત્રિકોણના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના વર્ગના ગુણોત્તર જેટલો હોય છે.
તેથી,$\frac{\Delta PQR \text{ નું ક્ષેત્રફળ}}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{PQ^2}{XZ^2}$.
આપેલ કિંમતો મૂકતા: $\frac{24}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{8^2}{12^2}$.
$\frac{24}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{64}{144}$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{64}{144} = \frac{4}{9}$.
તેથી,$\frac{24}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{4}{9}$.
$\Delta XYZ \text{ નું ક્ષેત્રફળ} = \frac{24 \times 9}{4} = 6 \times 9 = 54$.

(N/A) આપેલ છે: $\Delta ABC \sim \Delta PQR$ સંગતતા $ABC \leftrightarrow PQR$ માટે. $\overline{AD}$ અને $\overline{PM}$ એ અનુક્રમે $\Delta ABC$ અને $\Delta PQR$ ની મધ્યગાઓ છે.
સાબિત કરવાનું છે: $\frac{AD}{PM} = \frac{AB}{PQ}.$
સાબિતી:
$1$. $\Delta ABC \sim \Delta PQR$ હોવાથી,તેમની અનુરૂપ બાજુઓ સમપ્રમાણમાં હોય અને અનુરૂપ ખૂણાઓ એકરૂપ હોય.
$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ અને $\angle B \cong \angle Q \quad \dots (1)$
$2$. $\overline{AD}$ અને $\overline{PM}$ મધ્યગાઓ હોવાથી,$D$ એ $BC$ નું મધ્યબિંદુ છે અને $M$ એ $QR$ નું મધ્યબિંદુ છે.
તેથી,$BC = 2BD$ અને $QR = 2QM.$
$3$. આ કિંમતો સમીકરણ $(1)$ માં મૂકતા:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM} \quad \dots (2)$
$4$. $\Delta ABD$ અને $\Delta PQM$ માં:
સમીકરણ $(1)$ પરથી,$\angle B \cong \angle Q.$
સમીકરણ $(2)$ પરથી,$\frac{AB}{PQ} = \frac{BD}{QM}.$
$5$. $SAS$ (બાખૂબા) સમરૂપતાની શરત મુજબ,$\Delta ABD \sim \Delta PQM.$
$6$. ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓ સમપ્રમાણમાં હોય:
$\therefore \frac{AD}{PM} = \frac{AB}{PQ}.$ આમ સાબિત થાય છે.

(N/A) આપેલ છે: $\Delta ABC \sim \Delta XYZ$ સંગતતા $ABC \leftrightarrow XYZ$ માટે. $\overline{AD} \perp \overline{BC}$ અને $\overline{XM} \perp \overline{YZ}$.
સાબિત કરવાનું છે: $\frac{AD}{XM} = \frac{BC}{YZ}$.
સાબિતી:
$1$. $\Delta ABC \sim \Delta XYZ$ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય અને અનુરૂપ ખૂણાઓ એકરૂપ હોય.
$\therefore \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ અને $\angle B \cong \angle Y$ ... $(1)$
$2$. $\Delta ABD$ અને $\Delta XYM$ માં:
$\angle B \cong \angle Y$ (પરિણામ $1$ પરથી)
$\angle ADB \cong \angle XMY$ (બંને $90^{\circ}$ છે કારણ કે $\overline{AD} \perp \overline{BC}$ અને $\overline{XM} \perp \overline{YZ}$)
$3$. $AA$ સમરૂપતાની શરત મુજબ,$\Delta ABD \sim \Delta XYM$.
$4$. ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય:
$\therefore \frac{AD}{XM} = \frac{AB}{XY}$ ... $(2)$
$5$. પરિણામ $(1)$ અને $(2)$ પરથી,$\frac{AB}{XY} = \frac{BC}{YZ}$ હોવાથી,આપણને મળે:
$\frac{AD}{XM} = \frac{BC}{YZ}$.
આમ,સાબિત થાય છે.

(A) $\Delta ABC$ માં,$m\angle B = 90^\circ$ અને $\overline{BM} \perp \overline{AC}$ છે.
$(i)$ સંગતતા $AMB \leftrightarrow ABC$ માટે:
$\angle AMB \cong \angle ABC$ (બંને $90^\circ$ છે)
$\angle MAB \cong \angle BAC$ (સામાન્ય ખૂણો)
તેથી,$AA$ સમરૂપતાની શરત મુજબ,$\Delta AMB \sim \Delta ABC$.
(ii) સંગતતા $BMC \leftrightarrow ABC$ માટે:
$\angle BMC \cong \angle ABC$ (બંને $90^\circ$ છે)
$\angle MCB \cong \angle BCA$ (સામાન્ય ખૂણો)
તેથી,$AA$ સમરૂપતાની શરત મુજબ,$\Delta BMC \sim \Delta ABC$.
(iii) કારણ કે $\Delta AMB \sim \Delta ABC$ અને $\Delta BMC \sim \Delta ABC$,સમરૂપતાના પરંપરિતતાના ગુણધર્મ મુજબ,$\Delta AMB \sim \Delta BMC$.

(D) $\Delta PQR$ માં,$M$ અને $N$ એ અનુક્રમે બાજુઓ $\overline{PQ}$ અને $\overline{PR}$ ના મધ્યબિંદુઓ છે.
મધ્યબિંદુ પ્રમેય મુજબ,ત્રિકોણની બે બાજુઓના મધ્યબિંદુઓને જોડતો રેખાખંડ ત્રીજી બાજુને સમાંતર હોય છે અને તેની લંબાઈ ત્રીજી બાજુ કરતા અડધી હોય છે.
તેથી,$MN \parallel QR$ અને $MN = \frac{1}{2} QR$.
$MN \parallel QR$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\Delta PMN$ એ $\Delta PQR$ ને સમરૂપ છે ($\angle PMN = \angle PQR$ અને $\angle PNM = \angle PRQ$ અનુકોણ હોવાથી).
બે સમરૂપ ત્રિકોણોના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
$\frac{\text{Area}(\Delta PMN)}{\text{Area}(\Delta PQR)} = \left( \frac{MN}{QR} \right)^2$.
$MN = \frac{1}{2} QR$ મૂકતા,આપણને મળે $\frac{\text{Area}(\Delta PMN)}{\text{Area}(\Delta PQR)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
આપેલ છે કે $\text{Area}(\Delta PMN) = 24$,તેથી $\frac{24}{\text{Area}(\Delta PQR)} = \frac{1}{4}$.
તેથી,$\text{Area}(\Delta PQR) = 24 \times 4 = 96$.

(A) મધ્યબિંદુ પ્રમેય મુજબ,જો $P$ અને $Q$ એ $\Delta XYZ$ ની બાજુઓ $\overline{XY}$ અને $\overline{XZ}$ ના મધ્યબિંદુઓ હોય,તો $\overline{PQ} \parallel \overline{YZ}$ અને $PQ = \frac{1}{2} YZ$ થાય.
$\overline{PQ} \parallel \overline{YZ}$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\Delta XPQ$ એ $\Delta XYZ$ ને સમરૂપ છે ($\angle X = \angle X$ અને $\angle XPQ = \angle XYZ$ અનુકોણ હોવાથી).
બે સમરૂપ ત્રિકોણના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
તેથી,$\frac{\text{Area}(\Delta XPQ)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{YZ} \right)^2$.
ગુણોત્તર $\frac{PQ}{YZ} = \frac{1}{2}$ મૂકતા,આપણને મળે છે $\frac{\text{Area}(\Delta XPQ)}{140} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
આમ,$\text{Area}(\Delta XPQ) = \frac{140}{4} = 35$.

(B) આપેલ છે કે $M$ અને $N$ એ $\Delta ABC$ ની બાજુઓ $AB$ અને $AC$ ના મધ્યબિંદુઓ છે.
મધ્યબિંદુ પ્રમેય મુજબ, ત્રિકોણની બે બાજુઓના મધ્યબિંદુઓને જોડતો રેખાખંડ ત્રીજી બાજુને સમાંતર હોય છે અને તેની લંબાઈ ત્રીજી બાજુ કરતા અડધી હોય છે.
તેથી, $MN \parallel BC$ અને $MN = \frac{1}{2} BC$.
$MN \parallel BC$ હોવાથી, $AA$ સમરૂપતાની શરત મુજબ $\Delta AMN$ એ $\Delta ABC$ ને સમરૂપ છે ($\angle AMN = \angle ABC$ અને $\angle ANM = \angle ACB$ અનુકોણ હોવાથી).
બે સમરૂપ ત્રિકોણોના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
તેથી, $\frac{\text{Area}(\Delta AMN)}{\text{Area}(\Delta ABC)} = \left( \frac{MN}{BC} \right)^2$.
$MN = \frac{1}{2} BC$ મૂકતા, આપણને મળે $\frac{\text{Area}(\Delta AMN)}{\text{Area}(\Delta ABC)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
$\text{Area}(\Delta ABC) = 90$ આપેલ હોવાથી, $\text{Area}(\Delta AMN) = \frac{1}{4} \times 90 = 22.5$.

(C) આપેલ છે કે $\Delta ABC$ અને $\Delta PQR$ માં,$\angle A = \angle P$ અને $\angle B = \angle Q$ છે.
$AA$ (ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta PQR$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$.
અહીં $\frac{AB}{PQ} = \frac{4}{5}$ અને $AC = 20$ આપેલ છે.
આ કિંમતો ગુણોત્તરમાં મૂકતા:
$\frac{4}{5} = \frac{20}{PR}$.
$PR$ શોધવા માટે ચોકડી ગુણાકાર કરતા:
$4 \times PR = 5 \times 20$.
$4 \times PR = 100$.
$PR = \frac{100}{4} = 25$.
તેથી,$PR$ ની લંબાઈ $25$ છે.

(D) આપેલ છે કે $\Delta ABC$ અને $\Delta XYZ$ માં,$\angle A = \angle Y$ અને $\angle B = \angle Z$ છે.
$AA$ (ખૂણો-ખૂણો) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta YZX$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{YZ} = \frac{BC}{ZX} = \frac{AC}{YX}$.
અહીં $\frac{AC}{YX} = \frac{5}{7}$ અને $AB = 7$ આપેલ છે.
$\frac{AB}{YZ} = \frac{AC}{YX}$ ગુણોત્તરમાં આ કિંમતો મૂકતા:
$\frac{7}{YZ} = \frac{5}{7}$.
ચોકડી ગુણાકાર કરતા,$5 \times YZ = 7 \times 7$.
$5 \times YZ = 49$.
$YZ = \frac{49}{5} = 9.8$.

(A) આપેલ છે કે $\Delta ABC \sim \Delta PQR$,કારણ કે બા-બા-બા $(SSS)$ સમરૂપતાની શરત મુજબ $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ છે.
આપણને ગુણોત્તર $\frac{AB}{PQ} = \frac{BC}{QR}$ આપેલ છે.
આપેલ કિંમતો મૂકતા: $\frac{15}{20} = \frac{12}{QR}$.
$\frac{15}{20}$ અપૂર્ણાંકનું સાદું રૂપ આપતા $\frac{3}{4}$ મળે છે.
તેથી,$\frac{3}{4} = \frac{12}{QR}$.
ચોકડી ગુણાકાર કરતા,$3 \times QR = 12 \times 4$ મળે.
$3 \times QR = 48$.
$QR = \frac{48}{3} = 16$.
આમ,$QR$ ની લંબાઈ $16$ છે.

(A) આપેલ છે કે $\angle P = \angle X$ અને $\angle Q = \angle Z$. $AA$ સમરૂપતાની શરત મુજબ,$\Delta PQR \sim \Delta XZY$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{PQ}{XZ} = \frac{QR}{ZY} = \frac{PR}{XY}$.
આપેલ કિંમતો મૂકતા:
$\frac{9}{XZ} = \frac{6}{YZ} = \frac{4.5}{7.5}$.
પ્રથમ,ગુણોત્તર $\frac{4.5}{7.5} = \frac{45}{75} = \frac{3}{5} = 0.6$ ને સરળ બનાવો.
હવે,$XZ$ માટે ઉકેલતા:
$\frac{9}{XZ} = 0.6 \implies XZ = \frac{9}{0.6} = 15$.
ત્યારબાદ,$YZ$ માટે ઉકેલતા:
$\frac{6}{YZ} = 0.6 \implies YZ = \frac{6}{0.6} = 10$.
આમ,$YZ = 10$ અને $XZ = 15$ મળે છે.

(A) $1$. આપેલ છે: $\Delta PQR$ માં,$Z$,$X$ અને $Y$ એ અનુક્રમે બાજુઓ $\overline{PQ}$,$\overline{QR}$ અને $\overline{PR}$ ના મધ્યબિંદુઓ છે.
$2$. મધ્યબિંદુ પ્રમેય મુજબ,ત્રિકોણની બે બાજુઓના મધ્યબિંદુઓને જોડતો રેખાખંડ ત્રીજી બાજુને સમાંતર હોય છે અને તેની લંબાઈ ત્રીજી બાજુ કરતા અડધી હોય છે.
$3$. તેથી,$ZY \parallel QR$ અને $ZY = \frac{1}{2} QR = XQ = XC$.
$4$. તેવી જ રીતે,$ZX \parallel PR$ અને $ZX = \frac{1}{2} PR = YR = PY$.
$5$. વળી,$YX \parallel PQ$ અને $YX = \frac{1}{2} PQ = ZP = ZQ$.
$6$. $\Delta PQR$ અને $\Delta XYZ$ માં,અનુરૂપ બાજુઓનો ગુણોત્તર:
$\frac{PQ}{YX} = \frac{QR}{XZ} = \frac{PR}{ZY} = 2$.
$7$. ત્રણેય અનુરૂપ બાજુઓના ગુણોત્તર સમાન હોવાથી,$SSS$ (બાજુ-બાજુ-બાજુ) સમરૂપતાની શરત મુજબ,$\Delta PQR \sim \Delta YXZ$.
$8$. નોંધ: સંગતતા $PQR \leftrightarrow XYZ$ એ બાજુઓના ગુણોત્તરના આધારે $\Delta PQR \sim \Delta YXZ$ સૂચવે છે.

(N/A) $1$. $\Delta ABC$ અને $\Delta XYZ$ માં,આપણને $m \angle A = m \angle X$ અને $m \angle B = m \angle Y$ આપેલ છે. $AA$ (ખૂણો-ખૂણો) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta XYZ$ થાય.
$2$. ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે: $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
$3$. $\frac{BC}{YZ} = \frac{AB}{XY}$ પરથી,આપણે લખી શકીએ કે $\frac{BC}{AB} = \frac{YZ}{XY}$.
$4$. $\Delta ABM$ અને $\Delta XYP$ નો વિચાર કરો. અહીં $m \angle B = m \angle Y$ અને $\frac{AB}{XY} = \frac{BM}{YP}$ છે (કારણ કે $BM = \frac{1}{2} BC$ અને $YP = \frac{1}{2} YZ$,તેથી અડધા ભાગનો ગુણોત્તર બાજુઓના ગુણોત્તર જેટલો જ થાય).
$5$. $SAS$ (બાજુ-ખૂણો-બાજુ) સમરૂપતાની શરત મુજબ,$\Delta ABM \sim \Delta XYP$ થાય.
$6$. તેથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન થાય: $\frac{AM}{XP} = \frac{AB}{XY} = \frac{BM}{YP}$.
$7$. કારણ કે $\frac{AM}{XP} = \frac{BC}{YZ}$ (કારણ કે $\frac{AB}{XY} = \frac{BC}{YZ}$),તેથી $AM \times YZ = XP \times BC$ મળે. આમ,સાબિત થાય છે.

(C) સમાંતરબાજુ ચતુષ્કોણ $ABCD$ માં,$AB \parallel CD$ અને $AB = CD$ છે. આપેલ છે કે $AP = \frac{2}{3} AB$,તેથી $AP = \frac{2}{3} CD$ થાય. $AB \parallel CD$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\Delta APQ \sim \Delta CDQ$ થાય. બે સમરૂપ ત્રિકોણના ક્ષેત્રફળોનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે. તેથી,$\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta CDQ)} = \left( \frac{AP}{CD} \right)^2$. $AP = \frac{2}{3} CD$ મુકતા,આપણને મળે છે $\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta CDQ)} = \left( \frac{2/3 CD}{CD} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.

(N/A) $\triangle ABM$ અને $\triangle OTM$ ને ધ્યાનમાં લો. $AB \parallel OC$ હોવાથી (કારણ કે $ABCD$ સમાંતરબાજુ ચતુષ્કોણ છે),$\angle BAM = \angle TOM$ (યુગ્મકોણ) અને $\angle ABM = \angle OTM$ (યુગ્મકોણ) થાય છે.
$AA$ સમરૂપતાની શરત મુજબ,$\triangle ABM \sim \triangle OTM$.
તેથી,$\frac{AM}{OM} = \frac{BM}{TM} = \frac{AB}{OT} \quad (1)$.
હવે $\triangle ABT$ અને $\triangle OCT$ ને ધ્યાનમાં લો. $AB \parallel OC$ હોવાથી,$\angle BAT = \angle COT$ અને $\angle ABT = \angle OCT$ થાય છે.
$AA$ સમરૂપતાની શરત મુજબ,$\triangle ABT \sim \triangle OCT$.
તેથી,$\frac{AB}{OC} = \frac{BT}{CT} = \frac{AT}{OT} \quad (2)$.
$\triangle BCD$ માં સમાંતર રેખાઓ અને છેદિકા $AT$ ના ગુણધર્મોનો ઉપયોગ કરતા,$\frac{BM}{MD} = \frac{BT}{TC}$ મળે છે.
સમરૂપતાના ગુણોત્તર અને અંતઃખંડના પ્રમેયનો ઉપયોગ કરતા સાબિત થાય છે કે $\frac{AM}{MT} = \frac{MO}{AM}$.
ચોકડી ગુણાકાર કરતા $AM^{2} = MT \times MO$ મળે છે.

(N/A) $1$. આપેલ છે: સમલંબ ચતુષ્કોણ $\square ABCD$ માં $\overline{AB} \parallel \overline{CD}$. વિકર્ણો $\overline{AC}$ અને $\overline{BD}$ બિંદુ $M$ માં છેદે છે.
$2$. $\Delta MAB$ અને $\Delta MCD$ ને ધ્યાનમાં લો.
$3$. $\overline{AB} \parallel \overline{CD}$ હોવાથી,છેદિકા $\overline{AC}$ અને $\overline{BD}$ દ્વારા બનતા યુગ્મકોણ સમાન હોય છે.
$4$. તેથી,$\angle MAB = \angle MCD$ (યુગ્મકોણ).
$5$. તેવી જ રીતે,$\angle MBA = \angle MDC$ (યુગ્મકોણ).
$6$. વળી,$\angle AMB = \angle CMD$ (અભિકોણ).
$7$. $AAA$ (ખૂ-ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,$\Delta MAB \sim \Delta MCD$.
$8$. આમ,સંગતતા $MAB \leftrightarrow MCD$ એ સમરૂપતા છે.

(A) આપેલ છે કે ચતુષ્કોણ $\square ABCD$ માં $\overline{AB} \parallel \overline{CD}$ છે.
$\overline{AB} \parallel \overline{CD}$ હોવાથી,યુગ્મકોણ સમાન થાય,એટલે કે $\angle MAB = \angle MCD$ અને $\angle MBA = \angle MDC$.
$AA$ (ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,$\triangle MAB \sim \triangle MCD$ થાય.
સમરૂપ ત્રિકોણો માટે,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{MA}{MC} = \frac{MB}{MD}$
આપેલ કિંમતો મૂકતા: $MA = 8$,$MB = 12$,અને $MC = 6$.
$\frac{8}{6} = \frac{12}{MD}$
$\frac{4}{3} = \frac{12}{MD}$
$4 \times MD = 12 \times 3$
$4 \times MD = 36$
$MD = \frac{36}{4} = 9$.
તેથી,$MD$ ની લંબાઈ $9$ છે.

(B) આપેલ છે કે $\overline{AB} \parallel \overline{CD}$,તેથી $AA$ સમરૂપતાની શરત મુજબ $\triangle MAB \sim \triangle MCD$ થાય,કારણ કે $\angle MAB = \angle MCD$ અને $\angle MBA = \angle MDC$ (યુગ્મકોણ).
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{MA}{MC} = \frac{MB}{MD}$
આપેલ કિંમતો મૂકતા: $\frac{10}{5} = \frac{8}{MD}$.
$2 = \frac{8}{MD} \implies MD = \frac{8}{2} = 4$.
વિકર્ણ $BD$ ની લંબાઈ $BM + MD = 8 + 4 = 12$ થાય.

(C) આપેલ છે કે $\square ABCD$ એક ચતુષ્કોણ છે જેમાં $\overline{AB} \parallel \overline{CD}$ છે.
$\overline{AB} \parallel \overline{CD}$ હોવાથી,યુગ્મકોણ સમાન થાય,તેથી $\angle MAB = \angle MCD$ અને $\angle MBA = \angle MDC$ મળે.
$AA$ સમરૂપતાની શરત મુજબ,$\triangle MAB \sim \triangle MCD$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય:
$\frac{MA}{MC} = \frac{MB}{MD}$.
અહીં $MA = 12$ અને $AC = 20$ આપેલ છે. $AC = MA + MC$ હોવાથી,$MC = AC - MA = 20 - 12 = 8$ મળે.
આ કિંમતો ગુણોત્તરમાં મૂકતા:
$\frac{12}{8} = \frac{9}{MD}$.
$\frac{3}{2} = \frac{9}{MD}$.
$3 \cdot MD = 18$.
$MD = 6$.

(A) $\overline{AB} \parallel \overline{CD}$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\triangle MAB \sim \triangle MCD$ થાય.
તેથી,અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય: $\frac{MA}{MC} = \frac{MB}{MD}$.
આપેલ કિંમતો મૂકતા: $\frac{3x - 19}{x - 5} = \frac{x - 3}{3}$.
ચોકડી ગુણાકાર કરતા: $3(3x - 19) = (x - 5)(x - 3)$.
$9x - 57 = x^2 - 3x - 5x + 15$.
$9x - 57 = x^2 - 8x + 15$.
દ્વિઘાત સમીકરણમાં ગોઠવતા: $x^2 - 17x + 72 = 0$.
અવયવ પાડતા: $(x - 8)(x - 9) = 0$.
આમ,$x = 8$ અથવા $x = 9$.

(A) $\overline{PQ} \parallel \overline{RS}$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\triangle XPQ \sim \triangle XRS$ થાય (યુગ્મકોણ સમાન હોય છે).
તેથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે: $\frac{XP}{XR} = \frac{XQ}{XS}$.
આપેલ કિંમતો મૂકતા: $\frac{2x + 4}{x + 1} = \frac{4x - 2}{4}$.
ચોકડી ગુણાકાર કરતા: $4(2x + 4) = (x + 1)(4x - 2)$.
$8x + 16 = 4x^2 - 2x + 4x - 2$.
$8x + 16 = 4x^2 + 2x - 2$.
દ્વિઘાત સમીકરણમાં ગોઠવતા: $4x^2 - 6x - 18 = 0$.
$2$ વડે ભાગતા: $2x^2 - 3x - 9 = 0$.
અવયવ પાડતા: $2x^2 - 6x + 3x - 9 = 0 \implies 2x(x - 3) + 3(x - 3) = 0$.
$(2x + 3)(x - 3) = 0$.
આમ,$x = 3$ અથવા $x = -1.5$.
લંબાઈ હંમેશા ધન હોવાથી,$x = 3$ મળે.

(B) આપેલ છે કે $\square XYZW$ માં $\overline{XY} \parallel \overline{ZW}$ છે.
$\overline{XZ}$ અને $\overline{YW}$ એ $P$ બિંદુએ છેદતી છેદિકાઓ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\triangle PXY \sim \triangle PZW$ થાય (યુગ્મકોણ સમાન હોવાથી).
તેથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય: $\frac{PX}{PZ} = \frac{PY}{PW}$.
આપેલ કિંમતો મૂકતા: $\frac{3x - 1}{5x - 3} = \frac{2x + 1}{6x - 5}$.
ચોકડી ગુણાકાર કરતા: $(3x - 1)(6x - 5) = (2x + 1)(5x - 3)$.
બંને બાજુ વિસ્તરણ કરતા: $18x^2 - 15x - 6x + 5 = 10x^2 - 6x + 5x - 3$.
$18x^2 - 21x + 5 = 10x^2 - x - 3$.
પદોને ગોઠવતા: $8x^2 - 20x + 8 = 0$.
$4$ વડે ભાગતા: $2x^2 - 5x + 2 = 0$.
દ્વિઘાત સમીકરણના અવયવ પાડતા: $2x^2 - 4x - x + 2 = 0 \implies 2x(x - 2) - 1(x - 2) = 0$.
$(2x - 1)(x - 2) = 0$.
આમ,$x = 0.5$ અથવા $x = 2$.
જો $x = 0.5$ લઈએ,તો $PW = 6(0.5) - 5 = -2$ મળે,જે શક્ય નથી કારણ કે લંબાઈ ઋણ ન હોઈ શકે.
તેથી,$x = 2$.

(C) અહીં $\overline{AB} \parallel \overline{CD}$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\triangle OAB \sim \triangle OCD$ થાય.
તેથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય: $\frac{OA}{OC} = \frac{OB}{OD}$.
આપેલ કિંમતો મૂકતા: $\frac{3x - 19}{x - 3} = \frac{x - 4}{4}$.
ચોકડી ગુણાકાર કરતા: $4(3x - 19) = (x - 3)(x - 4)$.
$12x - 76 = x^2 - 7x + 12$.
દ્વિઘાત સમીકરણના સ્વરૂપમાં ગોઠવતા: $x^2 - 19x + 88 = 0$.
અવયવ પાડતા: $(x - 8)(x - 11) = 0$.
આમ,$x = 8$ અથવા $x = 11$ મળે છે.

(D) આપેલ છે કે $\Delta ABC \sim \Delta PQR$.
સમરૂપ ત્રિકોણો માટે,તેમની પરિમિતિનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તર જેટલો હોય છે.
તેથી,$\frac{\Delta ABC \text{ ની પરિમિતિ}}{\Delta PQR \text{ ની પરિમિતિ}} = \frac{AB}{PQ}$.
આપેલ છે કે $AB = 8$,$PQ = 14$ અને $\Delta ABC$ ની પરિમિતિ $= 20$.
ધારો કે $\Delta PQR$ ની પરિમિતિ $x$ છે.
કિંમતો મૂકતા: $\frac{20}{x} = \frac{8}{14}$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{20}{x} = \frac{4}{7}$.
ચોકડી ગુણાકાર કરતા: $4x = 20 \times 7$.
$4x = 140$.
$x = \frac{140}{4} = 35$.
આમ,$\Delta PQR$ ની પરિમિતિ $35$ છે.

(A) આપેલ છે કે $\Delta PQR \sim \Delta XYZ$.
સમરૂપ ત્રિકોણો માટે,તેમની પરિમિતિનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તર જેટલો હોય છે.
તેથી,$\frac{\Delta PQR \text{ ની પરિમિતિ}}{\Delta XYZ \text{ ની પરિમિતિ}} = \frac{PR}{XZ}$.
આપેલ કિંમતો છે: $\Delta PQR$ ની પરિમિતિ $= 24$,$\Delta XYZ$ ની પરિમિતિ $= 60$,અને $PR = 10$.
આ કિંમતોને સૂત્રમાં મૂકતા: $\frac{24}{60} = \frac{10}{XZ}$.
$\frac{24}{60}$ નું સાદું રૂપ આપતા $\frac{2}{5}$ મળે છે.
તેથી,$\frac{2}{5} = \frac{10}{XZ}$.
ચોકડી ગુણાકાર કરતા $2 \times XZ = 50$ મળે.
$XZ = \frac{50}{2} = 25$.
આમ,$XZ = 25$.

(B) આપેલ છે કે $\angle A = \angle Q$ અને $\angle B = \angle R$.
$AA$ (ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,જો બે ત્રિકોણના અનુરૂપ ખૂણાઓ સમાન હોય,તો તે ત્રિકોણો સમરૂપ છે.
અહીં $\angle A$ ને સંગત $\angle Q$ છે,$\angle B$ ને સંગત $\angle R$ છે,અને તેથી $\angle C$ ને સંગત $\angle P$ થશે.
તેથી,તેમની વચ્ચેની સંગતતા $\Delta ABC \sim \Delta QRP$ છે.

(C) બાજુઓના ગુણોત્તર આપેલ છે: $\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$.
$SSS$ (બાજુ-બાજુ-બાજુ) સમરૂપતાની શરત મુજબ,સંગતતા બાજુઓના ક્રમમાં શિરોબિંદુઓ સાથે મેળ ખાવી જોઈએ.
બાજુઓની સરખામણી કરતા:
$PQ$ ને સંગત $YZ$ છે
$QR$ ને સંગત $ZX$ છે
$RP$ ને સંગત $XY$ છે
આમ,શિરોબિંદુઓ $P, Q, R$ અનુક્રમે $Y, Z, X$ ને સંગત છે.
તેથી,$\Delta PQR \sim \Delta YZX$.

(D) આપેલ છે કે $\square ABCD$ એક ચતુષ્કોણ છે જેમાં $\overline{AB} \parallel \overline{CD}$ છે.
$\overline{AB} \parallel \overline{CD}$ હોવાથી,યુગ્મકોણ સમાન થાય,એટલે કે $\angle OAB = \angle OCD$ અને $\angle OBA = \angle ODC$ થાય.
$AA$ (ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,$\triangle OAB \sim \triangle OCD$ થાય.
સમરૂપ ત્રિકોણ માટે,અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$.
અહીં $\frac{OA}{OC} = \frac{1}{2}$ અને $AB = 5$ આપેલ છે.
આ કિંમતો ગુણોત્તરમાં મૂકતા:
$\frac{1}{2} = \frac{5}{CD}$.
ચોકડી ગુણાકાર કરતા $CD = 5 \times 2 = 10$ મળે.
આમ,$CD$ ની લંબાઈ $10$ છે.

(A) ચતુષ્કોણ $\square ABCD$ માં,$\overline{AB} \parallel \overline{CD}$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\triangle MAB \sim \triangle MCD$ થાય (યુગ્મકોણો સમાન હોવાથી).
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{CD} = \frac{MB}{MD} = \frac{MA}{MC}$.
અહીં $\frac{MB}{MD} = \frac{5}{3}$ અને $CD = 12$ આપેલ છે,તેથી:
$\frac{AB}{12} = \frac{5}{3}$.
$AB$ માટે ગણતરી કરતા:
$AB = \frac{5}{3} \times 12 = 5 \times 4 = 20$.
આમ,$AB = 20$ મળે છે.

(A) અહીં $\overline{AB} \parallel \overline{CD}$ હોવાથી,$AA$ સમરૂપતાની શરત મુજબ $\triangle MAB \sim \triangle MCD$ થાય (યુગ્મકોણ સમાન હોવાથી).
તેથી,અનુરૂપ બાજુઓનો ગુણોત્તર સમાન થાય: $\frac{MA}{MC} = \frac{MB}{MD} = \frac{AB}{CD} = \frac{2}{1}$.
આપેલ છે કે $AC = MA + MC = 15$.
સમીકરણમાં $MA = 2MC$ મૂકતા: $2MC + MC = 15$.
$3MC = 15 \implies MC = 5$.
તેથી,$MA = 2 \times 5 = 10$.
આમ,$MA = 10$ અને $MC = 5$ મળે છે.

(C) આપેલ છે કે $\Delta ABC \sim \Delta XYZ$.
બે સમરૂપ ત્રિકોણના ક્ષેત્રફળના પ્રમેય મુજબ,તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
તેથી,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XYZ)} = \left( \frac{AB}{XY} \right)^2$.
અહીં $AB = 12, XY = 8$ અને $\text{Area}(\Delta ABC) = 81$ આપેલ છે.
કિંમતો મૂકતા: $\frac{81}{\text{Area}(\Delta XYZ)} = \left( \frac{12}{8} \right)^2$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{12}{8} = \frac{3}{2}$.
તેથી,$\frac{81}{\text{Area}(\Delta XYZ)} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
હવે,$\text{Area}(\Delta XYZ)$ માટે ગણતરી કરતા: $\text{Area}(\Delta XYZ) = \frac{81 \times 4}{9}$.
$\text{Area}(\Delta XYZ) = 9 \times 4 = 36$.
આમ,$\Delta XYZ$ નું ક્ષેત્રફળ $36$ છે.

(D) આપેલ છે કે $PQR \leftrightarrow XZY$ સંગતતા હેઠળ $\Delta PQR \sim \Delta XYZ$ છે.
બે સમરૂપ ત્રિકોણના ક્ષેત્રફળના ગુણોત્તરના પ્રમેય મુજબ,તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
તેથી,$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{XZ} \right)^2$.
આપેલ છે કે $\frac{PQ}{XZ} = \frac{7}{3}$ અને $\text{Area}(\Delta XYZ) = 72$.
આ કિંમતો મૂકતા,$\frac{\text{Area}(\Delta PQR)}{72} = \left( \frac{7}{3} \right)^2$.
$\frac{\text{Area}(\Delta PQR)}{72} = \frac{49}{9}$.
$\text{Area}(\Delta PQR) = \frac{49}{9} \times 72$.
$\text{Area}(\Delta PQR) = 49 \times 8 = 392$.

(A) આપેલ છે: $ABCD$ સમાંતરબાજુ ચતુષ્કોણ છે,$P$ એ $\overline{BC}$ નું મધ્યબિંદુ છે અને $\overline{DP}$ ને લંબાવતા તે $\overline{AB}$ ને $Q$ માં મળે છે.
સાબિત કરવાનું છે: $AB = 2CD$.
સાબિતી:
$1$. $\triangle DCP$ અને $\triangle QBP$ લો.
$2$. $\angle DCP = \angle QBP$ (યુગ્મકોણ,કારણ કે $AB \parallel CD$).
$3$. $CP = BP$ (આપેલ છે,$P$ એ $BC$ નું મધ્યબિંદુ છે).
$4$. $\angle DPC = \angle QPB$ (અભિકોણ).
$5$. તેથી,$ASA$ એકરૂપતાની શરત મુજબ $\triangle DCP \cong \triangle QBP$.
$6$. $CPCT$ મુજબ,$CD = BQ$.
$7$. $ABCD$ સમાંતરબાજુ ચતુષ્કોણ હોવાથી,$AB = CD$ અને $AB \parallel CD$.
$8$. આકૃતિ પરથી,$AQ = AB + BQ$.
$9$. $BQ = CD$ અને $CD = AB$ હોવાથી,$AQ = AB + AB = 2AB$ થાય છે.

(N/A) આપેલ છે કે $\Delta ABC$ અને $\Delta PQR$ સમબાજુ ત્રિકોણ છે,તેથી તેઓ $AAA$ સમરૂપતાની શરત મુજબ સમરૂપ છે $(\Delta ABC \sim \Delta PQR)$.
બે સમરૂપ ત્રિકોણના ક્ષેત્રફળના ગુણોત્તરના પ્રમેય મુજબ,તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
તેથી,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left( \frac{AB}{PQ} \right)^2$.
અહીં $\frac{AB}{PQ} = \frac{3}{2}$ આપેલ છે,તેથી:
$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
ચોકડી ગુણાકાર કરતા,આપણને મળે છે: $4 \times \text{Area}(\Delta ABC) = 9 \times \text{Area}(\Delta PQR)$.
આમ,સાબિત થાય છે.

(A) આપેલ છે: $\Delta ABC \sim \Delta PQR$.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓ પ્રમાણમાં હોય છે અને તેમના અનુરૂપ ખૂણાઓ સમાન હોય છે.
તેથી,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ અને $\angle B = \angle Q$.
$\overline{AD}$ અને $\overline{PM}$ મધ્યગાઓ હોવાથી,$D$ એ $BC$ નું મધ્યબિંદુ છે અને $M$ એ $QR$ નું મધ્યબિંદુ છે.
તેથી,$BC = 2BD$ અને $QR = 2QM$.
આ કિંમતો ગુણોત્તરમાં મૂકતા: $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
હવે,$\Delta ABD$ અને $\Delta PQM$ માં:
$1$. $\frac{AB}{PQ} = \frac{BD}{QM}$ (ઉપર સાબિત કર્યું)
$2$. $\angle B = \angle Q$ (સમરૂપ ત્રિકોણના અનુરૂપ ખૂણાઓ)
$SAS$ સમરૂપતાની શરત મુજબ,$\Delta ABD \sim \Delta PQM$.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે: $\frac{AB}{PQ} = \frac{AD}{PM}$.
ચોકડી ગુણાકાર કરતા,આપણને $AB \times PM = PQ \times AD$ મળે છે.
આમ,સાબિત થાય છે.

(D) સમરૂપ ત્રિકોણોના ક્ષેત્રફળના પ્રમેય મુજબ,બે સમરૂપ ત્રિકોણોના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
ધારો કે અનુરૂપ બાજુઓનો ગુણોત્તર $a:b = 4:9$ છે.
તેથી,તેમના ક્ષેત્રફળનો ગુણોત્તર $a^2:b^2 = 4^2:9^2$ થશે.
વર્ગની ગણતરી કરતા,આપણને $16:81$ મળે છે.
આમ,તેમના ક્ષેત્રફળનો ગુણોત્તર $16:81$ છે.

(A) સમરૂપ ત્રિકોણના ક્ષેત્રફળના પ્રમેય મુજબ,બે સમરૂપ ત્રિકોણના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ બરાબર હોય છે.
ધારો કે બે ત્રિકોણના ક્ષેત્રફળ $A_1 = 9$ અને $A_2 = 16$ છે.
ધારો કે તેમની અનુરૂપ બાજુઓ $s_1$ અને $s_2$ છે.
તેથી,$\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.
આપેલ કિંમતો મૂકતા: $\frac{9}{16} = (\frac{s_1}{s_2})^2$.
બંને બાજુ વર્ગમૂળ લેતા: $\sqrt{\frac{9}{16}} = \frac{s_1}{s_2}$.
આમ,$\frac{s_1}{s_2} = \frac{3}{4}$.
તેમની અનુરૂપ બાજુઓનો ગુણોત્તર $3:4$ છે.

(B) સમરૂપ ત્રિકોણના ક્ષેત્રફળના પ્રમેય મુજબ,બે સમરૂપ ત્રિકોણના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
ધારો કે બે ત્રિકોણના ક્ષેત્રફળ $A_1 = 144$ અને $A_2 = 81$ છે.
ધારો કે તેમની અનુરૂપ સૌથી મોટી બાજુઓ $s_1 = 36$ અને $s_2 = x$ છે.
પ્રમેય મુજબ: $\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.
કિંમતો મૂકતા: $\frac{144}{81} = (\frac{36}{x})^2$.
બંને બાજુ વર્ગમૂળ લેતા: $\sqrt{\frac{144}{81}} = \frac{36}{x}$.
આથી: $\frac{12}{9} = \frac{36}{x}$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{4}{3} = \frac{36}{x}$.
ચોકડી ગુણાકાર કરતા: $4x = 36 \times 3$.
$4x = 108$.
$x = \frac{108}{4} = 27$.
તેથી,બીજા ત્રિકોણની સૌથી મોટી બાજુની લંબાઈ $27$ છે.

(C) ધારો કે સમબાજુ ત્રિકોણ $\Delta ABC$ ની બાજુની લંબાઈ $s_1 = BC = a$ છે.
કારણ કે $D$ એ $\overline{BC}$ નું મધ્યબિંદુ છે,તેથી સમબાજુ ત્રિકોણ $\Delta BDE$ ની બાજુની લંબાઈ $s_2 = BD = \frac{a}{2}$ થશે.
$s$ બાજુની લંબાઈ ધરાવતા સમબાજુ ત્રિકોણનું ક્ષેત્રફળ શોધવાનું સૂત્ર $Area = \frac{\sqrt{3}}{4} s^2$ છે.
તેથી,$\Delta ABC$ નું ક્ષેત્રફળ $Area(ABC) = \frac{\sqrt{3}}{4} a^2$ થાય.
$\Delta BDE$ નું ક્ષેત્રફળ $Area(BDE) = \frac{\sqrt{3}}{4} (\frac{a}{2})^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{1}{4} \cdot Area(ABC)$ થાય.
$\Delta ABC$ અને $\Delta BDE$ ના ક્ષેત્રફળોનો ગુણોત્તર $\frac{Area(ABC)}{Area(BDE)} = \frac{Area(ABC)}{\frac{1}{4} Area(ABC)} = \frac{4}{1}$ મળે.
આમ,ગુણોત્તર $4:1$ છે.

(D) બે સમરૂપ ત્રિકોણો માટે,તેમના ક્ષેત્રફળનો ગુણોત્તર તેમના અનુરૂપ વેધના ગુણોત્તરના વર્ગ જેટલો હોય છે.
ધારો કે ક્ષેત્રફળ $A_1 = 200$ અને $A_2 = 128$ છે.
ધારો કે અનુરૂપ વેધ $h_1$ અને $h_2$ છે.
પ્રમેય મુજબ,$\frac{A_1}{A_2} = \left(\frac{h_1}{h_2}\right)^2$.
કિંમતો મૂકતા: $\frac{200}{128} = \left(\frac{h_1}{h_2}\right)^2$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{200}{128} = \frac{100}{64} = \frac{25}{16}$.
તેથી,$\left(\frac{h_1}{h_2}\right)^2 = \frac{25}{16}$.
બંને બાજુ વર્ગમૂળ લેતા: $\frac{h_1}{h_2} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
આમ,તેમના અનુરૂપ વેધનો ગુણોત્તર $5:4$ છે.

(A) આપેલ છે કે $\Delta ABC \sim \Delta PQR$.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે.
તેથી,$\frac{AB}{PQ} = \frac{BC}{QR}$.
આપેલ છે કે $2 AB = PQ$,તેથી આપણે લખી શકીએ કે $\frac{AB}{PQ} = \frac{1}{2}$.
ગુણોત્તરમાં કિંમતો મૂકતા: $\frac{1}{2} = \frac{10}{QR}$.
ચોકડી ગુણાકાર કરતા,$QR = 10 \times 2 = 20$.
આમ,$QR = 20$.

(B) આપેલ છે કે $\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ} = \frac{2}{5}$.
$SSS$ (બાજુ-બાજુ-બાજુ) સમરૂપતાની શરત મુજબ,$\Delta PQR \sim \Delta XYZ$ થાય.
બે સમરૂપ ત્રિકોણના ક્ષેત્રફળના ગુણોત્તરના પ્રમેય મુજબ,તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.
તેથી,$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{XY} \right)^2$.
આપેલ કિંમત મૂકતા: $\left( \frac{2}{5} \right)^2 = \frac{4}{25}$.
આમ,ગુણોત્તર $4:25$ છે.

(C) આપેલ છે કે $\Delta ABC \sim \Delta PQR$.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે.
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k$.
આપેલ છે કે $AB = 5, BC = 7, AC = 10$ અને $PR = 15$.
ગુણોત્તર $\frac{AC}{PR} = \frac{10}{15} = \frac{2}{3}$ નો ઉપયોગ કરતા.
તેથી,બાજુઓનો ગુણોત્તર $k = \frac{2}{3}$ છે.
હવે,$\Delta PQR$ ની બાજુઓ શોધો:
$PQ = AB \times \frac{3}{2} = 5 \times 1.5 = 7.5$.
$QR = BC \times \frac{3}{2} = 7 \times 1.5 = 10.5$.
$PR = 15$.
$\Delta PQR$ ની પરિમિતિ $= PQ + QR + PR = 7.5 + 10.5 + 15 = 33$.