Get the most accurate GSEB Solutions for Class 7 Mathematics Chapter 07 ત્રિકોણની એકરૂપતા here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 7 Mathematics. Our expert-created answers for Class 7 Mathematics are available for free download in PDF format.
Detailed Chapter 07 ત્રિકોણની એકરૂપતા GSEB Solutions for Class 7 Mathematics
For Class 7 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 7 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 07 ત્રિકોણની એકરૂપતા solutions will improve your exam performance.
Class 7 Mathematics Chapter 07 ત્રિકોણની એકરૂપતા GSEB Solutions PDF
Question 1. નીચેનામાં એકરૂપતાની કઈ શરતનો ઉપયોગ કરશો?
(a) પક્ષ: AC = DF
AB = DE
BC = EF
આથી, \( \triangle ABC \cong \triangle DEF \)
Answer: એકરૂપતાની બાબાબા શરતને આધારે
In simple words: This congruence is based on the SSS (Side-Side-Side) condition, where all three sides of one triangle are equal to the three corresponding sides of another triangle.
Exam Tip: Remember that SSS congruence means if three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.
Question 1.
(b) પક્ષ: ZX = RP
RQ = ZY
\( \angle PRQ = \angle XZY \)
આથી, \( \triangle PQR \cong \triangle XYZ \)
Answer: એકરૂપતાની બાખૂબા શરતને આધારે
In simple words: This congruence is based on the SAS (Side-Angle-Side) condition, which means if two sides and the included angle of one triangle are equal to the two corresponding sides and the included angle of another, the triangles are congruent.
Exam Tip: For SAS congruence, it's crucial that the angle is *between* the two sides being compared.
Question 1.
(c) પક્ષ: \( \angle MLN = \angle FGH \)
\( \angle NML = \angle GFH \)
ML = FG
આથી, \( \triangle LMN \cong \triangle GFH \)
Answer: એકરૂપતાની ખૂબાબૂ શરતને આધારે
In simple words: This congruence is based on the ASA (Angle-Side-Angle) condition, which states that if two angles and the included side of one triangle are equal to the two corresponding angles and the included side of another triangle, they are congruent.
Exam Tip: For ASA congruence, the side must be the one *between* the two angles, also known as the included side.
Question 1.
(d) પક્ષ: EB = DB
AE = BC
\( \angle A = \angle C = 90^\circ \)
આથી, \( \triangle ABE \cong \triangle CDB \)
Answer: એકરૂપતાની કાકબા શરતને આધારે
In simple words: This congruence is based on the RHS (Right angle-Hypotenuse-Side) condition, which applies to right-angled triangles. If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one side of another right-angled triangle, they are congruent.
Exam Tip: RHS congruence is only applicable for right-angled triangles. Ensure both triangles have a 90-degree angle, and the hypotenuse is one of the equal sides.
Question 2. તમારે સાબિત કરવું છે કે \( \triangle ART \cong \triangle PEN \)
(a) જો તમારે બાબાબા શરતનો ઉપયોગ કરવો હોય, તો તમારે
(i) AR = ... (ii) RT = ... (iii) AT = ... બતાવવું પડે.
(b) જો \( \angle T = \angle N \) આપેલ હોય અને બાખૂબા શરતનો ઉપયોગ કરવો હોય, તો
(i) RT =..... અને (ii) PN = .... હોવું જોઈએ.
(c) જો AT = PN આપેલ હોય અને તમારે ખૂબાબૂ શરતનો ઉપયોગ કરવો હોય, તો કયાં બે પરિણામો હોવાં જોઈએ?
(i) ? (ii) ?
Answer:
અહીં \( \triangle ART \cong \triangle PEN \) સાબિત કરવું છે.
(a) જો SSS શરતનો ઉપયોગ કરવો હોય, તો તમારે
(i) AR = PE (ii) RT = EN (iii) AT = PN બતાવવું પડે.
(b) જો \( \angle T = \angle N \) છે તથા SAS શરત સંતોષવી છે, તો –
(i) RT = EN અને (ii) PN = AT હોવું જોઈએ.
(c) જો AT = PN છે તથા ASA શરત સંતોષવી છે, તો –
(i) \( \angle RAT = \angle EPN \) (ii) \( \angle ATR = \angle PNE \) બે પરિણામો હોવાં જોઈએ.
In simple words: To prove triangle congruence, we need specific pairs of equal parts based on the chosen rule. For SSS, all three corresponding sides must be equal. For SAS, two sides and the included angle. For ASA, two angles and the included side.
Exam Tip: Understand the specific conditions for each congruence rule (SSS, SAS, ASA, RHS) to correctly identify the necessary equal parts.
Question 3. તમારે \( \triangle APY \cong \triangle AMQ \) સાબિત કરવાનું છે. નીચેની સાબિતીમાં ખૂટતાં કારણો આપો :
| પગલું | કારણ |
|---|---|
| (i) PM = QM | (i) ............ |
| (ii) \( \angle PMA = \angle QMA \) | (ii) ............ |
| (iii) AM = AM | (iii) ............ |
| (iv) \( \triangle AMP \cong \triangle AMQ \) | (iv) ............ |
Answer:
| પગલું | કારણ |
|---|---|
| (i) PM = QM | (i) આપેલ છે. |
| (ii) \( \angle PMA = \angle QMA \) | (ii) આપેલ છે. |
| (iii) AM = AM | (iii) સામાન્ય બાજુ |
| (iv) \( \triangle AMP \cong \triangle AMQ \) | (iv) બાખૂબા શરતને આધારે |
In simple words: We are completing a proof of congruence. The first two points are given information, the third point is a common side shared by both triangles, and the final conclusion uses the SAS (Side-Angle-Side) congruence rule.
Exam Tip: When filling in reasons for a proof, always refer back to the basic definitions and postulates (e.g., "Given," "Common side," "Vertically opposite angles," or specific congruence rules).
Question 4. \( \triangle ABC \) માં \( \angle A = 30^\circ, \angle B = 40^\circ \) અને \( \angle C = 110^\circ \)
\( \triangle PQR \) માં \( \angle P = 30^\circ, \angle Q = 40^\circ \) અને \( \angle R = 110^\circ \)
એક વિદ્યાર્થી કહે છે કે ખૂખૂખૂ શરત પ્રમાણે \( \triangle ABC \cong \triangle PQR \) છે.
શું એ સાચો છે? શા માટે? શા માટે નહીં?
Answer: ના, વિદ્યાર્થીનો જવાબ ખૂખૂબૂ શરત નીચે \( \triangle ABC \cong \triangle PQR \) છે એ ખોટો છે. કારણ: ખૂખૂબૂ એ એકરૂપતાની શરત નથી.
In simple words: No, the student's statement that the triangles are congruent by the AAA (Angle-Angle-Angle) condition is incorrect. The AAA rule does not prove congruence; it only shows that the triangles are similar, meaning they have the same shape but possibly different sizes.
Exam Tip: Remember that AAA (Angle-Angle-Angle) is a similarity criterion, not a congruence criterion. For congruence, at least one corresponding side must be equal.
Question 5. નીચે આપેલ આકૃતિમાં બે ત્રિકોણો એકરૂપ છે. અનુરૂપ અંગો નિશાનીથી દર્શાવેલાં છે. \( \triangle RAT = ... \) શું લખી શકાય?
Answer: \( \triangle RAT \) અને \( \triangle WON \) માં
RA = WO (આકૃતિમાં આપ્યું છે.)
\( \angle RAT = \angle WON \) (આકૃતિમાં આપ્યું છે.)
AT = ON (આકૃતિમાં આપ્યું છે.)
\( \implies \) સંગતતા RAT \( \rightarrow \) WON માટે બાખૂબા શરત સંતોષાય છે.
અહીં, R \( \rightarrow \) W, A \( \rightarrow \) O અને T \( \rightarrow \) N
\( \implies \triangle RAT \cong \triangle WON \)
આમ, \( \triangle RAT \cong \triangle WON \) લખી શકાય.
In simple words: Based on the markings in the figure, two sides (RA and AT) and the included angle (A) of \( \triangle RAT \) are congruent to the corresponding two sides (WO and ON) and the included angle (O) of \( \triangle WON \). This means the triangles are congruent by the SAS (Side-Angle-Side) rule.
Exam Tip: When determining congruence from a diagram, carefully observe the markings for equal sides (single, double, triple dashes) and equal angles (single, double arcs) to apply the correct congruence criterion.
Question 6. એકરૂપતાનું વિધાન પૂર્ણ કરોઃ \( \triangle QRS = ? \)
Answer: \( \triangle QRS \) અને \( \triangle TPQ \) માં
\( \angle Q = \angle T \), PT = QR અને ZT = \( \angle Q \) (આકૃતિમાં દર્શાવેલ છે.)
\( \implies \) સંગતતા QRS \( \rightarrow \) TPQ માટે ખૂબાબૂ શરત સંતોષાય છે.
અહીં, Q \( \rightarrow \) T, R \( \rightarrow \) P અને S \( \rightarrow \) Q
\( \implies \triangle QRS \cong \triangle TPQ \)
In simple words: By matching the corresponding angles and sides shown in the diagram, we can conclude that triangle QRS is congruent to triangle TPQ. This congruence is based on the ASA (Angle-Side-Angle) rule.
Exam Tip: When completing a congruence statement, ensure the order of vertices reflects the corresponding equal parts (e.g., if angle A = angle D, then A must correspond to D in the statement).
Question 7. ચોરસ ખાનાંવાળા કાગળ પર સમાન ક્ષેત્રફળવાળા બે એવા ત્રિકોણો
(i) જે ત્રિકોણો એકરૂપ હોય.
(ii) જે ત્રિકોણો એકરૂપ નથી.
તેમની પરિમિતિ વિશે શું કહી શકાય?
Answer:
(નોંધ: ત્રિકોણનું ક્ષેત્રફળ = \( \frac{1}{2} \times \) પાયો \( \times \) વેધ)
(i) અહીં \( \triangle ABC \) અને \( \triangle PCD \) એકરૂપ ત્રિકોણો છે તથા તેમનાં ક્ષેત્રફળ સરખાં છે.
\( \triangle ABC \) નું ક્ષેત્રફળ = \( \frac{1}{2} \times 4 \times 3 = 6 \) ચો સેમી
\( \triangle PCD \) નું ક્ષેત્રફળ = \( \frac{1}{2} \times 4 \times 3 = 6 \) ચો સેમી
આમ, \( \triangle ABC \) અને \( \triangle PCD \) નાં ક્ષેત્રફળો સરખાં છે.
હવે, \( \triangle ABC \) ની પરિમિતિ = (3 + 4 + 5) સેમી = 12 સેમી
\( \triangle PCD \) ની પરિમિતિ = (3 + 4 + 5) સેમી = 12 સેમી
\( \implies \triangle ABC \) ના પારમિતિ = \( \triangle PCD \) ની પરિમિતિ
(ii) અહીં \( \triangle XYZ \) અને \( \triangle MYZ \) એકરૂપ ત્રિકોણો નથી, પણ તેમનાં ક્ષેત્રફળ સરખાં છે.
\( \triangle XYZ \) નું ક્ષેત્રફળ = \( \frac{1}{2} \times 4 \times 3 = 6 \) ચો સેમી
\( \triangle MYZ \) નું ક્ષેત્રફળ = \( \frac{1}{2} \times 4 \times 3 = 6 \) ચો સેમી
આમ, \( \triangle XYZ \) અને \( \triangle MYZ \) નાં ક્ષેત્રફળો સરખાં છે.
હવે, આકૃતિમાં જોતાં MN = XY છે. \( \implies \) MN = XY = 3
\( \triangle XYZ \) ની પરિમિતિ = (3 + 4 + 5) સેમી = 12 સેમી
\( \triangle MYZ \) ની પરિમિતિ = (4 + 3.5 + 4) સેમી = 11.5 સેમી
અહીં બંને ત્રિકોણો એકરૂપ નથી.
\( \implies \triangle XYZ \) ની પરિમિતિ \( \neq \triangle MYZ \) ની પરિમિતિ
In simple words: When two triangles are congruent, their areas and perimeters are always equal. However, if two triangles only have the same area, they might not be congruent, and therefore, their perimeters could be different.
Exam Tip: Congruent figures always have equal areas and equal perimeters. Figures with equal areas do not necessarily have equal perimeters or be congruent.
Question 8. બે ત્રિકોણોની એવી કાચી આકૃતિ દોરો કે જેમાં એકરૂપ ભાગની પાંચ જોડીઓ હોય છતાં ત્રિકોણો એકરૂપ ન હોય.
Answer: પ્રશ્ન ખોટો છે. એકરૂપ ભાગ પાંચ હોય છતાં ત્રિકોણો એકરૂપ ન હોય તેવું ન બને. એટલે આવી આકૃતિ દોરી ન શકાય.
In simple words: This question is incorrect. If two triangles have five pairs of congruent corresponding parts, they must be congruent. It is not possible to draw such a figure where they have five equal parts but are not congruent.
Exam Tip: Remember the minimum conditions for congruence (SSS, SAS, ASA, RHS). If more than these minimum parts are congruent, the triangles will definitely be congruent.
Question 9. \( \triangle ABC \) અને \( \triangle PQR \) એરૂપ બને તે માટે અનુરૂપ અંગોની વધુ એક જોડી આપો. તમે કઈ શરતનો ઉપયોગ કર્યો?
Answer: અહીં \( \triangle ABC \) અને \( \triangle PQR \) એકરૂપ બતાવવા છે.
આકૃતિમાં \( \angle B = \angle Q \) અને \( \angle C = \angle R \) આપેલ છે.
\( \triangle ABC \) અને \( \triangle PQR \) ને એકરૂપતા માટે BC = QR જોડી જરૂરી છે.
આમ, ખૂબા શરત પ્રમાણે \( \triangle ABC \cong \triangle PQR \)
અહીં, ખૂબાબૂ શરતનો ઉપયોગ થાય છે.
In simple words: To make \( \triangle ABC \) and \( \triangle PQR \) congruent, given that \( \angle B = \angle Q \) and \( \angle C = \angle R \), we need the included side BC to be equal to QR. This would allow us to use the ASA (Angle-Side-Angle) congruence criterion.
Exam Tip: When given two angles, the missing part needed for ASA congruence is always the side *between* those two angles.
Question 10. સમજાવોઃ \( \triangle ABC = \triangle FED \) શા માટે છે?
Answer: \( \triangle ABC \) અને \( \triangle FED \) માં
\( \angle B = \angle E = 90^\circ \)
\( \angle A = \angle F \) (આપેલ છે.)
\( \angle C = \angle D \) (બંને ત્રિકોણોના ત્રીજા ખૂણા સરખા માપના હોય.)
BC = ED (આપેલ છે.)
અહીં સંગતતા ABC \( \rightarrow \) FED માટે ખૂબાબૂ શરત સંતોષાય છે.
અહીં, A \( \rightarrow \) F, B \( \rightarrow \) E અને C \( \rightarrow \) D છે.
\( \implies \triangle ABC \cong \triangle FED \)
In simple words: The two triangles \( \triangle ABC \) and \( \triangle FED \) are congruent by the ASA (Angle-Side-Angle) condition. We have a right angle at B and E, angle A equals angle F, and the side BC is equal to ED. Since the third angles (C and D) must also be equal, this fulfills the ASA requirement.
Exam Tip: If two angles of a triangle are equal to two angles of another triangle, their third angles must also be equal. This is helpful when using ASA or AAS congruence criteria.
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GSEB Solutions Class 7 Mathematics Chapter 07 ત્રિકોણની એકરૂપતા
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The complete and updated GSEB Class 7 Maths Solutions Chapter 7 ત્રિકોણની એકરૂપતા Exercise 7.2 is available for free on StudiesToday.com. These solutions for Class 7 Mathematics are as per latest GSEB curriculum.
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