Maharashtra Board Class 10 Maths Chapter 1 Similarity Set 1 Solutions

Question 1. Select the appropriate alternative.
(i) In ∆ABC and ∆PQR, in a one to one correspondence \( \frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ} \), then
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में एक त्रिभुज ABC दिखाया गया है। दूसरे चित्र में एक त्रिभुज PQR दिखाया गया है। ये दोनों त्रिभुज संगतता दर्शाने के लिए उपयोग किए गए हैं।
(A) ∆PQR – ∆ABC
(B) ∆PQR – ∆CAB
(C) ∆CBA – ∆PQR
(D) ∆BCA – ∆PQR
Answer: (B) ∆PQR – ∆CAB
In simple words: Given ratios of sides \( \frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ} \) imply that ∆ABC is similar to ∆QRP, which means it is also similar to ∆CAB. So, ∆PQR ~ ∆CAB is the correct correspondence.

🎯 Exam Tip: Pay close attention to the order of vertices when establishing similarity based on side ratios. Corresponding sides must be in the correct sequence.

Question 1.
(ii) If in ∆DEF and ∆PQR, ∠D = ∠Q, ∠R = ∠E, then which of the following statements is false?
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में त्रिभुज DEF दर्शाया गया है। दूसरे चित्र में त्रिभुज PQR दर्शाया गया है। दोनों त्रिभुज कोणों की समानता के आधार पर समरूपता को समझने में मदद करते हैं।
(A) \( \frac{EF}{PR} = \frac{DF}{PQ} \)
(B) \( \frac{DE}{PQ} = \frac{EF}{RP} \)
(C) \( \frac{DE}{QR} = \frac{DF}{PQ} \)
(D) \( \frac{EF}{DF} = \frac{RP}{QR} \)
Answer: (B) \( \frac{DE}{PQ} = \frac{EF}{RP} \)
∆DEF ~ ∆QRP ... [AA test of similarity]
\( \therefore \frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{PQ} \) ... [Corresponding sides of similar triangles]
In simple words: For the triangles to be similar, the corresponding angles must be equal, which leads to proportional sides. The given angle correspondence ∠D = ∠Q and ∠R = ∠E means ∆DEF ~ ∆QRP. We need to find the statement that does not follow this proportionality. Option (B) has incorrect side proportionality.

🎯 Exam Tip: When given angle conditions for similarity, correctly identify the corresponding vertices to set up the proportional side ratios. Any deviation in this correspondence will lead to a false statement.

Question 1.
(iii) In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE, then which of the statements regarding the two triangles is true?
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में त्रिभुज ABC दिखाया गया है। दूसरे चित्र में त्रिभुज DEF दर्शाया गया है। ये दोनों त्रिभुज कोणों की समानता और भुजाओं के अनुपात के आधार पर समरूपता और सर्वांगसमता को समझने में सहायक हैं।
(A) The triangles are not congruent and not similar.
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.
Answer: (B) The triangles are similar but not congruent.
In simple words: The given conditions ∠B = ∠E and ∠F = ∠C imply that the triangles are similar by AA test. However, since AB = 3DE, the sides are not equal, only proportional, meaning they cannot be congruent.

🎯 Exam Tip: Similarity is established by equal angles or proportional sides, while congruence requires both equal angles and equal sides. A ratio of 3:1 for sides confirms similarity but rules out congruence.

Question 1.
(iv) ∆ABC and ∆DEF are equilateral triangles, A(∆ABC) : A(∆DEF) = 1 : 2. If AB = 4, then what is length of DE?
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में एक छोटा समबाहु त्रिभुज ABC दिखाया गया है। दूसरे चित्र में एक बड़ा समबाहु त्रिभुज DEF दर्शाया गया है। यह चित्र त्रिभुजों के क्षेत्रफल और भुजाओं के बीच संबंध को दर्शाता है।
(A) \( 2\sqrt{2} \)
(B) 4
(C) 8
(D) \( 4\sqrt{2} \)
Answer: (D) \( 4\sqrt{2} \)
Refer Q. 6 Practice Set 1.4
In simple words: Since both are equilateral triangles, they are similar. The ratio of their areas is equal to the square of the ratio of their corresponding sides. Using this, we can find the length of DE.

🎯 Exam Tip: Remember the property that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. This is a fundamental concept for such problems.

Question 1.
(v) In the adjoining figure, seg XY || seg BC, then which of the following statements is true?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसके भीतर एक रेखाखंड XY है जो भुजा BC के समानांतर है। बिंदु X भुजा AB पर है और बिंदु Y भुजा AC पर है, जिससे दो समरूप त्रिभुज बनते हैं।
(A) \( \frac{AB}{AX} = \frac{AC}{AY} \)
(B) \( \frac{AX}{XB} = \frac{AY}{AC} \)
(C) \( \frac{AX}{YC} = \frac{AY}{XB} \)
(D) \( \frac{AB}{YC} = \frac{AC}{XB} \)
Answer: (A) \( \frac{AB}{AX} = \frac{AC}{AY} \)
∆ABC ~ ∆AXY ... [AA test of similarity]
\( \therefore \frac{AB}{AX} = \frac{AC}{AY} \) ...[Corresponding sides of similar triangles]
\( \therefore \frac{AB}{AC} = \frac{AX}{AY} \) ... [Alternendo]
In simple words: When a line segment is parallel to one side of a triangle and intersects the other two sides, it creates a smaller triangle that is similar to the original triangle. This similarity leads to proportional sides.

🎯 Exam Tip: The Basic Proportionality Theorem (BPT) and its converse are crucial here. When a line is parallel to a side, the ratios of the parts of the other two sides are equal, and the whole sides are proportional. Option (A) directly reflects this property.

Question 2. In ∆ABC, B-D-C and BD = 7, BC = 20, then find following ratios.
(i) \( \frac{A(\Delta ABD)}{A(\Delta ADC)} \)
(ii) \( \frac{A(\Delta ABD)}{A(\Delta ABC)} \)
(iii) \( \frac{A(\Delta ADC)}{A(\Delta ABC)} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें एक बिंदु D भुजा BC पर स्थित है। बिंदु A से भुजा BC पर एक लंब AE खींचा गया है। यह त्रिभुज के आधार और ऊँचाई का उपयोग करके क्षेत्रफल के अनुपात को समझाने में मदद करता है।
Solution:
Draw AE \( \perp \) BC, B – E – C.
BC = BD + DC [B – D – C]
\( \therefore \) 20 = 7 + DC
\( \therefore \) DC = 20 - 7 = 13
(i) ∆ABD and ∆ADC have same height AE.
\( \frac{A(\Delta ABD)}{A(\Delta ADC)} = \frac{BD}{DC} \) [Triangles having equal height]
\( \therefore \frac{A(\Delta ABD)}{A(\Delta ADC)} = \frac{7}{13} \)
(ii) ∆ABD and ∆ABC have same height AE.
\( \frac{A(\Delta ABD)}{A(\Delta ABC)} = \frac{BD}{BC} \) [Triangles having equal height]
\( \therefore \frac{A(\Delta ABD)}{A(\Delta ABC)} = \frac{7}{20} \)
(iii) ∆ADC and ∆ABC have same height AE.
\( \frac{A(\Delta ADC)}{A(\Delta ABC)} = \frac{DC}{BC} \) [Triangles having equal height]
\( \therefore \frac{A(\Delta ADC)}{A(\Delta ABC)} = \frac{13}{20} \)
In simple words: When triangles share the same height, the ratio of their areas is equal to the ratio of their bases. We first find the length of DC, then apply this property to find the ratios of the areas.

🎯 Exam Tip: Remember that the ratio of areas of triangles with equal heights is directly proportional to the ratio of their corresponding bases. Clearly label all segment lengths to avoid calculation errors.

Question 3. Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm, then what is the corresponding base of the bigger triangle?
Solution:
Let A\(_{1}\) and A\(_{2}\) be the areas of two triangles. Let b\(_{1}\) and b\(_{2}\) be their corresponding bases.
A\(_{1}\) : A\(_{2}\) = 2 : 3
\( \therefore \frac{A_1}{A_2} = \frac{2}{3} \)
We know that for triangles having equal height,
\( \frac{A_1}{A_2} = \frac{b_1}{b_2} \) [Triangles having equal height]
\( \therefore \frac{2}{3} = \frac{6}{b_2} \)
\( \therefore b_2 = \frac{6 \times 3}{2} \)
\( \therefore b_2 = 9 \) cm
\( \therefore \) The corresponding base of the bigger triangle is 9 cm.
In simple words: For triangles with the same height, the ratio of their areas is the same as the ratio of their bases. Using the given area ratio and the smaller base, we can calculate the larger base.

🎯 Exam Tip: This question tests a direct application of the theorem regarding areas of triangles with equal heights. Ensure you set up the proportion correctly between areas and bases.

Question 4. In the adjoining figure, ∠ABC = ∠DCB = 90°, AB = 6, DC = 8, then \( \frac{A(\Delta ABC)}{A(\Delta DCB)} = ? \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समकोण त्रिभुज ABC और DCB को दर्शाता है, जो एक उभयनिष्ठ आधार BC साझा करते हैं। बिंदु A और D शीर्ष हैं और कोण ABC और DCB 90 डिग्री हैं। यह त्रिभुजों के क्षेत्रफल और उनके आधार/ऊँचाई के संबंध को समझने में मदद करता है।
Solution:
∆ABC and ∆DCB have same base BC.
\( \therefore \frac{A(\Delta ABC)}{A(\Delta DCB)} = \frac{AB}{DC} \) [Triangles having equal base]
\( = \frac{6}{8} \)
\( \therefore \frac{A(\Delta ABC)}{A(\Delta DCB)} = \frac{3}{4} \)
In simple words: When two triangles share the same base, the ratio of their areas is equal to the ratio of their corresponding heights. In this case, AB and DC act as the heights.

🎯 Exam Tip: Recognize that if triangles share a common base, their area ratio simplifies to the ratio of their respective heights (or altitudes) to that base. Always identify the common base and the perpendicular heights accurately.

Question 5. In the adjoining figure, PM = 10 cm, A(∆PQS) = 100 sq. cm, A(∆QRS) = 110 sq. cm, then find NR.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो त्रिभुजों PQS और QRS को दर्शाता है, जो एक उभयनिष्ठ आधार QS साझा करते हैं। PM त्रिभुज PQS की ऊँचाई है (QS पर लंब) और NR त्रिभुज QRS की ऊँचाई है (QS पर लंब)।
Solution:
∆PQS and ∆QRS have same base QS.
\( \frac{A(\Delta PQS)}{A(\Delta QRS)} = \frac{PM}{NR} \) [Triangles having equal base]
\( \therefore \frac{100}{110} = \frac{10}{NR} \)
\( \therefore NR = \frac{110 \times 10}{100} \)
\( \therefore NR = 11 \) cm
In simple words: Since triangles PQS and QRS share the same base QS, the ratio of their areas is equal to the ratio of their corresponding heights (PM and NR). Using the given areas and PM, we can find NR.

🎯 Exam Tip: This problem is a direct application of the principle that for triangles with a common base, their areas are proportional to their heights. Ensure proper substitution into the ratio formula.

Question 6. ∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio \( \frac{A(\Delta MNT)}{A(\Delta QRS)} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में त्रिभुज MNT दर्शाया गया है, जिसमें शीर्ष T से आधार MN पर एक लंब की लंबाई 5 है। दूसरे चित्र में त्रिभुज QRS दर्शाया गया है, जिसमें शीर्ष S से आधार QR पर एक लंब की लंबाई 9 है। यह समरूप त्रिभुजों के क्षेत्रफलों और उनकी संगत ऊँचाइयों के अनुपात को दर्शाता है।
Solution:
∆MNT ~ ∆QRS [Given]
\( \therefore \) ∠M \( \approx \) ∠Q (i) [Corresponding angles of similar triangles]
In ∆MLT and ∆QPS,
∠M = ∠Q [From (i)]
∠MLT = ∠QPS [Each angle is of measure 90°]
\( \therefore \) ∆MLT ~ ∆QPS [AA test of similarity]
\( \therefore \frac{MT}{QS} = \frac{TL}{SP} \) [Corresponding sides of similar triangles]
\( \therefore \frac{MT}{QS} = \frac{5}{9} \)
Now, ∆MNT ~ ∆QRS (ii) [Given]
\( \frac{A(\Delta MNT)}{A(\Delta QRS)} = \frac{MT^2}{QS^2} \) [Theorem of areas of similar triangles]
\( = (\frac{MT}{QS})^2 \)
\( = (\frac{5}{9})^2 \) [From (ii)]
\( \therefore \frac{A(\Delta MNT)}{A(\Delta QRS)} = \frac{25}{81} \)
In simple words: For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding altitudes. Given the lengths of the altitudes, we simply square their ratio to find the ratio of the areas.

🎯 Exam Tip: The theorem stating that the ratio of areas of similar triangles is equal to the square of the ratio of their corresponding altitudes (or medians, or side lengths) is key here. Make sure to square the ratio correctly.

Question 7. In the adjoining figure, A – D – C and B – E – C. seg DE || side AB. If AD = 5, DC = 3, BC = 6.4, then find BE.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें एक रेखाखंड DE भुजा AB के समानांतर है। बिंदु D भुजा AC पर है और बिंदु E भुजा BC पर है। यह आरेख आधारभूत समानुपातिकता प्रमेय को समझने में मदद करता है।
Solution:
In ∆ABC,
seg DE || side AB [Given]
\( \therefore \frac{AD}{DC} = \frac{BE}{EC} \) [Basic proportionality theorem]
\( \therefore \frac{5}{3} = \frac{x}{6.4-x} \)
\( \therefore 3x = 5(6.4 - x) \)
\( \therefore 3x = 32 - 5x \)
\( \therefore 8x = 32 \)
\( \therefore x = \frac{32}{8} = 4 \)
\( \therefore BE = 4 \) units
In simple words: Since DE is parallel to AB, by the Basic Proportionality Theorem, the line divides the other two sides proportionally. We set up the ratio of the segments and solve for the unknown length BE.

🎯 Exam Tip: The Basic Proportionality Theorem (Thales Theorem) is fundamental. If a line is parallel to one side of a triangle and intersects the other two sides, it divides the two sides in the same ratio. Be careful with algebraic manipulation.

Question 8. In the adjoining figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280, then find PQ, QR and RS.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र चार समानांतर रेखाखंडों PA, QB, RC और SD को दर्शाता है, जो सभी एक तिर्यक रेखा AD पर लंबवत हैं। एक अन्य तिर्यक रेखा PS इन समानांतर रेखाखंडों को काटती है। यह तीन समानांतर रेखाओं और उनकी तिर्यक रेखाओं के गुणों को समझने में मदद करता है।
Solution:
seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. [Given]
\( \therefore \) seg PA || seg QB || seg RC || seg SD (i) [Lines perpendicular to the same line are parallel to each other]
Let the value of PQ be x and that of QR be y.
PS = PQ + QS [P – Q – S]
\( \therefore \) 280 = x + QS
\( \therefore \) QS = 280 – x (ii)
Now, seg PA || seg QB || seg SD [From (i)]
\( \therefore \frac{AB}{BD} = \frac{PQ}{QS} \) [Property of three parallel lines and their transversals]
\( \therefore \frac{AB}{BC+CD} = \frac{PQ}{QS} \) [B - C - D]
\( \therefore \frac{60}{70+80} = \frac{x}{280-x} \)
\( \therefore \frac{60}{150} = \frac{x}{280-x} \)
\( \therefore \frac{2}{5} = \frac{x}{280-x} \)
\( \therefore 5x = 2(280 - x) \)
\( \therefore 5x = 560 - 2x \)
\( \therefore 7x = 560 \)
\( \therefore x = \frac{560}{7} = 80 \)
\( \therefore PQ = 80 \) units
QS = 280 - x [From (ii)]
= 280 - 80
= 200 units
But, QS = QR + RS [Q – R – S]
\( \therefore \) 200 = y + RS
\( \therefore \) RS = 200 – y (iii)
Now, seg QB || seg RC || seg SD [From (i)]
\( \therefore \frac{BC}{CD} = \frac{QR}{RS} \) [Property of three parallel lines and their transversals]
\( \therefore \frac{70}{80} = \frac{y}{200-y} \)
\( \therefore \frac{7}{8} = \frac{y}{200-y} \)
\( \therefore 8y = 7(200 – y) \)
\( \therefore 8y = 1400 – 7y \)
\( \therefore 15y = 1400 \)
\( \therefore y = \frac{1400}{15} = \frac{280}{3} \)
\( \therefore QR = \frac{280}{3} \) units
RS = 200 - y [From (iii)]
\( = 200 - \frac{280}{3} \)
\( = \frac{200 \times 3 - 280}{3} \)
\( = \frac{600 - 280}{3} \)
\( \therefore RS = \frac{320}{3} \) units
In simple words: When parallel lines are intersected by transversals, they cut off proportional intercepts on the transversals. We use this property twice, first to find PQ, then to find QR and RS.

🎯 Exam Tip: This problem involves the "Property of three parallel lines and their transversals." Break down the problem into smaller segments, use algebra to solve for unknowns, and ensure consistent application of the theorem.

Question 9. In ∆PQR, seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR
Complete the proof by filling in the boxes.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज PQR को दर्शाता है, जिसमें PM एक माध्यिका है। रेखाखंड MX कोण PMQ का समद्विभाजक है और MY कोण PMR का समद्विभाजक है। बिंदु X भुजा PQ पर है और बिंदु Y भुजा PR पर है।
Solution:
Proof:
In ∆PMQ, ray MX is bisector of ∠PMQ.
\( \therefore \frac{MP}{MQ} = \frac{PX}{XQ} \) (i) [Theorem of angle bisector]
In ∆PMR, ray MY is bisector of ∠PMR.
\( \therefore \frac{MP}{MR} = \frac{PY}{YR} \) (ii) [Theorem of angle bisector]
But, \( \frac{MP}{MQ} = \frac{MP}{MR} \) [M is the midpoint of QR, hence MQ = MR]
\( \therefore \frac{PX}{XQ} = \frac{PY}{YR} \)
\( \therefore \) XY || QR [Converse of basic proportionality theorem]
In simple words: We apply the angle bisector theorem in triangles PMQ and PMR. Since PM is a median, MQ = MR, leading to equal ratios. Then, by the converse of the Basic Proportionality Theorem, XY is parallel to QR.

🎯 Exam Tip: This proof relies on two key theorems: the Angle Bisector Theorem and the Converse of the Basic Proportionality Theorem. Remember that a median implies equal base segments, which is critical for equating the ratios.

Question 10. In the adjoining figure, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6, then find \( \frac{AX}{XY} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें कोण B और कोण C के समद्विभाजक बिंदु X पर मिलते हैं। रेखा AX भुजा BC को बिंदु Y पर काटती है। यह आरेख त्रिभुज के कोण समद्विभाजक प्रमेय के अनुप्रयोग को समझने में मदद करता है।
Solution:
Let the value of BY be x.
BC = BY + YC [B – Y – C]
\( \therefore 6 = x + YC \)
\( \therefore YC = 6 - x \)
in ∆BAY, ray BX bisects ∠B. [Given]
\( \therefore \frac{AB}{BY} = \frac{AX}{XY} \) (i) [Property of angle bisector of a triangle]
Also, in ∆CAY, ray CX bisects ∠C. [Given]
\( \therefore \frac{AC}{YC} = \frac{AX}{XY} \) (ii) [Property of angle bisector of a triangle]
\( \therefore \frac{AB}{BY} = \frac{AC}{YC} \) [From (i) and (ii)]
\( \therefore \frac{5}{x} = \frac{4}{6-x} \)
\( \therefore 5(6-x) = 4x \)
\( \therefore 30-5x = 4x \)
\( \therefore 9x = 30 \)
\( \therefore x = \frac{30}{9} = \frac{10}{3} \)
Now, \( \frac{AX}{XY} = \frac{AB}{BY} \) [Substituting the value of x in equation (i)]
\( = \frac{5}{\frac{10}{3}} \)
\( = \frac{5 \times 3}{10} \)
\( \therefore \frac{AX}{XY} = \frac{3}{2} \)
In simple words: The angle bisector theorem states that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. By applying this theorem twice and solving for the unknown length, we can find the desired ratio.

🎯 Exam Tip: This problem involves applying the Angle Bisector Theorem twice. First, to find the ratio of the sides of the base, and then to find the ratio of AX to XY. Systematically label variables and solve algebraic equations.

Question 11. In ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that \( \frac{AP}{PD} = \frac{PC}{BP} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समलंब चतुर्भुज ABCD को दर्शाता है, जिसमें भुजा AD भुजा BC के समानांतर है। विकर्ण AC और BD बिंदु P पर एक दूसरे को काटते हैं।
Solution:
Proof:
seg AD || seg BC and BD is their transversal. [Given]
\( \therefore \) ∠DBC = ∠BDA [Alternate angles]
\( \therefore \) ∠PBC = ∠PDA (i) [D – P – B]
In ∆PBC and ∆PDA,
∠PBC = ∠PDA [From (i)]
∠BPC = ∠DPA [Vertically opposite angles]
\( \therefore \) ∆PBC ~ ∆PDA [AA test of similarity]
\( \therefore \frac{BP}{PD} = \frac{PC}{AP} \) [Corresponding sides of similar triangles]
\( \therefore \frac{AP}{PD} = \frac{PC}{BP} \) [By alternendo]
In simple words: Since AD is parallel to BC, alternate angles formed by the transversal BD are equal. This leads to the similarity of triangles PBC and PDA. From similar triangles, corresponding sides are proportional, which proves the required ratio.

🎯 Exam Tip: For problems involving parallel lines and transversals in quadrilaterals (like trapezoids), look for alternate interior angles or corresponding angles to establish triangle similarity. Vertically opposite angles are also a common condition.

Question 12. In the adjoining figure, XY || seg AC. If 2 AX = 3 BX and XY = 9, complete the activity to find the value of AC.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें एक रेखाखंड XY भुजा AC के समानांतर है। बिंदु X भुजा AB पर है और बिंदु Y भुजा BC पर है।
Solution:
2 AX = 3 BX [Given]
\( \therefore \frac{AX}{BX} = \frac{3}{2} \)
\( \therefore \frac{AX+BX}{BX} = \frac{3+2}{2} \) [By componendo]
\( \therefore \frac{AB}{BX} = \frac{5}{2} \) (i) [A-X-B]
In ∆BCA and ∆BYX,
∠BCA \( \approx \) ∠BYX }
∠BAC \( \approx \) ∠BXY } [Corresponding angles]
\( \therefore \) ∆BCA ~ ∆BYX [By AA test of similarity]
\( \therefore \frac{BA}{BX} = \frac{AC}{XY} \) [Corresponding sides of similar triangles]
\( \therefore \frac{5}{2} = \frac{AC}{9} \) [From (i)]
\( \therefore AC = \frac{9 \times 5}{2} \)
\( \therefore AC = 22.5 \) units
In simple words: Since XY is parallel to AC, triangles BYX and BCA are similar. This means their corresponding sides are proportional. We use the given ratio of AX to BX, along with componendo, to find the ratio of AB to BX. Then, we apply the similarity of triangles to find AC.

🎯 Exam Tip: Recognize that parallel lines create similar triangles. Utilize the property of similar triangles where the ratio of corresponding sides is equal. The componendo property helps in relating parts of a side to the whole side.

Question 13. In the adjoining figure, the vertices of square DEFG are on the sides of ∆ABC. If ∠A = 90°, then prove that DE\(^{2}\) = BD × EC.
(Hint: Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समकोण त्रिभुज ABC को दर्शाता है (∠A = 90°), जिसके भीतर एक वर्ग DEFG बनाया गया है। वर्ग के शीर्ष D और E भुजा BC पर हैं, G भुजा AB पर है, और F भुजा AC पर है।
Solution:
Proof:
JDEFG is a square.
\( \therefore \) DE = EF = GF = GD (i) [Sides of a square]
∠GDE = ∠DEF = 90° [Angles of a square]
\( \therefore \) seg GD \( \perp \) side BC, seg FE \( \perp \) side BC (ii)
In ∆BAC and ∆BDG,
∠BAC = ∠BDG [From (ii), each angle is of measure 90°]
∠ABC = ∠DBG [Common angle]
\( \therefore \) ∆BAC ~ ∆BDG (iii) [AA test of similarity]
In ∆BAC and ∆FEC,
∠BAC = ∠FEC [From (ii), each angle is measure 90°]
∠ACB = ∠ECF [Common angle]
\( \therefore \) ∆BAC ~ ∆FEC (iv) [AA test of similarity]
\( \therefore \) ∆BDG ~ ∆FEC [From (iii) and (iv)]
\( \therefore \frac{BD}{FE} = \frac{GD}{EC} \) (v) [Corresponding sides of similar triangles]
\( \therefore \frac{BD}{DE} = \frac{DE}{EC} \) [From (i) and (v)]
\( \therefore DE^2 = BD \times EC \)
In simple words: We establish similarity between ∆BDG and ∆BAC, and ∆FEC and ∆BAC using the AA test. This implies ∆BDG ~ ∆FEC. Using the property of similar triangles and that all sides of a square are equal (GD = FE = DE), we can derive the desired relationship.

🎯 Exam Tip: This problem involves multiple similarity conditions. First, show that the smaller triangles (BDG and FEC) are similar to the larger triangle (ABC). Then, use transitivity of similarity to relate ∆BDG and ∆FEC. Finally, substitute equal sides of the square to prove the relationship.