Maharashtra Board Class 10 Maths Chapter 1 Similarity Set 1.4 Solutions

Question 1. The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.
Answer:
Solution:
Let the corresponding sides of similar triangles be \(S_1\) and \(S_2\).
Let \(A_1\) and \(A_2\) be their corresponding areas.
\( \therefore S_1 : S_2 = 3 : 5 \)
\( \therefore \frac{S_1}{S_2} = \frac{3}{5} \)
\( \frac{A_1}{A_2} = \frac{S_1^2}{S_2^2} \) [Theorem of areas of similar triangles]
\( = \left(\frac{S_1}{S_2}\right)^2 \)
\( = \left(\frac{3}{5}\right)^2 \) [From (i)]
\( = \frac{9}{25} \)
\( \therefore \) Ratio of areas of similar triangles = \(9 : 25\)
In simple words: The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides. Given the side ratio 3:5, squaring this gives the area ratio as 9:25.

🎯 Exam Tip: Remember the fundamental theorem stating that the ratio of areas of similar triangles is the square of the ratio of their corresponding sides. This is a crucial concept for similar triangle problems.

Question 2. If \( \triangle ABC \sim \triangle PQR \) and AB : PQ = 2:3, then fill in the blanks.
Answer:
Solution:
\( \frac{A(\triangle ABC)}{A(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{2^2}{3^2} = \frac{4}{9} \) [Theorem of areas of similar triangles]
In simple words: For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. Given the side ratio of 2:3, the area ratio becomes \(2^2 : 3^2\), which is 4:9.

🎯 Exam Tip: Understanding how to apply the area ratio theorem directly to fill-in-the-blanks questions quickly shows mastery of the concept. Ensure to square both numerator and denominator.

Question 3. If \( \triangle ABC \sim \triangle PQR \), \( A(\triangle ABC) = 80 \), \( A(\triangle PQR) = 125 \), then fill in the blanks.
Answer:
Solution:
\( \frac{A(\triangle ABC)}{A(\triangle PQR)} = \frac{80}{125} = \frac{16}{25} \) (i) [Given]
\( \therefore \frac{A(\triangle ABC)}{A(\triangle PQR)} = \frac{AB^2}{PQ^2} \) (ii) [Theorem of areas of similar triangles]
\( \therefore \frac{AB^2}{PQ^2} = \frac{16}{25} \) [From (i) and (ii)]
\( \therefore \frac{AB}{PQ} = \frac{4}{5} \) [Taking square root of both sides]
In simple words: Using the given areas and the theorem relating area ratios to the square of side ratios, we first simplify the area ratio (80/125 to 16/25). Then, taking the square root of this ratio gives the corresponding side ratio of 4:5.

🎯 Exam Tip: When given areas and asked for side ratios, remember to simplify the area ratio first before taking the square root to find the side ratio. Show each step clearly.

Question 4. \( \triangle LMN \sim \triangle PQR \), \( 9 \times A(\triangle PQR) = 16 \times A(\triangle LMN) \). If QR = 20, then find MN.
Answer:
Solution:
\( 9 \times A(\triangle PQR) = 16 \times A(\triangle LMN) \) [Given]
\( \therefore \frac{A(\triangle LMN)}{A(\triangle PQR)} = \frac{9}{16} \) (i)
Now, \( \triangle LMN \sim \triangle PQR \) [Given]
\( \therefore \frac{A(\triangle LMN)}{A(\triangle PQR)} = \frac{MN^2}{QR^2} \) (ii) [Theorem of areas of similar triangles]
\( \therefore \frac{MN^2}{QR^2} = \frac{9}{16} \) [From (i) and (ii)]
\( \therefore \frac{MN}{QR} = \frac{3}{4} \) [Taking square root of both sides]
\( \therefore \frac{MN}{20} = \frac{3}{4} \)
\( \therefore MN = \frac{20 \times 3}{4} \)
\( \therefore MN = 15 \) units
In simple words: First, rearrange the given area equation to find the ratio of areas of the two similar triangles, which is 9:16. Then, use the theorem that the ratio of areas is the square of the ratio of corresponding sides. Taking the square root gives the side ratio as 3:4. Finally, substitute the given side length QR to find MN.

🎯 Exam Tip: Pay close attention to setting up the ratio of areas correctly from the given equation. Misplacing numerator and denominator can lead to incorrect results. Clearly show the steps for taking the square root.

Question 5. Areas of two similar triangles are 225 sq. cm. and 81 sq. cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
Answer:
Solution:
Let the areas of two similar triangles be \(A_1\) and \(A_2\).
\(A_1 = 225\) sq. cm. \(A_2 = 81\) sq. cm.
Let the corresponding sides of triangles be \(S_1\) and \(S_2\) respectively.
\(S_2 = 12\) cm (Side of the smaller triangle)
\( \frac{A_1}{A_2} = \frac{S_1^2}{S_2^2} \) [Theorem of areas of similar triangles]
\( \therefore \frac{225}{81} = \frac{S_1^2}{12^2} \)
\( \therefore S_1^2 = \frac{225 \times 12^2}{81} \)
\( \therefore S_1 = \sqrt{\frac{225 \times 144}{81}} \)
\( \therefore S_1 = \frac{15 \times 12}{9} \)
\( \therefore S_1 = 20 \) cm
\( \therefore \) The length of the corresponding side of the bigger triangle is 20 cm.
In simple words: We apply the theorem that the ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides. By setting up the proportion with the given areas and the side of the smaller triangle, we can solve for the unknown side of the bigger triangle.

🎯 Exam Tip: Identify which area belongs to which side. If \(A_1\) is bigger, its corresponding side \(S_1\) will also be bigger. Be careful with calculations involving squares and square roots.

Question 6. \( \triangle ABC \) and \( \triangle DEF \) are equilateral triangles. If \( A(\triangle ABC) : A(\triangle DEF) = 1:2 \) and AB = 4, find DE.
Answer:
Solution:
In \( \triangle ABC \) and \( \triangle DEF \),
\( \angle A = \angle D \)
\( \angle B = \angle E \)
\( \} \) [Each angle is of measure \( 60^\circ \)]
\( \therefore \triangle ABC \sim \triangle DEF \) [AA test of similarity]
\( \therefore \frac{A(\triangle ABC)}{A(\triangle DEF)} = \frac{AB^2}{DE^2} \) [Theorem of areas of similar triangles]
\( \therefore \frac{1}{2} = \frac{4^2}{DE^2} \)
\( \therefore DE^2 = 4^2 \times 2 \)
\( \therefore DE = \sqrt{16 \times 2} \)
\( \therefore DE = 4\sqrt{2} \) units
In simple words: Since both triangles are equilateral, they are similar by the AA test. The ratio of their areas is given as 1:2. Using the theorem that the ratio of areas is the square of the ratio of corresponding sides, we set up an equation with AB=4 to find DE.

🎯 Exam Tip: Recognize that all equilateral triangles are similar. This simplifies the first step. Be precise with algebraic manipulations, especially when dealing with square roots.

Question 7. In the adjoining figure, seg PQ || seg DE, \( A(\triangle PQF) = 20 \) sq. units, PF = 2 DP, then find A (J DPQE) by completing the following activity.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिकोणीय आकृति है जिसमें एक बड़ा त्रिभुज FDE है और उसके भीतर एक छोटा त्रिभुज FPQ है। PQ भुजा DE भुजा के समानांतर है, और F बिंदु दोनों त्रिभुजों का एक साझा शीर्ष है। बिंदु P, FD पर है और बिंदु Q, FE पर है। यह आरेख समरूप त्रिभुजों और उनके क्षेत्रों के बीच संबंध को समझने में मदद करता है।
Solution:
\( A(\triangle PQF) = 20 \) sq.units, PF = 2 DP, [Given]
Let us assume DP = x.
\( \therefore \) PF = 2x
DF = DP + PF
\( \therefore DF = x + 2x = 3x \)
In \( \triangle FDE \) and \( \triangle FPQ \),
\( \angle FDE = \angle FPQ \) [Corresponding angles]
\( \angle FED = \angle FQP \) [Corresponding angles]
\( \therefore \triangle FDE \sim \triangle FPQ \) [AA test of similarity]
\( \therefore \frac{A(\triangle FDE)}{A(\triangle FPQ)} = \frac{DF^2}{PF^2} \) [Theorem of areas of similar triangles]
\( \therefore \frac{A(\triangle FDE)}{A(\triangle FPQ)} = \frac{(3x)^2}{(2x)^2} = \frac{9x^2}{4x^2} = \frac{9}{4} \)
\( \therefore A(\triangle FDE) = \frac{9}{4} \times A(\triangle FPQ) \)
\( = \frac{9}{4} \times 20 = 9 \times 5 = 45 \) sq. units
\( A(DPQE) = A(\triangle FDE) - A(\triangle FPQ) \)
\( = 45 - 20 \)
\( = 25 \) sq. units
In simple words: First, establish the relationship between DF, DP, and PF (DF = 3x if DP = x and PF = 2x). Then, prove that triangles FDE and FPQ are similar using the AA test (due to parallel lines and corresponding angles). Apply the theorem that the ratio of their areas is the square of the ratio of corresponding sides to find A(ΔFDE). Finally, subtract A(ΔFPQ) from A(ΔFDE) to find the area of the trapezium DPQE.

🎯 Exam Tip: When dealing with figures involving parallel lines, always look for corresponding angles to prove similarity. Be meticulous in setting up ratios for sides and areas, especially when one side is a multiple of another segment.