(i) \(\angle\)A = \(\angle\)P = 60°
\(\angle\)B = \(\angle\)Q = 80°
\(\angle\)C = \(\angle\)R = 40°
Therefore, ∆ABC ∼ ∆PQR [By AAA similarity criterion]
\(\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}\)
(ii) ΔABC∼ΔQRP [By SSS similarity criterion]
(iii) The given triangles are not similar as the corresponding sides are not proportional.
(iv) The given triangles are not similar as the corresponding sides are not proportional.
(v) The given triangles are not similar as the corresponding sides are not proportional.
(vi) In ∆DEF,
\(\angle\)D +\(\angle\)E +\(\angle\)F = 180º (Sum of the measures of the angles of a triangle is 180º.)
70º + 80º +\(\angle\)F = 180º \(\angle\)F = 30º Similarly, in ∆PQR, \(\angle\)P +\(\angle\)Q +\(\angle\)R = 180º (Sum of the measures of the angles of a triangle is 180º.)
\(\angle\)P + 80º +30º = 180º
\(\angle\)P = 70º
In ∆DEF and ∆PQR,
\(\angle\)D = \(\angle\)P (Each 70°)
\(\angle\)E = \(\angle\)Q (Each 80°)
\(\angle\)F = \(\angle\)R (Each 30°)
∴ ∆DEF ∼ ∆PQR [By AAA similarity criterion]
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आप अदिति / आदित्य हैं। आपकी दादीजी को खेलों में अत्यधिक रुचि है। ओलंपिक खेल-2024 में भारत के प्रदर्शन के बारे में जानकारी देते हुए लगभग 100 शब्दों में पत्र लिखिए।
There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.