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State and proof area theorem

Answer» Theorems on the area of similar trianglesTheorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.To prove this theorem, consider two similar triangles ΔABC and ΔPQR;\xa0As, Area of triangle =\xa01/2\xa0× Base × HeightTo find the area of ΔABC and ΔPQR, draw the altitudes AD and PE from the vertex A and P of ΔABC and ΔPQR as shown in the figure given below:Now, area of ΔABC =\xa01/2\xa0× BC × ADarea of ΔPQR =\xa01/2\xa0× QR × PEThe ratio of the areas of both the triangles can now be given as:area\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0= (1/2×BC×AD) / (1/2×QR×PE)⇒\xa0area\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0=\xa0BC\xa0×\xa0AD / QR\xa0×\xa0PE\xa0……………. (1)Now in ∆ABD and ∆PQE, it can be seen that:∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)∠ADB = ∠PEQ (Since both the angles are 90°)From AA criterion of similarity ∆ADB ~ ∆PEQ⇒\xa0AD/PE\xa0=\xa0AB/PQ …………….(2)Since it is known that ΔABC~ ΔPQR,AB/PQ\xa0=\xa0BC/QR\xa0=\xa0AC/PR\xa0…………….(3)Substituting this value in equation (1), we getarea\xa0of\xa0ΔABC/area\xa0of\xa0ΔPQR\xa0=\xa0AB/PQ\xa0×\xa0AD/PEUsing equation (2), we can writearea\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0=\xa0ABPQ\xa0×\xa0ADPE⇒area\xa0of\xa0ΔABCarea\xa0of\xa0ΔPQR\xa0=(AB/PQ)2Also from equation (3),area\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0=\xa0(AB/PQ)2\xa0=(BC/QR)2\xa0=\xa0(CA/RP)2This proves the theorem\xa0Note I could not post triangles pl draw yourself


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