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Question
ΔABC ~ ΔPBR, BC = 8 cm, AC = 10 cm , ∠B = 90°, `"BC"/"BR" = 5/4` then construct ∆ABC and ΔPBR
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Solution
Analysis: In ∆ABC, ∠B = 90° ......[Given]
∴ AC2 = AB2 + BC2 ......[Pythagoras theorem]
∴ 102 = AB2 + 82
∴ AB2 = 100 – 64
∴ AB2 = 36
∴ AB = 6 cm ......[Taking square root of both sides]
Steps of construction:
- Draw seg BC of length 8 cm.
- Take ∠B as 90° and draw an arc of 6 cm on it. Name the point as A.
- Join seg AC to obtain ∆ABC.
- Draw ray BX such that ∠CBX is an acute angle.
- Locate points B1, B2, B3, B4, B5 on ray BX such that, BB1 = B1B2 = B2B3 = B3B4 = B4B5.
- Join point C and B5.
- Through point B4 draw a line parallel to seg CB5 which intersects seg BC at point R.
- Draw a line parallel to AC through R to intersect line AB at point P.
∆PBR is the required triangle similar to ∆ABC.
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