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Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. |
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Answer» Solution :Steps of Construction : 1. DRAW a line segment BC = 4 cm. 2. From B. draw a line AB = 3 cm which making right angle at B. 3. Join `AC. DeltaABC` is the given right triangle. 4. From B, draw an acute `angleCBY` downwards. 5. On BY, take five POINTS `B_(1),B_(2),B_(3),B_(4)andB_(5)` such that`BB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)=B_(4)B_(5).` 6. Join `B_(3)C.` 7. From point `B_(5),` draw `B_(5)C'"||"B_(3)C` producijng BC to C'. 8. Form point C', draw `C'A"||"CA` PRODUCING BA to A' Hence, `DeltaA'BC'` is the required triangle. Justification : By construction, `B_(5)C'"||"B_(3)C` `therefore""(BC)/(C C')-3/2` `Now,""(BC')/(BC)=(BC+C C')/(BC)=1+(C C')/(BC)=1+2/3=5/3` `Also,""C'A'"||"CA` `therefore""DeltaABC~DeltaA'BC'` `and""(A'B)/(AB)=(BC')/(BC)=(A'C')/(CA)=5/3.` 
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