Construction of a Similar Triangle, When One Vertex Is Common
Question 6 Dr...
Question
Question 6 Draw a triangle ABC with side BC=7cm,∠B=45∘,∠A=105∘. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ΔABC.
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Solution
Sum of all side of triangle =180∘ ∴∠A+∠B+∠C=180∘ ∠C=180∘−150∘=30∘ Steps of Construction: Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of 45∘ with BC. Step III: At C, a ray making an angle of 30∘ with BC is drawn intersecting the previous ray at A. Thus, ∠A=105∘. Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A. Step V 4 points B1,B2,B3andB4 are marked at equal distances on BX. Step VI: B3C is joined and B4C′ is made parallel to B3C. Step VII: C'A' is made parallel CA. Thus, A'BC' is the required triangle. Justification: ∠B=45∘ (Common) ∠C=∠C′ ΔAB′C′ ~ ΔABC by AA similarity condition. ∴BCBC′=ABA′B′=ACA′C′ also, BCBC′=BB3BB4=34 ⇒AB=43AB′,BC=43BC′andAC=43A′C′