Q28 of 47 Page 1

Construct a triangle ABC with sides BC = 7 cm, ∠ B = 45° and ∠ A =105°. Then construct a triangle whose sides are 3/4 times the corresponding sides of Δ ABC.

We know that the sum of all angles of a triangle = 180°

∴ ∠A + ∠B + ∠C = 180°


⇒ ∠C + 45° + 105° = 180°


∴ ∠C = 180° - 150° = 30°


Construction Steps:


Step 1: Draw BC = 7 cm.


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Step 2: At point B, draw a ray at ∠B = 45°.


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Step 3: At C, draw a ray making an angle of 30° with BC intersecting the previous ray at A.


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Step 4: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.


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Step 5: Locate 4 points B1, B2, B3 and B4 on BX so that BB1 = B1 B2 = B2 B3 = B3 B4.


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Step 6: Join B4C and draw a line through B3 || B4C to intersect BC at C′.


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Step 7: Draw a line through C′ parallel to the line CA to intersect BA at A.


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Thus, ΔA’BC’ is the required triangle.


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