Question

Construct a ΔABC in which AB+AC=5.6 cm, BC=4.5 cm, and ∠B = 45°.

Answer 1

Steps of construction:

Step 1: Draw a line segment BC of 4.5  cm 

Step 2: At B, draw an angle XBC of 45°

Step 3: With center B and radius 5.6 , cm draw an arc which intersects BX at D.

Step 4: Join DC

Step 5: Draw the perpendicular bisector of DC which intersects BD at A.

Step 6: Join AC

Therefore Δ ABC is a required triangle.

Answer 2

ΔABC is Constructed with AB+AC=5.6 cm, BC=4.5 cm, and ∠B = 45°.

Step-by-step explanation:

Step 1 :  Draw  a line segment BC = 4.5 cm

Step 2 : Using Protector Draw a angle of 45° at B  on BC  

Step 3 : Take a width of 5.6 cm from B on angle drawn in step 2 and label it D

Step 4 : join CD

Step 5 : sing compass taking width slightly less than CD and taking C as center , draw arcs on both sides  of CD

Step 6 : keeping compass width same taking D as center draw arcs such that it intersects arcs drawn in step 5 at X & Y

Step 7 : join XY and draw a line intersecting BD at A

Step 8 : join AC

ΔABC is required triangle with AB+AC=5.6 cm, BC=4.5 cm, and ∠B = 45°.

( explanation : AD = AC ) as A is on perpendicular bisector og CD

AB + AD = 5.6  => AB + AC = 5.6 cm

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