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