From a semicircular region 0ABCD, a triangle ABD in which AB = 3cm and BD = 4cm is removed. Find the perimeter of the remaining figure.
Answers
The perimeter of the remaining figure is 14.85 cm.
Step-by-step explanation:
Referring to the given figure, we have
OABCD is a semicircle
ABD is a triangle where AB = 3 cm and BD = 4 cm
We know that, angle subtended anywhere in circumference in a semi-circle is a right angle
∴ ∠ABD = 90°
⇒ ∆ABD is a right ∆
Now, applying Pythagoras theorem in ΔABD,
AD² = BD²+AB²
⇒ AD² = 4² + 3²
⇒ AD = √[16+9] = √25
⇒ AD = 5 cm ← diameter of the semicircle
∴ The radius of the semicircle, r = diameter/2 = 5/2 = 2.5 cm
Now,
The circumference of semi-circle OABCD is given by,
= [π x Radius] + diameter
= [3.14 x 2.5] + 5
= 12.85 cm
It is given that the Δ ABD is removed
Thus,
The circumference/perimeter of the remaining figure is given by,
= [12.85 - 5] + AB + BD
= 7.85 + 3 + 4
= 14.85 cm
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