Question

from a semicircle region OABCD a triangle ABD in which AB= 3 cm and BD= 4 cm is removed find the perimeter of the remaining figure

Answer 1

Please find the solved answer along with the diagram attached. 
We have : AB = 3 CM; BD = 4 CM
We know that, angle in a semi-circle is a right angle
So, ∠ABD = 90°  
So, ∆ABD is a right ∆
Now, AD² = BD²+AB² =AD² = 4² + 3² = 16+9 = 25
=AD = 5 cm So, diameter, AD = 5 cm
Now, circumference of semi-circle OABCD = Π x Radius + diameter = 3.14x2.5+5 = 12.85 cm
Now, when the triangle ABD is removed then the figure becomes (attached file)
So, circumference of the remaining figure = 7.85+AB+BD = 7.85+3+4 = 14.85 cm

c16ba073a0cf98faa11cca7ea55e0833.docx

Answer 2

Math

5 points

Please find the solved answer along with the diagram attached. 

We have :

AB = 3 CM; BD = 4 CM

We know that, angle in a semi-circle is a

right angle

So, ∠ABD

= 90°

 

So, ∆ABD is a right ∆

Now, AD² = BD²+AB²

=AD² = 4² + 3² = 16+9 = 25

=AD = 5 cm

So, diameter, AD = 5 cm

Now, circumference of semi-circle OABCD =

Π x Radius + diameter = 3.14x2.5+5 = 12.85 cm

Now, when the triangle ABD is removed then

the figure becomes (attached file)

So, circumference of the remaining figure =

7.85+AB+BD = 7.85+3+4 = 14.85 cm

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