Question

please hrlp me solve this question fast of maths. hope it will help on your studys too,so solve it and give answer fast i will surely mark it as brainliest to all who give answers
Q No. 2]B Q1 and Q2

Answer 1

◆◆◆◆◆step by step explanation◆◆◆◆◆

2.b.1)
Perimeter of semi circle(protracter)=108

perimeter of semicircle=1/2×2πr

perimeter=πr

【Putting value of perimeter】

108=22/7×r

108×7/22=r

34.36=r

Diameter= 2×radius

Diameter=2×34.36

Diameter=68.72cm

2.B.2)
In the triangle it is given that D is the mid-point of BC so,
BD=CD
If D is the mid-point of BC so we can carry it as median.i.e.

(ar)ABD=(ar)ADC

(ar)ABD=1/2of (ar)ABC

We know centroid divide median in ratio of 2 : 1  , As G is centroid so AG :  DG = 2 : 1

Given : D is a mid point of BC so AD is median and we know median divide triangle into  two equal area triangle , So

Area of ∆ ABD  =  Area of ∆ ACD = Area of ∆ ABC2                             --- ( 1 )

We know centroid divide median in ratio of 2 : 1  , As G is centroid so AG :  DG = 2 : 1

Then ,

Area of ∆ AGB = 2/3of Area of ∆ ABD , Substitute value from equation 1 we get

Area of ∆ AGB = 2/3of Area of ∆ ABC2

Area of ∆ AGB = Area of ∆ ABC3                                                   --- ( 2 )

And

Area of ∆ BGD = 13of Area of ∆ ABD , Substitute value from equation 1 we get

Area of ∆ BGD = 1/3of Area of ∆ ABC2

Area of ∆ BGD = Area of ∆ ABC/6                                      --- ( 3 )
 Area of ∆ BGDArea of ∆ ABD  , Substitute value from equation 1 and 3 and get

Area of ∆BGDArea of ∆ ABD = Area of ∆ ABC/6Area of ∆ ABC/2 = Area of ∆ ABC/6×2/Area of ∆ ABC = 1/3