Question

Can someone pls help ​

Answer 1

Answer:

square so all sides are equal

bc^2+mc^2=bm^2

16x^2+4x^2=20x^2=2root5x

nd^2+dm^2=nm^2

x^2+4x^2=5x2=root5x

then:-

1/2*b*h=1/2*2root5x*root5x

=1/2*10x^2=5x^2

Answer 2

Answer:  5x^2

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Work Shown:

ND = x

AN = 3x

MD = 2x

MC = 2x

BC = 4x

AB = 4x

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P = Area of square ABCD

P = (AB)^2

P = (4x)^2

P = 16x^2

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Q = Area of triangle ABN

Q = (1/2)*base*height

Q = (1/2)*AN*AB

Q = (1/2)*3x*4x

Q = 6x^2

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R = Area of triangle MBC

R = (1/2)*base*height

R = (1/2)*BC*MC

R = (1/2)*4x*2x

R = 4x^2

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S = Area of triangle MND

S = (1/2)*base*height

S = (1/2)*ND*MD

S = (1/2)*x*2x

S = x^2

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T = Area of triangle BMN

T = P - Q - R - S

T = 16x^2 - 6x^2 - 4x^2 - x^2

T = 5x^2

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An alternative method is to use the pythagorean theorem to find the lengths of BM and MN. Then you can directly compute the area of triangle BMN. You should find that BM = sqrt(20)*x and MN = sqrt(5)*x.