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In rectangle ABCD, the angle bisector of ∠A intersects side DC at point M and the angle bisector of ∠C intersects side AB at point N. The length of CM is equal to the length of AM and it is 6 in longer than the length of DM . Find the perimeter of quadrilateral ANCM.

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frika
AM is an angle A bisector and as ABCD is rectangular, m∠A=90^{0}, then m∠DAM=45^{0}.
Consider right triangle ADM. m∠D=90^{0} (because ABCD is rectangle), so m∠DMA=180^{0}-90^{0}-45^{0}=45^{0}. Therefore, ΔADM is isoscales and that's why AD=DM. Let x be the length of DM. Since ΔMC is 6 in longer than DM, MC has length x+6. The length of AM is equal to the length of CM, so AM=x+6.
Return to the triangle ADM, by the Pythagorean theorem AM^2=AD^2+DM^2.
(x+6)^2=x^2+x^2
x^2+12x+36=2x^2
x^2-12x-36=0
D=(-12)^2-4*(-36)=144+144=288
[tex] \sqrt{D} = \sqrt{288} =12 \sqrt{2} [/tex]
[tex] x_{1} = \frac{12+12 \sqrt{2} }{2} =6+6 \sqrt{2} [/tex]
[tex] x_{2} = \frac{12-12 \sqrt{2} }{2} =6-6 \sqrt{2} [/tex] - incorect answer, because x_{2}<0.
The perimeter of quadrilateral ANCM is equal to 4(x+6)=[tex]4(6+6 \sqrt{2}+6 )=48+24 \sqrt{2} [/tex].


ashishdwivedilVT

The perimeter of quadrilateral ANMC is 24√2 +48.

What is a rectangle?

The rectangle is a quadrilateral having each angle as 90 degrees and opposites sides are equal.

From the attached diagram,

DM = x

MC = x+6

Since, ∠AMD = 45degree

so, AD=DM =x

It is given that

AM = MC = x+6

From Pythagoras Theorem,

AM² =AD²+DM²

[tex](x+6)^{2} = x^{2} +x^{2}[/tex]

[tex]x = 6(\sqrt{2} +1)[/tex]

So, the perimeter of quadrilateral ANMC = 4(x+6) = 4(6√2+12)

=24√2 +48

Therefore, the perimeter of quadrilateral ANMC is 24√2 +48.

To get more about rectangles visit:

https://brainly.com/question/25292087

Ver imagen ashishdwivedilVT