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# In the figure above, what is the perimeter of ∆ ABC in terms of m?

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Math Expert
Joined: 02 Sep 2009
Posts: 40929

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In the figure above, what is the perimeter of ∆ ABC in terms of m? [#permalink]

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12 Aug 2017, 16:22
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In the figure above, what is the perimeter of ∆ ABC in terms of m?

(A) 10m
(B) 15m
(C) 17m
(D) 7m + 3√2 m
(E) 12m + 3√2 m

[Reveal] Spoiler:
Attachment:

2017-08-13_0320.png [ 5.35 KiB | Viewed 333 times ]

The OA will be automatically revealed on Saturday 19th of August 2017 04:22:27 PM Pacific Time Zone
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Kudos [?]: 118693 [0], given: 12006

Senior Manager
Joined: 22 May 2016
Posts: 476

Kudos [?]: 113 [0], given: 444

In the figure above, what is the perimeter of ∆ ABC in terms of m? [#permalink]

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12 Aug 2017, 16:55
Bunuel wrote:

In the figure above, what is the perimeter of ∆ ABC in terms of m?

(A) 10m
(B) 15m
(C) 17m
(D) 7m + 3√2 m
(E) 12m + 3√2 m

[Reveal] Spoiler:
Attachment:
2017-08-13_0320.png

Let X be the point where B intersects side AC

Left triangle ABX is right isosceles (45-45-90) with side ratio $$x: x: x\sqrt{2}$$

3m corresponds with x
Hypotenuse, side AB, therefore is 3$$\sqrt{2}$$m

Triangle BCX, on the right, is 3-4-5 triangle. Hypotenuse, side BC, is 5m

Perimeter = sum of three side lengths

5m + 7m + 3$$\sqrt{2}$$m =

12m + 3$$\sqrt{2}$$m

Kudos [?]: 113 [0], given: 444

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Re: In the figure above, what is the perimeter of ∆ ABC in terms of m? [#permalink]

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12 Aug 2017, 18:07

The triangle to the left is an isosceles right triangle which has
sides in the ratio $$1:1:\sqrt{2}$$.
Similarly, the triangle is a right angled triangle with sides
in the ratio of the Pythagorean triplet(3:4:5)

Hence, the hypotenuse of the triangle on the left is $$3\sqrt{2}$$
where as the hypotenuse of the triangle to the right is 5m.

Hence, the perimeter of the triangle is $$5m+4m+3m$$+$$3\sqrt{2}$$m = $$12m$$ + $$3\sqrt{2}$$m
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Kudos [?]: 344 [0], given: 15

Intern
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Re: In the figure above, what is the perimeter of ∆ ABC in terms of m? [#permalink]

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12 Aug 2017, 22:00
let in the triangle ABC, where the line from B meets line AC be E.

So,
in triangle BCE,hypotenuse BC =(3^2+4^2)^1/2=(9+16)^1/2=25^1/2=5
Here 5M, as the lengths are given terms of M.

for the triangle,
ABE,

AB =(3^2+3^2)^1/2
=(18)^1/2
=3√2
I.E 3√2M

HENCE PERIMETER: BC +AB +AC=5M+3√2M+3M+4M=(12+3√2)M

Kudos [?]: 0 [0], given: 3

Re: In the figure above, what is the perimeter of ∆ ABC in terms of m?   [#permalink] 12 Aug 2017, 22:00
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