right triangles 2

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by aneesh.kg » Mon Jun 11, 2012 10:58 am
Q1:
If the triangles are similar, then each pair corresponding sides are in the same ratio.

(AB + BC + AC) = 24
and
XZ = (4/3)AC, then
ZY = (4/3)BC
YX = (4/3)AB
or
XZ + ZY + YX = 4/3 *(AB + BC + CA) = 4/3*(24) = 32

[spoiler](C)[/spoiler] is correct.
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by aneesh.kg » Mon Jun 11, 2012 11:00 am
Q2:
Required distance = length of the hypotenuse of the right-angled triangle, in which
Distance due south and Distance due east are the two perpendicular sides.

Length of Hypotenuse = (10^2 + 24^2)^0.5 = 26

[spoiler](B)[/spoiler] is correct.
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by aneesh.kg » Mon Jun 11, 2012 11:03 am
Area = (1/2)*base*height
One of the perpendicular sides are given, and the other can be found using the Pythagoras' theorem.
Other perpendicular side = 15

Area = (1/2)(8)(15) = 60 square units

[spoiler](C)[/spoiler] is correct.

Some common Pythagorean triplets:
3-4-5
6-8-10
12-5-13
8-15-17
24-10-26
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by dimochka » Mon Jun 11, 2012 11:07 am
1. You know the perimeter is 24, and based on the measurements given, the triangles are in a 4:3 ratio, which means that all sides are also in a 4:3 ratio. Since perimeter is the sum of the lengths of all sides, the perimeter is also in a 4:3 ratio, and the answer is [spoiler]24*(4/3)=32[/spoiler]

2. You should use the pythagorean theorem to figure it out. The two legs of the triangle are 10 and 24.
10^2+24^2 = C^2 --> C = [spoiler]sqrt(100+576) = 26[/spoiler]
If you know some of the pythagorean triples, you may be able to skip the calculations. Most students should know [3,4,5], but another useful one is [5,12,13]. If you notice, the measurements in question #2 are exactly twice that.

3. Again use the pythagorean theorem to figure out the third side: B = [spoiler]sqrt(C^2-A^2) = 15[/spoiler].
Then find the area of the triangle: [spoiler]8*15/2 = 60[/spoiler]