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Find the missing side of each triangle. i need help!!!

Nezavi [6.7K]

Answer:

√161 yd

Step-by-step explanation:

Hi there!

We're given a right triangle (notice the right angle), the measure of one of the legs (one of the sides that make up the right angle) as 8 yd, and the hypotenuse (the side opposite to the right angle) as 15 yd.

We need to find x, which is the other leg in the triangle

If we are given 2 sides and need to find a third, we can use Pythagorean Theorem, which states that a²+b²=c² if a and b are legs and c is the hypotenuse

In this case, a=8, b=x, and c=15

So let's substitute those values into the formula

(8)²+x²=(15)²

raise everything to the second power

64+x²=225

subtract 64 from both sides

x²=161

take the square root of both x and 161

x=√161, x=-√161

-√161 cannot work in this case, as distance cannot be negative.

The question asks for the answer to be in simplified radical form, but √161 cannot be simplified down further

so the answer is <u>√161 yd</u>

Hope this helps! :)

3 0

3 years ago

Would appreciate the help 24 points

andre [41]

Answer:

13) Rectangle

14) parallelogram

15) co ordinates areD (0,a) andE (c,b)

16) co ordinates are D(0,b) and E(a,0)

17)(-1,1) (4,1) (-1,4)

18)(0,0) (a,0) (a,a) (0,a)

Step-by-step explanation:

<h3>13)</h3><h3>Rectangle</h3>

The opposite sides of a rectangle are parallel and equal

left height = right height

(by using distance formula)

(0,3),(2,-2) = (5,5),(7,0)

√(2-0)²+(-2-3)² = √(7-5)²+(0-5)²

√(2)²+(-5)² = √(2)²+(-5)²

√4+25 = √(2)²+(-5)²

√29 = √29 hence equal

<h3>14)</h3><h3>parallelogram</h3>

The opposite sides of a rectangle are not equal

left height = right height

(by using distance formula)

(0,0),(3,-4) = (3,4),(7,0)

√(3-0)²+(-4-0)² = √(7-3)²+(0-4)²

√(3)²+(-4)² = √(5)²+(-4)²

√9+16 = √(25)+(16)

√(25) ≠√41

5 ≠ √41

<h3>15)</h3><h3>co ordinates are D(0,a) and E(c,b)</h3><h3>16)</h3><h3>co ordinates are D(0,b) and E(a,0) </h3>

since rhombus have equal opposite sides

6 0

3 years ago

Please help me out :P

Veseljchak [2.6K]

cos θ = $\frac{-4\sqrt{65} }{65}$ , sin θ = $\frac{-7\sqrt{65} }{65}$ , cot θ = 4/7, sec θ = $\frac{-\sqrt{65} }{4}$ , cosec θ = $\frac{-\sqrt{65} }{7}$