Find the length of the shorter leg of a right triangle if the longer leg is 15 feet more than the shorter leg and the hypotenuse is 15 feet less than twice the shorter leg

45

Step-by-step explanation:

Let the length of the shorter leg be represented as x.

Let the length of the longer let let be represented as x+15

While the length of the hypotenuse be 2x-15.

From the knowledge of phythagoras theorem:

x^2+(x+15)^2 = (2x-15)^2

x^2+x^2+15x+15x+225 = 4x^2-30x-30x+225

2x^2+30x+225 = 4x^2-60x+225

2x^2-4x^2+30x+60x+225-225=0

-2x^2+90x+0= 0

2x^2-90x = 0

Divide through by 2

x^2-45x=0

x(x-45)=0

x = 0 or x-45=0

Considering x-45=0

x-45=0

x = 45

If x is for the shorter leg,

The length for the longer leg is: x+15

= 45+15

= 60

For the hypotenuse = 2x-15

= 2×45-15

= 90-15

= 75

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

answ5

step-by-step explanation:

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