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Setler [38]
2 years ago
6

If the square of the hypotenuse of an isosceles right triangle is 128 cm2, find the length of each side.​

Mathematics
2 answers:
kotegsom [21]2 years ago
5 0
<h3>⋆Solution:</h3>

<u>Let the two equal side of right angled isosceles triangle, right angled at Q be k cm.</u>

⋆ Given: h2 = 128

<u>So, we get</u>

\small\bold{PR2 = PQ2 + QR2}

\small\bold{h2 = k2 + k2 }

\small\bold\pink{ ⇒ }\small\bold{ 128 = 2k2 }

\small\bold\pink{ ⇒ }\small\bold{\frac{128}{2} = k2 }

\small\bold\pink{ ⇒ }\small\bold{ 64 = k2}

\small\bold\pink{ ⇒ }\small\bold{ \sqrt{64} = k}

\small\bold\pink{ ⇒ }\small\bold{8 = k }

<u>Therefore, length of each side is 8 cm.</u>

denis-greek [22]2 years ago
3 0

Answer:

<u>8 cm</u>

Step-by-step explanation:

Let the length be l.

Then, by Pythagorean Theorem,

l² + l² = 128

2l² = 128

l² = 64

l = <u>8 cm</u>

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The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be y. The magnitude of 40\text{ m/s} will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 60^{\circ}, we have:

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Now that we've found the vertical component of the velocity and launch, we can use kinematics equation v_f^2=v_i^2+2a\Delta y to solve this problem, where v_f/v_i is final and initial velocity, respectively, a is acceleration, and \Delta y is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 9.8\:\mathrm{m/s^2}. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

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