Downside 1 :

Discover the size of the hypotenuse of the proper triangle proven beneath. Inform whether or not the aspect lengths type a Pythagorean triple.

Downside 2 :

Discover the size of the leg of the proper triangle proven beneath.

Downside 3 :

Discover the realm of the triangle to the closest tenth of a meter.

Downside 4 :

One of many diagonals of a rectangle is 20 cm lengthy. If the distinction between its size and width is 4 cm, then discover the size and width of the rectangle.

## Solutions

1. Reply :

By Pythagorean Theorem, we have now

(Hypotenuse)^{2} = (leg)^{2} + (leg)^{2}

Substitute.

x^{2} = 5^{2} + 12^{2}

Simplify.

x^{2} = 25 + 144

x^{2} = 169

Take sq. root on both sides.

√x^{2} = √169

x = 13

So, the size of the hypotenuse is 13 models.

As a result of 13^{2} = 12^{2} + 5^{2}, they type a Pythagorean triple.

2. Reply :

By Pythagorean Theorem, we have now

(Hypotenuse)^{2} = (leg)^{2} + (leg)^{2}

Substitute.

14^{2} = 7^{2} + x^{2}

Simplify.

196 = 49 + x^{2}

Subtract 49 from both sides.

147 = x^{2}

Take sq. root on both sides.

√147 = √x^{2}

√147 = x

Use product property.

√49 ⋅ √3 = x

Simplify the unconventional.

7√3 = x

Therefore, the required aspect size is 7√3 models.

3. Reply :

We’re provided that the bottom of the triangle is 10 meters, however we have no idea the peak h.

As a result of the triangle is isosceles, it may be divided into two congruent proper triangles with the given dimensions. Use the Pythagorean Theorem to search out the worth of h.

By Pythagorean Theorem, we have now

7^{2} = 5^{2} + h^{2}

Simplify.

49 = 25 + h^{2}

Subtract 25 from both sides.

24 = h^{2}

Take sq. root on both sides.

√24 = √h^{2}

√24 = h

Now discover the realm of the unique triangle.

Space = 1/2 ⋅ b ⋅ h

Substitute.

Space = 1/2 ⋅ 10 ⋅ √24

Use calculator to approximate.

Space = 1/2 ⋅ 10 ⋅ √24

Space ≈ 24.5 m^{2}

Therefore, the realm of the triangle is about 24.5 m^{2}.

4. Reply :

Let x be the size of the rectangle.

For the reason that distinction between the size and width is 4 cm, the width of the rectangle needs to be both (x + 4) cm or (x – 4) cm.

We are able to take the width of the rectangle as (x + 4) cm.

Draw a sketch.

Within the image above, contemplate the proper triangle ABC.

By The Pythagorean Theorem,

x^{2} + (x + 4)^{2}^{ }= 20^{2}

x^{2} + (x + 4)(x + 4)^{ }= 400

x^{2 }+ x^{2 }+ 4x + 4x + 16^{ }= 400

2x^{2 }+ 8x + 16 = 400

Subtract 400 from both sides.

2x^{2 }+ 8x – 384^{ }= 0

Divide either side by 2.

x^{2 }+ 4x – 192^{ }= 0

Issue and remedy.

(x + 16)(x – 12) = 0

x + 16 = 0 or x – 12 = 0

x = -16 or x = 12

For the reason that size of a rectangle can by no means be a damaging worth, we are able to ignore x = -16.

So, the worth of x is 12.

Then,

x + 4 = 12 + 4

x + 4 = 16

Subsequently,

size = 12 cm

width = 16 cm

Kindly mail your suggestions to v4formath@gmail.com

We at all times recognize your suggestions.

©All rights reserved. onlinemath4all.com