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The hypotenuse of a right angled triangle is 2√13 cm . If the smaller side is increased by 2 cm and the larger side is increased

White raven [17]

Statement:

The hypotenuse of a right angled triangle is 2√13 cm. If the smaller side is increased by 2 cm and the larger side is increased by 3 cm, the new hypotenuse will be √117 cm.

To find out:

The length of the larger side of the right angled triangle.

Solution:

Let us consider x as the smaller side and y as the larger side.

Then, in the right angled triangle,

x² + y² = (2√13)² ...(I) [By Pythagoras Theorem]

Now, if the smaller side is increased by 2 cm, then the smaller side will be (x + 2).

And if the larger side is increased by 3 cm, then the larger side will be (y + 3).

Then, in the new right angled triangle,

(x + 2)² + (y + 3)² = (√117)² [By Pythagoras Theorem]

or, x² + 2 × 2 × x + 2² + y² + 2 × 3 × y + 3² = (√117)²

or, x² + 4x + 4 + y² + 6y + 9 = (√117)²

or, x² + y² + 4x + 6y + 13 = (√117)²

Now, put the value of x² + y² from equation (I),

or, (2√13)² + 4x + 6y + 13 = (√117)²

or, (2 × 2 × √13 × √13) + 4x + 6y + 13 = (√117 × √117)

or, 52 + 4x + 6y + 13 = 117

or, 4x + 6y = 117 - 52 - 13

or, 4x + 6y = 52

or, 4x = 52 - 6y

or, x = $\frac {(52 - 6y)}{4}$ ...(II)

Now, put the value of x of equation (II) in (I),

x² + y² = (2√13)²

or, $\frac {(2704-624y +36y²)}{16}$ + y² = 52

or, $\frac {(2704-624y +36y² + 16y²)}{16}$ = 52

or, 52y²-624y + 2704 = 52 × 16

or, 52y² - 624y + 2704 - 832 = 0

or, 52y² - 624y + 1872 = 0

or, 52(y² - 12y + 36) = 0

or, y²-12y +36 = 0 ÷ 52

or, y²-12y +36 = 0

or, (y)² - 2 × 6 × y + (6)² = 0

or, (y - 6)² = 0

or, y - 6=0

or, y = 6

We have taken y as the length of the larger side of the right angled triangle.

So, the length of the larger side is 6 cm.

Answer:

6 cm

Hope you could understand.

If you have any query, feel free to ask.

6 0

3 years ago

Kylie and Ethan have each saved some money. I’m their savings, each coin is worth leas then 50 cents and each bill is less then

timofeeve [1]

Answer:

Kylie could have a $5 bill and seven quarters while Ethan could have two $1 bills and forty pennies.

Step-by-step explanation:

Since Kylie has seven quarters, that is seven multiplied by the value of the quarter, a quarter is 25 cents. So 7 x 25 = 125 OR $1 and 25 cents. Leaving Kylie with a total of $7 and 25 cents with the addition of the $5 bill.

Ethan has two $1 bills and forty pennies, forty multiplied by the value of the penny, a penny is 1 cent. So 40 x 1 = 40 OR 40 cents. Leaving Ethan with $2 and 40 cents with the addition of the two $1 bills from earlier.

Ethan has more bills and coins than Kylie and yet still has less money than Kylie.

7 0

4 years ago

What is the area of the shaded region?

Mekhanik [1.2K]

9514 1404 393

Answer:

434 -49π ≈ 280.1 cm²

Step-by-step explanation:

The shaded area is the difference between the enclosing rectangle area and the circle area.

The rectangle is 14 cm high and 31 cm wide, so has an area of ...

A = WH

A = (31 cm)(14 cm) = 434 cm²

The circle area is given by ...

A = πr²

A = π(7 cm)² = 49π cm²

__

The shaded area is the difference of these, so is ...

shaded area = rectangle - circle

= (434 - 49π) cm² ≈ 280.1 cm²

4 0

3 years ago

⚠URGENT!!!⚠ solve for m.

Zinaida [17]

Answer:

Step-by-step explanation:

Remark

Read the following carefully.

There is a beautiful theorem that has to do with the endpoints of two angles sharing the same endpoints.

To be a little clearer, I hope, that makes < BAC = <BDC because both angles have B and C as their endpoints inside the circle. Make sure you understand that statement before moving on.

For this problem <BDC = <CAB = 33 degrees.

That means that ADC = 37 + 33 = 70

Solution

<ADC and CBA are opposite angles.

That means that they add to 180

From the above statement in the Remark section <ADC = 37 + 33 = 70 degrees <ABD + <DBC = <ABC = m + 71

<ABC + ADC = 180

m + 71 + 70 = 180 Combine

m + 141 = 180 Subtract 141 from both sides.

m+141-141= 180 - 141 Combine

m = 39

Answer: m = 39

3 0

2 years ago

Solve the simultaneous equations 4x+3y=1 4x^2+3xy+y^2=2

Fantom [35]

Notice that the second equation,

4x ² + 3xy + y ² = 2

can be written as

x (4x + 3y) + y ² = 2

and the first equation says 4x + 3y = 1, so this reduces to

x + y ² = 2

Solve the first equation for x :

4x + 3y = 1

4x = 1 - 3y

x = (1 - 3y)/4

Substitute this into the reduced second equation to get a quadratic equation in y, which happens to be easily factorized and solved:

(1 - 3y)/4 + y ² = 2

1 - 3y + 4y ² = 8

4y ² - 3y - 7 = 0

(4y - 7) (y + 1) = 0

4y - 7 = 0 or y + 1 = 0

y = 7/4 or y = -1

Solve for x :

x = (1 - 3 (7/4))/4 or x = (1 - 3 (-1))/4

x = -17/16 or x = 1

So the two solutions are (x, y) = (-17/16, 7/4) and (x, y) = (1, -1).

8 0

3 years ago