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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

2 years ago

-6/20+150% Please help I need the answer quickly.

tia_tia [17]

Is this

-6 ÷ 20 + 150%

or is -6/20 a fraction?

7 0

3 years ago

Read 2 more answers

A secant and a tangent to a circle intersect in a 42-degree angle. The two arcs of the circle intercepted by the secant and the

Amanda [17]

Answer:

The measure of the third arc is $150\°$

Step-by-step explanation:

step 1

we know that

The measurement of the external angle is the semi-difference of the arcs which comprises

in this problem

Let

x----> the greater arc of the circle intercepted by the secant and the tangent

y----> the smaller arc of the circle intercepted by the secant and the tangent

$42\°=\frac{1}{2}(x-y)$

$84\°=(x-y)$ ----> equation A

$\frac{x}{y}=\frac{7}{3}$

$x=\frac{7}{3}y$ -----> equation B

Substitute equation B in equation A and solve for y

$84\°=(\frac{7}{3}y-y)$

$84\°=(\frac{4}{3}y)$

$y=3*84\°/4=63\°$

Find the value of x

$x=\frac{7}{3}(63\°)=147\°$

step 2

Find the measure of the third arc

Let

z------> the measure of the third arc

we know that

$x+y+z=360\°$ -----> complete circle

substitute the values and solve for z

$147\°+63\°+z=360\°$

$z=360\°-(147\°+63\°)=150\°$

3 0

2 years ago

Assume V and W are finite-dimensional vector spaces and T is a linear transformation from V to W, T: Upper V right arrow Upper

scZoUnD [109]

Answer:

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.

Step-by-step explanation:

Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that

T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)

so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.

Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that

c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0

T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0

T(c_1v_1+ c_2v_2 ... c_pv_p) = 0

But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.

8 0

2 years ago

In a certain mid-size city, 90% of the high school graduates are from public schools

Marat540 [252]

E.) .53

I just took this so I hope it helps

8 0

3 years ago