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Gwar [14]
4 years ago
14

Prove that a triangle with the sides a - 1 cm to root under a ​

Mathematics
1 answer:
vivado [14]4 years ago
4 0

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.  

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm  

Hence,

{AB}^2 = (a -1)^2 

Now We know that

(a- b)^2 = a^2+b^2 - 2ab

So;

{AB}^2= a^2 + 1^2 -2\times a \times1

{AB}^2 = a^2 + 1 -2a

Now;

{BC}^2 = (2\sqrt{a})^2= 4a

Also;

{CA}^2 = (a + 1)^2

Now We know that

(a+ b)^2 = a^2+b^2 + 2ab

{CA}^2= a^2 + 1^2 +2\times a \times1

{CA}^2 = a^2 + 1 +2a

{CA}^2 = AB^2 + BC^2

[By Pythagoras theorem]

a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a

Hence, {CA}^2 = AB^2 + BC^2

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

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Suppose T is a transformation from ℝ2 to ℝ2. Find the matrix A that induces T if T is reflection over the line y=1/2x
trapecia [35]

Answer:

A = \left[\begin{array}{cc}1&\frac{4}{5}\\\frac{4}{3}&-\frac{3}{5}\end{array}\right]

Step-by-step explanation:

We have to see how the canonical vectors are transformed throught T. Lets first define T in any basis.

Since T is a reflection, then any element of the line y = x/2 if fixed by T. Therefore T(2,1) = (2,1).

On the other hand, any vector perpendicular to the line direction should be sent to its opposite value. We can take, for example, (-1,2) (note that the scalar product (2,1) * (-1,2) = -2+2 = 0). As a consecuence T(-1,2) = (1,-2). We have

  • T(2,1) = (2,1)
  • T(-1,2) = (1,-2)

By summing the first vector with the double of the second one we get, using linearity

T(0,5) = T( (2,1) + 2(-1,2)) = T(2,1) + 2T(-1,2) = (2,1) + 2(1,-2) = (4,-3)

Hence, T(0,1) = (4/5,-3/5)

Now, we take the second vector and substract it the double of the first one (to kill the second variable)

T(-3,0) = T( (-1,2) - 2*(2,1) ) = T(-1,2) -2T(2,1) = (1,-2)-2(2,1) = (-3,-4)

Therefore, T(1,0) = (1,4/3)

The matrix A induced by  T has in its first column T(1,0) and in its second column T(0,1). We conclude that

A = \left[\begin{array}{cc}1&\frac{4}{5}\\\frac{4}{3}&-\frac{3}{5}\end{array}\right]

3 0
3 years ago
Find the area please
tatuchka [14]

Answer:

D

Step-by-step explanation:

Recall that the area of a triangle is given by:

\displaystyle A = \frac{1}{2} bh

In this case, the base is <em>x</em> and the height is <em>y</em>. Hence:

\displaystyle A = \frac{1}{2} xy

We can write the following ratios:

\displaystyle \sin \alpha = \frac{x}{12} \text{ and } \cos \alpha = \frac{y}{12}

Solve for <em>x</em> and <em>y</em>:

\displaystyle x = 12\sin \alpha \text{ and } y = 12\cos \alpha

Substitute:

\displaystyle A = \frac{1}{2}\left(12\sin \alpha\right)\left(12\cos \alpha\right)

And simplify. Hence:

\displaystyle A = 72\sin \alpha \cos\alpha

In conclusion, our answer is D.

6 0
3 years ago
What’s the missing part
Alchen [17]
It should be 15sqrt(3)/2
7 0
3 years ago
Read 2 more answers
(+ y + 3) (y + 1).
Pavlova-9 [17]

Answer:

y^2 + 4y + 3

Step-by-step explanation:

(y + 3)(y + 1)

(y*y) + (y*1) + (3*y) + (3*1)

y^2 + y + 3y + 3

y^2 + 4y + 3

Best of Luck!

8 0
3 years ago
Point G is the centroid of triangle ABC. AG = (5x+4) units and GF= (3x-1) units. What is A F?
Vika [28.1K]
<span>Because G is the centroid of the triangle, therefore AG = 2GF
That is, 5x + 4 = 2(3x - 1)
5x + 4 = 6x - 2
4 + 2 = 6x - 5x
6 = x
Therefore AG = 5*6 + 4 = 34
GF = 3*6 - 1 = 17
A.F = AG + GF = 34 + 17 = 51
Answer: 51</span>
6 0
3 years ago
Read 2 more answers
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