In the figure ,the equal sides of the iscoscles triangular are 5cm each,the base in 6 cm and the height is 4cm height.The length of the rectangle are all 10cm Calculate the TSA (total surface area )of the triangular prism

Answers

Answer 1

Answer:

Answer:

284 square cm.

Step-by-step explanation:

To calculate the total surface area (TSA) of the triangular prism formed by an isosceles triangle and a rectangle, you need to find the surface areas of both the triangular faces and the three rectangular faces and then add them together.

1. **Triangular Faces**:

You have an isosceles triangle with a base of 6 cm and a height of 4 cm. The total area of both triangular faces can be calculated using the formula for the area of a triangle:

Area of one triangular face = (1/2) * base * height

Area of one triangular face = (1/2) * 6 cm * 4 cm = 12 square cm

Since there are two identical triangular faces, the total area of both triangular faces is 2 * 12 square cm = 24 square cm.

2. **Rectangular Faces**:

You have a rectangular prism with dimensions 10 cm x 10 cm x 6 cm. There are three rectangular faces.

- The two rectangular faces with dimensions 10 cm x 10 cm have an area of 100 square cm each.

- The third rectangular face with dimensions 6 cm x 10 cm has an area of 60 square cm.

The total area of all three rectangular faces is 2 * 100 square cm + 60 square cm = 260 square cm.

3. **Total Surface Area (TSA)**:

Now, you can calculate the TSA of the triangular prism by adding the areas of the triangular faces and the areas of the rectangular faces:

TSA = Area of Triangular Faces + Area of Rectangular Faces

TSA = 24 square cm + 260 square cm = 284 square cm

So, the total surface area of the triangular prism is 284 square cm.

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Answers

Only E) is true. If I were to arbitrarily pick x = 2.2 and y =1.1, then I could show that A), B), C), and D) to be false, given x/y = 2.2/1.1 = 2 (which is an integer). However for E), if we define x/y = n where n is an integer, then by rearranging terms, we can show x = n*y where n still equals an integer and our original equation is still valid.

Identify the expression equivalent to 4(x + x + 7) − 2x + 8 − 4 by substituting x = 1 and x = 2.

Answers

4(x+x+7)-2x+8-4 X=1
4(1+1+7) -2(1) +8-4
4(9)+2
38

4(x+x+7)-2x+8-4 X=2
4(2+2+7) -2(2)+4
4(11)-4+4
44

Answer:

Step-by-step explanation:

So if we do 4(x+x+7)-2x+8-4 X=1

then 4(1+1+7) -2(1) +8-4

and 4(9)+2

it would be 38

then if we do 4(x+x+7)-2x+8-4 X=2

and 4(2+2+7) -2(2)+4

also 4(11)-4+4 it would be 44

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Answers

Answer:

Step-by-step explanation:

See attachment. The hypotenuse must be the longest line. Check using the Pythagorean Theorem.

The roots of x2 − ( ) + 34 are 5 ± 3i.

Answers

use quadratic formula
if you had ax^2+bx+c=0, then
x= $\frac{-b+/- \sqrt{b^(2)-4ac} }{2a}$
a=1
b=?
c=34
subsitute
$\frac{-b+/- \sqrt{b^(2)-4(1)(34)} }{2(1)}$ =5+/-3i
$\frac{-b+/- \sqrt{b^(2)-136} }{2}$ =5+/-3i
make 5+/-3 into fraction over 2,(10+/-6i)/2
$\frac{-b+/- \sqrt{b^(2)-136} }{2}$ =(10+/-6i)/2
multiply both sides by 2
$-b+/- \sqrt{b^(2)-136}$ =10+/-6i
we conclude that -b=10
b=-10

ok so equaton is
x^2-10x+34

The roots of x² -bx+34=0 are 5 ± 3i. Then b = ?

-------------------

Vieta's formulas :

5 - 3i + 5 + 3i = b ⇒ b =10

Now, we have the equation:

x² -10x + 34 = 0

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Answers

Question 1 is correct. AM = 4 and CM cong. AM therefore CM = 4.
Question 2 is correct. Triangles congruent by SAS --> BA cong BC = 12
Question 3 Line BM intersects point D
Questions 4-6 Attachement 1 is an example with these triangles and their circumcenters as det. by their perp. bisectors. As you can see:
The POC of a right triangle lies on the triangle, acute, inside, and obtuse, outside.
Question 7-8 First off: DF = 6, not BD.
We can clearly see 6 congruent triangles within this equilateral one.
BD corresponds to AD and = 11.5
So does DC.
Question 9-10 See attachement 2
Question 11 Sometimes -- see the right triangle in attachment 1
Question 12 Never -- See attachment 4, of course you can always just draw one for yourself. Just draw an acute, right, and obtuse triangle!
Question 13 Always -- See attahcment 3
Question 14 Never -- See attachment 3
Question 15 Sometimes -- see obtuse triangle in attachment 1

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An equilateral triangle has an apothem of 5 cm. Find the perimeter of the triangle to the nearest centimeter.

Answers

The apothem is the distance from the center to the midpoint of one of the sides of a regular polygon. You can make a right triangle with the apothem, the line from the midpoint to the corner and the line from the center to the corner. An equilateral triangle has 60 degree angles (180/3). The right triangle has half of one of those angles so 30 degrees. Now we have a 30-60-90 triangle where the short leg is 5cm. The long leg, which is also half of one side of the triangle is thus $5√(3)$ . A whole side of the triangle is $10√(3)$ . Multiply that by 3 to get the perimeter of $30√(3)$ .

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