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2 years ago

What is the length of the diagonal of a 10 cm by 15 cm rectangle?

Mathematics

2 answers:

padilas [110]2 years ago

6 0

We can use the Pythagorean theorem to solve for the diagonal. The diagonal is the same as the hypotenuse of a triangle.

Pythagorean Theorem: a² + b² = c²

Now, solve for the diagonal.

(10)² + (15)² = c²

100 + 225 = c²

325 = c²

√325 = √c²

18.0277... = c

Round if necessary.

18.0277... = 18.03

Therefore, the diagonal of the rectangle is approximately 18.03cm.

Best of Luck!

Ksivusya [100]2 years ago

4 0

Answer:

18.02 cm

Step-by-step explanation:

A diagonal makes a rectangle into two right triangles. So if we use the Pythagorean theorem we can find the hypotenuse which is the diagonals.

Remember, because we know the length and width, we know the two sides.

10^2 + 15^2 = c^2

100+225 = c^2

325 = c^2

18.027 ≈ c

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3 years ago

Triangle ABC has AB=5, BC=7, and AC=9. D is on AC with BD=5. Find the length of DC

e-lub [12.9K]

Answer:

$DC=\frac{8}{3}\ units$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The triangle ABD is an isosceles triangle

because

AB=BD

The segment BM is a perpendicular bisector segment AD

In the right triangle ABM

Applying the Pythagorean Theorem

$BM^2=AB^2-AM^2$

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$AB=5\ units\\AM=x\ units$

substitute

$BM^2=5^2-x^2$

$BM^2=25-x^2$ -----> equation A

In the right triangle BMC

Applying the Pythagorean Theorem

$BM^2=BC^2-MC^2$

we have

$BC=7\ units\\MC=AC-AM=(9-x)\ units$

substitute

$BM^2=7^2-(9-x)^2$

$BM^2=49-(81-18x+x^2)$

$BM^2=49-81+18x-x^2$

$BM^2=-x^2+18x-32$ ----> equation B

equate equation A and equation B

$-x^2+18x-32=25-x^2$

solve for x

$18x=25+32\\18x=57\\\\x=\frac{57}{18}$

Simplify

$x=\frac{19}{6}$

Find the length of DC

$DC=AC-2x$

substitute the given values

$DC=9-2(\frac{19}{6})$

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3 years ago

A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3296 tickets overall. It has s

rodikova [14]

Answer:

1022 of $10 tickets, 1326 of $20 tickets, 948 of VIP $30 tickets

Step-by-step explanation:

We can set-up a system of equations to find the number of each kind of ticket. We know there are the cheap tickets c, the medium tickets m and the VIP tickets v. Since 3296 tickets were purchased, then c+m+v=3296.

We also know that 304 more medium priced tickets have been sold then cheap tickets. We write 304+c=m.

Lastly, we know that a total of $65,180 was sold. We write 10c+20m+30v=65,180.

We will solve by substituting one equation into the other. We substitute m=304+c into c+m+v=3296. Simplify and isolate the variable v.

c+m+v=3296

c+(304+c)+v=3296

c+304+c+v=3296

2c+304-304+v=3296-304

2c-2c+v=2992-2c

v=2992-2c

We will now substitute this into the remaining equation with our first substitution.

10c+20m+30v=65,180

10c+20(304+c)+30(2992-2c)=65,180

10c+6080+20c+89760-60c=65180

-30c+95840=65180

-30c+95840-95840=65180-95840

-30c=-30660

c=1022

This means 1,022 tickets purchased were $10 tickets. We substitute this equation back into m=304+c to find m.

m=304+1022

m=1326

This means 1,326 were $20 tickets.

Lastly we know 948 VIP tickets were bought because 1022+1326+948=3296

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3 years ago