Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$6$ . Ребро куба $АВСDА_1В_1С_1D_1$ равно $4\,\text{см}$ . Через диагональ основания $BD$ под углом $45^\circ$ к плоскости основания проведена плоскость $BDK$ , пересекающая боковое ребро в точке $К$ . Найдите площадь треугольника $BDK$ .

View step-by-step help

View step-by-step help

Find derivatives of radical functions

Full solution

Q.

6

. Ребро куба

АВСDА_1В_1С_1D_1

равно

4\,\text{см}

. Через диагональ основания

BD

под углом

45^\circ

к плоскости основания проведена плоскость

BDK

, пересекающая боковое ребро в точке

К

. Найдите площадь треугольника

BDK

.

Find Diagonal Length: Determine the length of the diagonal BD of the base of the cube. $\newline$ Since the cube has an edge length of $4\,\text{cm}$ , we can use the Pythagorean theorem to find the length of the diagonal BD of the square base. The diagonal of a square is the hypotenuse of a right-angled triangle with the sides of the square as the other two sides. $\newline$ $BD = \sqrt{(AB^2 + AD^2)}$ $\newline$ $BD = \sqrt{(4^2 + 4^2)}$ $\newline$ $BD = \sqrt{(16 + 16)}$ $\newline$ $BD = \sqrt{32}$ $\newline$ $BD = 4\sqrt{2}\,\text{cm}$
Calculate Triangle Height: Determine the height of the triangle $BDK$ . Since the plane $BDK$ is at a $45^\circ$ angle to the base, and we know that the cube's edge is perpendicular to the base, the height of the triangle $BDK$ will be equal to the length of the edge of the cube, which is $4\,\text{cm}$ . Height $(BK) = 4\,\text{cm}$
Area Calculation: Calculate the area of triangle BDK. $\newline$ The area of a triangle is given by the formula: $\newline$ Area = $(1/2) \times \text{base} \times \text{height}$ $\newline$ In triangle BDK, BD is the base and BK is the height. $\newline$ Area(BDK) = $(1/2) \times BD \times BK$ $\newline$ Area(BDK) = $(1/2) \times 4\sqrt{2} \times 4$ $\newline$ Area(BDK) = $2 \times 4\sqrt{2}$ $\newline$ Area(BDK) = $8\sqrt{2}$ cm $^2$
Verify Answer: Verify that the area calculation is correct and that it answers the question prompt. $\newline$ We have found the area of triangle $BDK$ using the correct formula and the given dimensions of the cube and the angle of the plane. The calculation is correct and answers the question prompt.

More problems from Find derivatives of radical functions

Question

Find the derivative of

f(x)

\newline

f(x) = \cos(x)

\newline

f'(x) =

Posted 1 month ago

Question

Find the derivative of

f(x)

\newline

f(x) = \tan(x)

\newline

f'(x) =

Posted 2 months ago

Question

Find the derivative of

f(x)

\newline

f(x) = e^x

\newline

f'\left(x\right) =

Posted 2 months ago

Question

Find the derivative of

f(x)

\newline

f(x) = \ln(x)

\newline

f'(x) =

Posted 2 months ago

Question

Find the derivative of

f(x)

\newline

f(x) = \tan^{-1}{x}

\newline

f'(x) =

Posted 2 months ago

Question

Find the derivative of

f(x)

\newline

f(x) = x e^x

\newline

f'\left(x\right) =

Posted 2 months ago

Question

Find the derivative of

f(x)

\newline

f(x) = \frac{e^x}{x}

\newline

f'(x) =

Posted 2 months ago

Question

Find the derivative of

f(x)

\newline

f(x) = e^{(x + 1)}

\newline

f'\left(x\right) =

Posted 1 month ago

Question

Find the derivative of

f(x)

\newline

f(x) = \sqrt{x+3}

\newline

f'(x) =

Posted 2 months ago

Question

Find the derivative of

g(x)

\newline

g(x) = \ln(2x)

\newline

g^{\prime}(x) =

Posted 3 months ago