It is known that the cube $PQRS.TUVW$ has an edge length of $10$ cm. If $A$ is the point in the middle of $PQ$ and $B$ lies on $AR$ so that $WB$ is perpendicular to $AR$ , then the length of $WB$ is

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Q. It is known that the cube

PQRS.TUVW

has an edge length of

10

cm. If

A

is the point in the middle of

PQ

and

B

lies on

AR

so that

WB

is perpendicular to

AR

, then the length of

WB

Cube Dimensions: question_prompt: Find the length of $WB$ in the cube $PQRS.TUVW$ where $A$ is the midpoint of $PQ$ and $B$ is on $AR$ such that $WB$ is perpendicular to $AR$ .
Triangle PAB: The edge length of the cube is $10\,\text{cm}$ , so $PQ = 10\,\text{cm}$ . Since $A$ is the midpoint of $PQ$ , $PA = PQ/2 = 10\,\text{cm} / 2 = 5\,\text{cm}$ .
Finding $AB$ : In triangle $PAB$ , $AB$ is the hypotenuse of the right-angled triangle with $PA$ and $PB$ as the other two sides. Since $B$ lies on $AR$ , $PB$ is also the height of the cube, which is $10\,\text{cm}$ .
Calculating WA: To find $AB$ , we use the Pythagorean theorem: $AB^2 = PA^2 + PB^2$ . Substituting the known lengths, we get $AB^2 = 5^2 + 10^2 = 25 + 100 = 125$ .
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm.
Finding WB: Taking the square root of both sides to find $AB$ , we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find $WB$ . Since $B$ lies on $AR$ and $WB$ is perpendicular to $AR$ , triangle $WAB$ is also a right-angled triangle with $WB$ as one of the sides.
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find WB. Since B lies on AR and WB is perpendicular to AR, triangle WAB is also a right-angled triangle with WB as one of the sides. In triangle WAB, we have $WA$ as the hypotenuse and $AB$ and $WB$ as the other two sides. $WA$ is the diagonal of the cube's face, which can be found using the Pythagorean theorem in triangle WPA, where $WP$ and $PA$ are edges of the cube.
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find WB. Since B lies on AR and WB is perpendicular to AR, triangle WAB is also a right-angled triangle with WB as one of the sides. In triangle WAB, we have $WA$ as the hypotenuse and $AB$ and $WB$ as the other two sides. $WA$ is the diagonal of the cube's face, which can be found using the Pythagorean theorem in triangle WPA, where $WP$ and $PA$ are edges of the cube. Calculating $WA$ , we have $WA^2 = WP^2 + PA^2$ . Since $WP = PA = 10$ cm, $WA$ $0$ .
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find WB. Since B lies on AR and WB is perpendicular to AR, triangle WAB is also a right-angled triangle with WB as one of the sides. In triangle WAB, we have WA as the hypotenuse and AB and WB as the other two sides. WA is the diagonal of the cube's face, which can be found using the Pythagorean theorem in triangle WPA, where WP and PA are edges of the cube. Calculating WA, we have $WA^2 = WP^2 + PA^2$ . Since $WP = PA = 10$ cm, $WA^2 = 10^2 + 10^2 = 100 + 100 = 200$ . Taking the square root of both sides to find WA, we get $WA = \sqrt{200} = \sqrt{(100*2)} = 10\sqrt{2}$ cm.
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find WB. Since B lies on AR and WB is perpendicular to AR, triangle WAB is also a right-angled triangle with WB as one of the sides. In triangle WAB, we have WA as the hypotenuse and AB and WB as the other two sides. WA is the diagonal of the cube's face, which can be found using the Pythagorean theorem in triangle WPA, where WP and PA are edges of the cube. Calculating WA, we have $WA^2 = WP^2 + PA^2$ . Since $WP = PA = 10$ cm, $WA^2 = 10^2 + 10^2 = 100 + 100 = 200$ . Taking the square root of both sides to find WA, we get $WA = \sqrt{200} = \sqrt{(100*2)} = 10\sqrt{2}$ cm. Using the Pythagorean theorem again for triangle WAB, we have $WB^2 = WA^2 - AB^2$ . Substituting the known lengths, we get $WB^2 = (10\sqrt{2})^2 - (5\sqrt{5})^2$ .
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find WB. Since B lies on AR and WB is perpendicular to AR, triangle WAB is also a right-angled triangle with WB as one of the sides. In triangle WAB, we have $WA$ as the hypotenuse and $AB$ and $WB$ as the other two sides. $WA$ is the diagonal of the cube's face, which can be found using the Pythagorean theorem in triangle WPA, where $WP$ and $PA$ are edges of the cube. Calculating $WA$ , we have $WA^2 = WP^2 + PA^2$ . Since $WP = PA = 10$ cm, $WA$ $0$ . Taking the square root of both sides to find $WA$ , we get $WA$ $2$ cm. Using the Pythagorean theorem again for triangle WAB, we have $WA$ $3$ . Substituting the known lengths, we get $WA$ $4$ . Simplifying, we have $WA$ $5$ .
Finding WB: Taking the square root of both sides to find AB, we get $AB = \sqrt{125} = \sqrt{(25*5)} = 5\sqrt{5}$ cm. Now, we need to find WB. Since B lies on AR and WB is perpendicular to AR, triangle WAB is also a right-angled triangle with WB as one of the sides. In triangle WAB, we have $WA$ as the hypotenuse and $AB$ and $WB$ as the other two sides. $WA$ is the diagonal of the cube's face, which can be found using the Pythagorean theorem in triangle WPA, where $WP$ and $PA$ are edges of the cube. Calculating $WA$ , we have $WA^2 = WP^2 + PA^2$ . Since $WP = PA = 10$ cm, $WA$ $0$ . Taking the square root of both sides to find $WA$ , we get $WA$ $2$ cm. Using the Pythagorean theorem again for triangle WAB, we have $WA$ $3$ . Substituting the known lengths, we get $WA$ $4$ . Simplifying, we have $WA$ $5$ . Taking the square root of both sides to find WB, we get $WA$ $6$ cm.