+0  
 
+1
244
4
avatar+1442 

1:  Let triangle PQR be a triangle in the plane, and let S be a point outside the plane of triangle PQR, so that SPQR is a pyramid whose faces are all triangles. Suppose that every edge of SPQR has length 18 or 41, but no face of SPQR is equilateral. Then what is the surface area of SPQR?

 

2:  A rectangular box is 8 cm thick, and its square bases measure 32 cm by 32 cm. What is the distance, in centimeters, from the center point P of one square base to corner Q of the opposite base? Express your answer in simplest terms.

3:  Let ABCDEFGH be a cube of side length 5, as shown. Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1. The plane through C, P, and Q intersects line DH at R. Find DR.

 

Thank you so much!

AnonymousConfusedGuy  May 13, 2018
 #1
avatar+20115 
0

3:  Let ABCDEFGH be a cube of side length 5, as shown. Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1. The plane through C, P, and Q intersects line DH at R. Find DR.

 

see answer: https://web2.0calc.com/questions/cube-question-help

 

laugh

heureka  May 14, 2018
 #2
avatar+20115 
0

2:  A rectangular box is 8 cm thick, and its square bases measure 32 cm by 32 cm. What is the distance, in centimeters, from the center point P of one square base to corner Q of the opposite base? Express your answer in simplest terms.

 

\(\begin{array}{|rcll|} \hline PQ &=& \sqrt{(\frac{32}{2})^2+ (\frac{32}{2})^2 + 8^2 } \\ &=& \sqrt{16^2+ 16^2 + 8^2 } \\ &=& \sqrt{576 } \\ &=& 24\ cm \\ \hline \end{array}\)

 

laugh

heureka  May 14, 2018
edited by heureka  May 14, 2018
 #3
avatar+20115 
0

1:  Let triangle PQR be a triangle in the plane, and let S be a point outside the plane of triangle PQR, so that SPQR is a pyramid whose faces are all triangles. Suppose that every edge of SPQR has length 18 or 41, but no face of SPQR is equilateral. Then what is the surface area of SPQR?

 

Heron:

 

\(\begin{array}{|rcll|} \hline A &=& 4\times \sqrt{s(s-41)(s-41)(s-18)} \quad & | \quad s=\frac{41+41+18}{2}=50 \\ &=& 4\times \sqrt{50\cdot(50-41)^2\cdot (50-18)} \\ &=& 4\times \sqrt{50\cdot 9^2\cdot 32} \\ &=& 4\cdot9 \sqrt{50\cdot 32} \\ &=& 36\sqrt{1600} \\ &=& 36\cdot 40 \\ &=& 1440 \\ \hline \end{array}\)

 

laugh

heureka  May 14, 2018
edited by heureka  May 14, 2018
 #4
avatar+1442 
+1

Thanks so much for the responses Heureka!

AnonymousConfusedGuy  May 14, 2018

7 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.