+0  
 
0
8
3
avatar+85 

These questions are difficult
1.

2.

3.

 Mar 18, 2024
 #1
avatar+128794 
+1

1.

                  P

   9                           17

 

                                 Y

 

Q                X                   R

             10

 

Since PX  is a bisector     let  QX = 10 - m  and RX  = m

So

RX / PR  = QX / PQ

m / 17  =  (10 - m)  / 9

9m = 17(10 - m)

9m  = 170 - 17m

26m  = 170

m  = 170 / 26  =  85/13  =XR

 

Law of Cosines

PQ^2  = QR^2 + PR^2   - 2(QR * PR)cos (PRQ)

9^2  = 10^2  + 17^2  - 2(10 * 17) cos (PRQ)

[ 9^2 - 10^2  - 17^2  ] / [ 340 ] =  cos (PRQ)  = 77/85

sin (PRQ) =  sqrt [ 85^2 - 77^2 ] / 85 =  36/85

 

Law of Sines

XY / sin (PRQ)   = XR  / sin 90

XY  =  sin (PRQ) (XR)

XY  = (36/85) ( 85/13)  =  36 / 13

 

cool cool cool cool

 Mar 19, 2024
 #2
avatar+128794 
+1

2.

PQ = sqrt (2^2 + 2^2 )  = sqrt 8

PA = QA =  sqrt [ 7^2 + 5^2] =  sqrt [ 74]

 

Law of Cosines

 

PQ^2  = PA^2 QA^2  - 2(PA * QA) cos PAQ

8 = 74 + 74  - 2(74)cosPAQ

[ 8 - 2 * 74 ] / [ -2 * 74 ]  = cos PAQ  =  35 /   37

 

sin PAQ =  sqrt [ 37^2  - 35^2 ] /  37  =   sqrt [144] / 37  =  12 /  37

 

cool cool cool

 Mar 19, 2024
 #3
avatar+128794 
+1

3. Not too hard

Area of  AMD + BNC =  (1/2) * 1  =  1/2

Area of MBND  =  Area of square  - 1/2 =    1/2

Shaded area = (1/2)[MBND ]   =  1/2 * 1/2 =  1/4

 

coolcoolcool

 Mar 19, 2024

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