+0  
 
0
7
3
avatar+85 

Can you solve these?
1.

2.

3.

4.

 Mar 25, 2024
 #1
avatar+128802 
+1

3. By  a property of medians ON  = (1/3)QN =  (1/3)(12)  = 4

 

NR  =  PR/2  =  14 / 2 =  7

 

And triangle  ONR is right  such that

 

sqrt [ NR^2  + ON^2 ] =   sqrt [ 7^2  + 4^2  ]  = sqrt [ 65]  =  OR

 

cool cool cool

 Mar 26, 2024
 #2
avatar+128802 
+1

4.  

Semi-perimeter of PQR    = [ 17 + 10 + 9 ] / 2 =   18

Area of PQR =  sqrt [ 18 * 1 * 8 * 9 ]   = sqrt [ 1296 ]  =  36

Find altitude SQ

36  = (1/2)PR * SQ

72  = 17 * SQ

SQ = 72 / 17

 

Let XR = x      Let QX = (10 - x)

By an angle bisector property

XR / PR   = QX / PQ

x / 17  = (10 - x) / 9

9x  = 17 (10 - x)

9x = 170 - 17x

26x  = 170

x = 170 / 26   =   85  / 13 =  XR

 

Triangles XYR and QSR  are similar

XY /XR   = QS / QR

XY / (85/13)  = (72/17)  /10

 

XY  = (85/13)(72/17)  / 10   =  36 / 13

 

 

cool cool cool

 Mar 26, 2024
 #3
avatar+128802 
+1

2.   H is the  orthocenter of the triangle  Draw CH  and extend it  to F where CF is the third altitude of the  triangle

 

Angle AHE  =  180   - 128 =  52

Angle AEH  = 90

Angle HAE  =  90  - 52 = 38

 

Angle BAH  =  28

 

Angle BAC =  Angle BAH + Angle HAE  =  28 + 38  = 66 

 

Angle CFA =  90

Angle BAC =  66

Angle FCA =  180 - 90 - 66   = 24

 

But Angle FCA = Angle HCA =  24

 

cool cool cool 

 Mar 26, 2024

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