Answer:
NO = 10 and OP = 18
Step-by-step explanation:
Given: KM = 48 , LD = 16 , NO : OP = 5 : 9 and NOPS is rectangle.
To find: Value of NO and OP
Let the value of NO and OP = 5x and 9x
frst we prove ΔKLD is similar to ΔKON and ΔMLD is similar to ΔMPS
In ΔKLD and ΔKON
∠KDL = ∠KNO = 90° ( corresponding angles )
∠DKL = ∠NKO ( common Angle )
⇒ ΔKLD is similar to ΔKON by AA similarity rule.
⇒ [tex]\frac{KD}{KN}=\frac{LD}{ON}[/tex]
subtract 1 from both sides
⇒ [tex]\frac{KD}{KN}-1=\frac{LD}{ON}-1[/tex]
⇒ [tex]\frac{KD-KN}{KN}=\frac{LD-ON}{ON}[/tex]
by substituting value from figure,
⇒ [tex]\frac{ND}{KN}=\frac{16-5x}{5x}[/tex]
⇒ [tex]ND=(\frac{16-5x}{5x})\times KN[/tex] ..................... (1)
In ΔMLD and ΔMPS
∠MDL = ∠MSP = 90° ( corresponding angles )
∠DML = ∠SMP ( common Angle )
⇒ ΔMLD is similar to ΔMPS by AA similarity rule.
⇒ [tex]\frac{MD}{SM}=\frac{LD}{PS}[/tex]
subtract 1 from both sides
⇒ [tex]\frac{MD}{SM}-1=\frac{LD}{PS}-1[/tex]
⇒ [tex]\frac{MD-SM}{SM}=\frac{LD-PS}{PS}[/tex]
by sustituting value from figure,
⇒ [tex]\frac{DS}{SM}=\frac{16-5x}{5x}[/tex]
⇒ [tex]DS=(\frac{16-5x}{5x})\times SM[/tex] ..................... (2)
Add eqn. (1) & (2), we get
[tex]ND+DS=(\frac{16-5x}{5x})\times KN + (\frac{16-5x}{5x})\times SM[/tex]
[tex]NS=(\frac{16-5x}{5x})(KN+SM)[/tex]
[tex]NS=(\frac{16-5x}{5x})(KM-NS)[/tex] ( from figure KN + SM = KM - NS)
substitute given values,
[tex]9x=(\frac{16-5x}{5x})(48-9x)[/tex]
[tex]9x \times 5x=16\times48+16\times(-9x)-5x\times48-5x\times(-9x)[/tex]
[tex]45x^2=768-144x-240x+45x^2[/tex]
[tex]45x^2-45x^2=768-384x[/tex]
[tex]768-384x=0[/tex]
[tex]384x=768[/tex]
[tex]x=\frac{768}{384}[/tex]
x = 2
⇒ NO = 5x = 5 × 2 = 10
⇒ OP = 9x = 9 × 2 = 18
