Question

Given: △KLM, KM=48, LD=16 LD⊥KM, NOPS - rectangle NO:OP=5:9 Find: NO, OP

Answer 1

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

first 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.

subtract 1 from both sides

by substituting value from figure,

⇒  ..................... (1)

In ΔMLD and ΔMPS

∠MDL   =   ∠MSP = 90° ( corresponding angles )

∠DML    =   ∠SMP   ( common Angle )

⇒ ΔMLD is similar to ΔMPS by AA similarity rule.

subtract 1 from both sides

by sustituting value from figure,

⇒  ..................... (2)

Add eqn. (1) & (2), we get

( from figure KN + SM = KM - NS)

substitute given values,

x = 2

⇒ NO = 5x = 5 × 2 = 10

⇒ OP = 9x = 9 × 2 = 18

Answer 2

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.

⇒ $\frac{KD}{KN}=\frac{LD}{ON}$

subtract 1 from both sides

⇒ $\frac{KD}{KN}-1=\frac{LD}{ON}-1$

⇒ $\frac{KD-KN}{KN}=\frac{LD-ON}{ON}$

by substituting value from figure,

⇒ $\frac{ND}{KN}=\frac{16-5x}{5x}$

⇒ $ND=(\frac{16-5x}{5x})\times KN$ ..................... (1)

In ΔMLD and ΔMPS

   ∠MDL   =   ∠MSP = 90° ( corresponding angles )

   ∠DML    =   ∠SMP   ( common Angle )

⇒ ΔMLD is similar to ΔMPS by AA similarity rule.

⇒ $\frac{MD}{SM}=\frac{LD}{PS}$

subtract 1 from both sides

⇒ $\frac{MD}{SM}-1=\frac{LD}{PS}-1$

⇒ $\frac{MD-SM}{SM}=\frac{LD-PS}{PS}$

by sustituting value from figure,

⇒ $\frac{DS}{SM}=\frac{16-5x}{5x}$

⇒ $DS=(\frac{16-5x}{5x})\times SM$ ..................... (2)

Add eqn. (1) & (2), we get

$ND+DS=(\frac{16-5x}{5x})\times KN + (\frac{16-5x}{5x})\times SM$

$NS=(\frac{16-5x}{5x})(KN+SM)$

$NS=(\frac{16-5x}{5x})(KM-NS)$ ( from figure KN + SM = KM - NS)

substitute given values,

$9x=(\frac{16-5x}{5x})(48-9x)$

$9x \times 5x=16\times48+16\times(-9x)-5x\times48-5x\times(-9x)$

$45x^2=768-144x-240x+45x^2$

$45x^2-45x^2=768-384x$

$768-384x=0$

$384x=768$

$x=\frac{768}{384}$

x = 2

⇒ NO = 5x = 5 × 2 = 10

⇒ OP = 9x = 9 × 2 = 18