Question

.ΔPQR~ΔXYZ, PQ=2,QR=3,PR=5 and perimeterof Δxyz=20 then arΔPQR/arΔXYZ= 

1/2

2/4

1/4

1/9​

Answer 1

Given :

• ∆PQR $\sim$ ∆ XYZ
• PQ = 2
• QR = 3
• PR = 5
• Perimeter (∆ XYZ) = 20

To find:

• ar ( ∆ PQR) / ar (∆ XYZ)

Solution :

$\because$ ∆ PQR $\sim$ ∆ XYZ ,

$\tt \dfrac{PQ}{XY} = \dfrac{QR}{YZ} = \dfrac{PR}{XZ}$

PQ : QR : PR = 2 : 3 : 5

•°• XY : YZ : XZ = 2:3: 5

• Given : Perimeter of ∆ XYZ = 20

Let the common ratio between sides be x .

Therefore ,

• XY = 2x
• YZ = 3x
• XZ = 5x

•°• XY + YZ + XZ = 20

2x + 3x + 5x = 20

10x = 20

x = 20/10

x = 2

Therefore ,

• XY = 2 × 2 = 4
• YZ = 3 × 2 = 6
• XZ = 5 × 2 = 10

Now , we know that the ratio of area of similar triangles is equal to the square of the ratios of the corresponding sides.

ar ( ∆PQR) / ar (∆XYZ) = PQ² / XY²

= 2² / 4²

= 4/ 16

= 1/4

$\boxed{\therefore \sf \dfrac{ar(\Delta PQR)}{ar (\Delta XYZ)} = \dfrac{1}{4}}$