Math, asked by maahira17, 1 year ago

In the given figure, given that ∆ABC ∼ ∆PQR and quad ABCD ∼ quad PQRS. Determine the value of x, y, z in each case.

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Answered by nikitasingh79
33

SOLUTION :  

(i) Given :  

∆ABC ∼ ∆PQR , AB = 12 , BC= 7 , AC = 10 , PQ = 9 , QR = x , PR = 9  

AB/PQ = BC/QR = AC/PR

[since triangles are similar, hence corresponding sides will be proportional]

12/9 = 7/x = 10/y

12/9 = 7/x  

12x = 9 ×7

x = (9×7)/12 = (3 × 7) /4

x = 21/4  

12/9 = 10/y

12y = 10 × 9

y = (10 × 9)/12 =( 5 × 9 )/6 = 5 × 3 /2

y = 15/2  

Hence, the values of x = 21/4 & y = 15/2.

(ii)  

Given :  

∆ABCD ∼ ∆PQRS , AB = 16 , BC= 50 , DC = 50/3 , AD = 20, PQ = x , QR = y , RS = z, PS = 7

AB/PQ = BC/QR = CD/RS =DA/SP

[since quadrilaterals are similar, hence corresponding sides will be proportional]

20/7 = 16/x = 50/y = 50/3z

20/7 = 16/x

20x = 16× 7  

x = (16×7)/20

x = 4×7 / 5

x = 28/5

20/7 = 50/y

20y = 50 × 7

y =(50×7)/20

y = 35 /2

20/7 = 50/3z

20 × 3z = 50 × 7

60z = 350

z = 350/60

z = 35/6

Hence, the values of x = 28/5 , y = 35/2 & z = 35/6.

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