**1. In the adjacent figure, two triangles are similar find the length of the missing side**

Solution:

Let the triangles be ABC and PQR

^{BC}/_{QR} = ^{AC}/_{PR}

^{5}/_{X} = ^{13}/_{39}

_{13X = 5 X 39}

_{X = }^{5X39}/_{13}^{ }= 5 X3 = 15

**What number is to 12 is 5 is 30?**

Solution

Let x be the number

x:12 :: 5 : 30

30x = 12×5

x = ^{12×5}/_{30 } = 2

**Solve the following properties:**

**(i). x : 5 = 3 : 6**

**(ii) 4 : y = 16 : 20**

**(iii) 2 : 3 = y : 9**

**(iv) 13 : 2 = 6.5 : x**

**(v) 2 : π = x : ^{22}/_{7}**

Solution:

(i). x : 5 = 3 : 6

6x = 5 x 3

6x = 15

x = ^{15}/_{6}

(ii) 4 : y = 16 : 20

4×20 = 16y

y = ^{4×20}/_{16}

y = 5

(iii) 2 : 3 = y : 9

2×9 = 3y

y = ^{2×9}/_{3} = 2×3 = 6

(iv) 13 : 2 = 6.5 : x

13x = 2 x 6.5

13x = 13

x = ^{13}/_{13} = 1

(v) 2 : π = x : ^{22}/_{7}

2x^{22}/_{7} = πx

x = (2x^{22}/_{7}) /_{π} =(2x^{22}/_{7}) /(^{22}/_{7})

x = 2

**find the mean proportion to :**

**(i) 8, 16**

**(ii) 0.3, 2.7**

**(ii)16 ^{2}/_{3} , 6**

**(iv) 1.25, 0.45**

Solution:

(i) 8, 16

Let x be the mean proportion to 8 and 16

Then ^{8}/_{x} = ^{x}/_{16}

x^{2} = 8 x 16 = 128

x = √128 = √(64×2) = 8√2

(ii) 0.3, 2.7

Let x be the mean proportion to 0.3 and 2.7

Then ^{0.3}/_{x} = ^{x}/_{2.7}

x^{2} = 0.3 x 2.7 = 0.81

x = √(0.81) = 0.9

(ii)16^{2}/_{3} , 6

Let x be the mean proportion to 16^{2}/_{3} and 6

Then (16^{2}/_{3}) /_{x} = ^{x}/_{6}

x^{2} = 16^{2}/_{3} x 6 = ^{50}/_{3} x 6 = 100

x = √100 = 10

(iv) 1.25, 0.45

Let x be the mean proportion to 1.25 and 0.45

Then ^{1.25}/_{x} = ^{x}/_{0.45}

x^{2} = 1.25 x 0.45

x = √(1.25 x 0.45) = √(1.25 x 0.45)x^{√(100×100)}/ _{√(100×100)}

= ^{√(125×45)}/_{√(100×100)} = ^{√(25x5x5x9)}/_{√(100×100)} = ^{5x5x3}/_{10×10}^{ }= ^{3}/_{4}

**Find the fourth proportion for the following:**

**(i) 2.8, 14, 3.5**

**(ii) 3 ^{1}/_{3}, 1^{2}/_{3}, 2^{1}/_{2}**

**(iii)1 ^{5}/_{7}, 2^{3}/_{4}, 3^{3}/_{5}**

Solution:

(i) 2.8, 14, 3.5

Let x be the fourth proportion

Then, 2.8 : 14 :: 3.5 : x

2.8x = 14×3.5

x = ^{14×3.5}/_{2.8} = 17.5

(ii) 3^{1}/_{3}, 1^{2}/_{3}, 2^{1}/_{2}

Let x be the fourth proportion

Then, 3^{1}/_{3}: 1^{2}/_{3} :: 2^{1}/_{2}: x

^{10}/_{3} :^{5}/_{3} : : ^{5}/_{2} : x

^{10}/_{3}x = ^{5}/_{3} x ^{5}/_{2}

^{10}/_{3}x = ^{25}/_{6}

x = ^{25}/_{6} x ^{3}/_{10} = ^{75}/_{60} = ^{15}/_{12} = ^{5}/_{4}

(iii)1^{5}/_{7}, 2^{3}/_{14}, 3^{3}/_{5}

Let x be the fourth proportion

Then, 1^{5}/_{7}: 2^{3}/_{14}:: 3^{3}/_{5}: x

^{12}/_{7} :^{31}/_{14} : : ^{18}/_{5} : x

^{12}/_{7 }x = ^{31}/_{14} x ^{18}/_{5}

^{12}/_{7 }x = ^{31×18}/_{14×5}

x = ^{31×18}/_{14×5} x ^{7}/_{12} = ^{31×3}/_{5×4} = ^{93}/_{20} = 4^{13}/_{20}

**Find the third proportion to:**

**(i) 12, 16**

**(ii) 4.5, 6**

**(iii) 5 ^{1}/_{2} , 16^{1}/_{2}**

Solution:

(i) 12, 16

Let x be the third proportion

Then 16:12 :: x : 16

12x = 16 x16

x = ^{16×16}/_{12} = ^{64}/_{3} = 21^{1}/_{3}

(ii) 4.5, 6

Let x be the third proportion

Then 6:4.5 :: x : 6

4.5x = 6 x6

x = ^{6×6}/_{4.5} = ^{36}/_{4.5} = ^{360}/_{45} = 8

(iii) 5^{1}/_{2} , 16^{1}/_{2}

Let x be the third proportion

Then 16^{1}/_{2}: 5^{1}/_{2}:: x : 16^{1}/_{2}

^{11}/_{2} x = ^{33}/_{2} x ^{33}/_{2}

x = ^{33}/_{2} x ^{33}/_{2} x ^{2}/_{11}= ^{33×3}/_{2} = ^{99}/_{2} = 49^{1}/_{2}

- I
**n a map**^{1}/_{4 }cm represents 25km, if two cities are 2^{1}/_{2}c apart on the map, what is the actual distance between them?

Solution:

Let 2^{1}/_{2 }cm represemts x km

^{1}/_{4}cm: 25km :: 2^{1}/_{2}cm : x km

^{1}/_{4} x x = 25 x 2^{1}/_{2}

^{x}/_{4} = ^{25 x5 }/_{2}

x = ^{25×5}/_{2} x 4 = 25x5x2 = 250km

**Suppose 30out of 500 components for a compter were found defective. At this aret how many defective components would he found in 1600 components?**

Solution:

Number of defective components in 500 components = 30

Let x be the number of defecrive components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = ^{30×1600}/_{500 }=96

## 1 thought on “Ratio and Proportions Exercise 2.4.2 – Class IX”

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