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The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

#### Options

20 cm

30 cm

40 cm

50 cm

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#### Solution

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter* *is given by,

`A = sqrt(s(s-a)(s-b)(s-c))`, where

`s = (a+b+c)/2`

Therefore the area of a triangle, say *A* having sides 50 cm, 78 cm and 112 cm is given by

a = 50 cm ; b = 78 cm ; c = 112 cm

`s = (50+78+112)/2`

`s = 240/2`

s = 120 cm

`A = sqrt(120 (120-50)(120-78)(120-112))`

`A = sqrt((120(70)(42)(8))`

`A = sqrt(2822400)`

A = 1680 cm^{2}

The area of a triangle, having *p* as the altitude will be,

Where, *A =* 1680 cm^{2}

We have to find the smallest altitude, so will substitute the value of the base *AC *with the length of each side one by one and find the smallest altitude distance i.e. *p*

**Case 1 **

**AC = 50 cm **

**1680 = `1/2 (50 xx p)`**

**1680 `xx 2 = 50 xx p `**

**`p = (1680 xx 2)/50`**

**`p = 67.2 cm `**

**Case 2**

**`AC = 78 cm `**

**`1680 = 1/2 (78 xx p)`**

**`1680 xx 2 = 78 xx p `**

**`p = (1680xx2)/78`**

**p = 43 cm **

**Case 3**

**Ac = 112 cm **

**`1680 =1/2 (112 xx p)`**

**`1680 xx 2 = 112 xx p `**

**`p = (1680 xx 2 )/112`**

** p = 30 cm **

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