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Mathematics

Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.

Mensuration

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Answer

Consider 34 cm, 20 cm and 42 cm as the sides of triangle.

a = 34 cm, b = 20 cm and c = 42 cm

We know that,

Semi perimeter (s) = (a+b+c)2\dfrac{(a + b + c)}{2}

Substituting the values we get,

s = (34+20+42)2=962\dfrac{(34 + 20 + 42)}{2} = \dfrac{96}{2} = 48 cm.

Area of triangle = s(sa)(sb)(sc)\sqrt{s(s - a)(s - b)(s - c)}

Substituting values we get,

A=48(4834)(4820)(4842)=48×14×28×6=112896=336 cm2.A = \sqrt{48(48 - 34)(48 - 20)(48 - 42)} \\[1em] = \sqrt{48 \times 14 \times 28 \times 6} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.

Here the shortest side of the triangle is 20 cm. Let height = h cm be the corresponding altitude.

We know that,

Area of triangle = 12\dfrac{1}{2} × base × height

Substituting the values we get,

⇒ 336 = 12\dfrac{1}{2} × 20 × h

⇒ h = 336×220\dfrac{336 \times 2}{20}

⇒ h = 33610\dfrac{336}{10}

⇒ h = 33.6 cm.

Hence, area of triangle = 336 cm2 and length of the altitude corresponding to the shortest side = 33.6 cm.

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