Question

answer the question.!!!! ​

Answer 1

Given:-

• First Side = 4cm

• Second Side = 13cm

• Third Side = 15cm

To Find:-

• The Altitude ( height ) of the triangle.

Formulae used:-

Area of ∆ = $\sf{\sqrt{ (s) ( s - a ) ( s - b ) ( s - c )}}$

Where,

• s = Half of the Perimeter
• a, b, c, = Given Sides.

Now,

• → Perimeter = a + b + c

• → Perimeter = 4 + 13 + 15 → 32

• → s = $\sf{\dfrac{Perimeter}{2}}$

• → s = $\sf{\dfrac{32}{2}= 16 }$

Therefore,

• → $\sf{Area\:of\:triangle = \sqrt{ (s) ( s - a ) ( s - b ) ( s - c )}}$

• → $\sf{\sqrt{ ( 16 ) ( 16 - 4 ) ( 16 - 13 ) ( 16 - 15 )}}$

• → $\sf{ \sqrt{ ( 16 ) ( 12 ) ( 3 ) ( 1 )}}$

• → $\sf{\sqrt{ 576}}$

• → $\sf{ 24 }$

Thus, The Area of ∆ is 24cm

Now,

• → Area of ∆ = ½ × Base × Height

• → 24 = ½ × 15 × h

• → 24 × 2 = 15h

• → 48 = 15h

• → h = 48/15

• → h = 3.2cm

Hence, The Height of ∆ is 3.2cm.

Answer 2

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