Question

the side of a triangle are 50m,78mand 112m.find the s latest altitude​

Answer 1

ANSWER

let a=50cm

b=78cm

c=112

S=

2

a+b+c

=

2

50+78+112

=

2

240

=120

area =

s(s−a)(s−b)(s−c)

=

120(120−50)(120−78)(120−112)

=

2822400

=1680cm

2

longest side = 120cm

corresponding to this side as the base the altitude will be smallest

area =

2

1

×base×altitude=1680

=

2

1

×112×altitude=1680

altitude = 30cm

Answer 2

Given

The sides of a triangle are 50 m, 78 m, and 112 m respectively.

\huge{To\: Find}ToFind

(1) The area of the triangle.

(2) The length of perpendicular corresponding to the side with base 112 m

\huge{Solution}Solution

\mathcal{We\: divide \: answer\: in\: two\: parts}Wedivideanswerintwoparts

\bf{\pink{First \:part \: -}}Firstpart−

The sides of the triangle as given are - 50 m, 78 m and 112 m

Let the first side be 'a'

Second side be 'b'

Third side = 'c'

Then,

a = 50 m

b = 78 m

c = 112 m

________________________

Then,

Perimeter (P) = a + b + c

= 50 + 78 + 112 m

= 240 m

And,

Semi perimeter (s) = perimeter/2

= 240/2

= 120 m

_______________________

Now, by the heron's formula, which is used to find the area of a triangle, we get,

Area = √[s(s-a)(s-b)(s-c)]

= √[120(120-50)(120-78)(120-112)]

= √[120(70)(42)(8)]

= √[2×2×2×3×5(2×5×7)(2×3×7)(2×2×2)]

=√[2×2×2×2×2×2×2×2×3×3×5×5×7×7]

= 2×2×2×2×3×5×7 m²

= 1680 m². [Equation 1]

That's the first answer.

_________________________

\bf{\pink{Second\: part\: -}}Secondpart−

Now, base = 112 m

Let the height be 'h'

We know that the area of any triangle is equal to half the product of its base and corresponding height (perpendicular).

Hence, the area of triangle

= ½ × 112 × h

= 56 h

But, by Equation 1, the area is 1680 m².

Hence, both areas are equal.

=> 56h = 1680

=> h = 1680/56

=> h = 30 m. [Equation 2]

That is your second answer.

________________________

\mathcal{\pink{Finally,}}Finally,

From equations 1 & 2, your answers are,

\boxed{\bold{\blue{\mathcal{(1)\: = \: 1680 \: m^2}}}}

(1)=1680m

2

\boxed{\bold{\green{\mathcal{(2)\: = \: 30 \: m}}}}

(2)=30m

______________________________

\huge{\bold{\red{\mathfrak{Thank\: you}}}}Thankyou