Question

48. The sides of the triangle are 45cm 60cm and 75cm Find the length drawn to the longest side from it opposite vertex?


a 27 cm

b.6.21 cm

c. 39 cm

d. 36 cm



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Answer 1

Given,

Sides of triangle = 45 cm, 60 cm, 75 cm

Let ABC is a triangle in which BC is the longest side and AD is the altitude drawn from A.

We have to find out the length of the altitude drawn to the longest side from its opposite vertex.

Firstly we will find out the area of the ΔABC.

AB(a) = 45 cm BC(b) = 75 cm AC(c) = 60 cm

So we use the Hero's formula to find out the area.

$area \: of \: triangle \: = \sqrt{s(s - a)(s - b)(s - c)}$

s = a + b +c/2

s= (45 + 75 + 60)/2

s = 90 cm

substituting s in heron formula,

we get area = 1350 cm square

we know that Area = 1/2 × base × altitude

here base = bc

altitude = Area × 2/base = (1350 × 2)/75 = 36 cm

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Answer 2

Answer:-

Option D)

Explanation:-

Given,

Sides of Triangle:-

• 45 cm
• 60 cm
• 75cm

To Find,

• The length drawn to the longest side of opposite vertex.

Solution,

Let ABC be the triangle in which BC is the longest side and AD is the altitude drawn from A.

Firstly, Let's find the area of ∆ABC by heron's formula.

$\star{ \underline {\boxed {\sf {\pmb{area \: of \: triangle \: \sqrt{s(s - a)(s - b)(s - c)} }}}}}$

Here,

• s = semi perimeter of the triangle
• a = AB = 45cm
• b = BC = 75cm
• c = AC = 60cm

Semi perimeter =

$\: \: \: \: \: \quad\sf{s = \frac{a + b + c}{2} } \\ \: \: \: \: \: \qquad \sf{ s = \frac{45 + 75 + 60}{2} } \\ \: \: \: \: \: \qquad\sf{s = \frac{180}{2}} = \cancel\frac{180}{2} \\ \: \: \: \: \: \: \qquad \star \underline {\boxed{ \tt {\pmb{s = 90cm}}}}$

Area of ∆ABC=

$\sf \sqrt{90(90 - 45)(90 - 75)(90 - 60)}$

$\sf \sqrt{90 \times 45 \times 15 \times 30} = \sqrt{182250}$

$= \sf1350 {cm}^{2}$

Again calculating the area by using formula:-

$\star {\underline{\boxed{\sf {\pmb {Area = \frac{1}{2} \times base \times height}}}}}$

Here,

Base = BC (75cm) and Height = AD

${ \large \sf \frac{1}{2}} \sf \times 75 \times AD= 1350$

$\sf75 \times AD = 1350 \times 2$

$\sf{AD = \frac{1350 \times 2}{75} } = \large\frac{ {}^{18} \cancel{1350} \times 2} {\cancel{75} }$

$\sf AD = 18 \times 2$

$\star {\underline{\boxed{\frak {\pmb {AD = 36cm}}}}}$

∴The length of the altitude drawn to the longest side from its opposite vertex is 36cm