Question

The sides of a triangle are 25cm, 39cm and 56cm respectively.
Find the Given terms :
( I ) Area of the Triangle.
( II ) The altitude to the longest side of the Triangle.​

Answer 1

Step-by-step explanation:

$\huge \underbrace \bold \blue {↓Question↓}$

The sides of a triangle are 25cm, 39cm and 56cm respectively.

Find the Given terms :

( I ) Area of the Triangle.

( II ) The altitude to the longest side of the Triangle.

$\huge \underbrace \bold \blue {↓Answer↓}$

i) Area of ∆ :

p = (25 + 39 + 56) ÷ 2 = 60

Area = √(60)(60-25)(60-39)(60-56)

Area = √176400

Area = 420 cm²

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ii) Find the altitude:

Area = 1/2 x base x height

420 = 1/2 x 56 x height

420 = 28 x height

height = 420 ÷ 28

height = 15 cm

_________________________________

Therefore, Area Of the ∆ will be 420cm²

and Altitude of ∆ will be 15cm.

_________________________________

Hope it will be Helpful

Thankyou

Answer 2

Question:

• ✭ The sides of a triangle are 25cm, 39cm and 56cm respectively. Find the Given terms :

• ❶ Area of the Triangle.
• ❷ The altitude to the longest side of the Triangle.

Answer:

• ❶ Area of triangle is 420 cm².
• ❷ The altitude to the longest side of the triangle is 15 cm.

Explanation for ❶ ::

Given that:

• Sides of triangle are 25 cm, 39 cm and 56 cm.

To Find:

• Area of the ∆?

Solution:

• Firstly let's calculate the semi perimeter (s) of ∆ ::

We know that,

✪ $\bf s = \dfrac{a + b + c}{2}$ ✪

• Where, s is semi perimeter and a, b, and c are sides of ∆. We have a = 25 cm, b = 39 cm and c = 56 cm.

According to the question putting all values in formula we get,

➻ $\sf s = \dfrac{25 + 39 + 56}{2}$

➻ $\sf s = {\cancel{\dfrac{120}{2}}}$

➻ $\bf \red{s = 60\:cm}$

• Now, let's find area of ∆ ::

According to heron's formula we know that,

✪ Area of ∆ = √[s(s - a)(s - b)(s - c)] ✪

According to the question putting all values in formula we get,

➻ Area of ∆ = √[60(60 - 25)(60 - 39)(60 - 56)]

➻ Area of ∆ = √(60 × 35 × 21 × 4)

➻ Area of ∆ = √176400

➻ Area of ∆ = √(420 × 420)

➻ $\bf\purple{Area\:of\:\triangle = 420\:cm^2}$

∴ Area of the triangle is 420 cm².

Explanation for ❷ ::

Given that:

• Sides of triangle are 25 cm, 39 cm and 56 cm.

To Find:

• The altitude to the longest side?

Solution:

We know that,

✪ $\bf Area\:of\:\triangle = \dfrac{1}{2}\:\times\:Base\:\times\:Altitude$ ✪

• We have to find altitude to longest side. So, here we have longest side = 56 cm. So, we take it as base.

According to the question putting all values in formula we get,

➻ $\sf 420 = \dfrac{1}{\cancel{2}}\:\times\:\cancel{56}\:\times\:Altitude$

➻ $\sf 420 = 1\:\times\:28\:\times\:Altitude$

➻ $\sf Altitude\:\times\:28 = 420$

➻ $\sf Altitude = {\cancel{\dfrac{420}{28}}}$

➻ $\bf\pink{Altitude = 15\:cm}$

∴ The altitude to the longest side of the triangle is 15 cm.

Learn more on brainly:

✧ Question ❶ ✧

• An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

✧ Answer ❶ ✧

• brainly.in/question/43884844

✧ Question ❷ ✧

• Using Heron's formula, find the area of an isosceles triangle whose perimeter is 16cm and base is 6 cm.

✧ Answer ❷ ✧

• brainly.in/question/43923177

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