Question

From the figure given below, prove that: ​
Please give full answer.

Answer 1

Step-by-step explanation:

To prove :-

• ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

$\underline{\huge{\sf Prove:}}$

Given,

Two triangles having anngles 1, 2, 3, 4, 5 and 6.

Let, First triangle be ∆ABC.

And, Second triangle be ∆PQR.

Here,

We know that,

Sum of all interior angles of triangle is $\blue{\sf \bold{180\degree}}.$

So,

In ∆ABC,

$\sf \angle A + \angle B + \angle C = 180 \degree$

So,

$\sf \angle 1 + \angle 2 + \angle 3 = 180 \degree$ ---(i)

In ∆PQR,

$\sf \angle P + \angle Q + \angle R = 180 \degree$

Thus,

$\sf \star \angle 4 + \angle 5 + \angle 6 = 180 \degree$ ---(ii)

Now, add equation (i) and (ii)

$\sf \longrightarrow \angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6= 180 \degree + 180 \degree$

$\longrightarrow \purple{\boxed{\sf \angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 = 360 \degree}\star}$

$\boxed{\sf Hence \: Proved.}$

Therefore,

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

________________________________

More about triangle.

Triangles are closed figures. They have three angles and three sides. Triangles are polygon.

Area of triangle = 1/2 × base × height

Perimeter of triangle = Sum of all sides.

Answer 2

Answer:

To proof :-

Sum of ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

Let's proof

Here, the figure is a mixture of two triangle.

Let, the first triangle be ABC

and second triangle be PQR

Now,

As we know that sum of interior angle of triangle is 180⁰.

Therefore,

➸∠A + ∠B + ∠C = 180

➜ ∠1 + ∠2 + ∠3 (EQ 1)

➸ ∠P + ∠Q + ∠R = 180

➜ ∠4+ ∠5 + ∠6 (EQ 2)

Add EQ 1 and 2

$\tt\angle \: 1 + \angle2 \: + \angle 3 + \angle4 + \angle5 + \angle6 = 360$

✺ Hence proved ✺