##### Triangles Part-1 || Chapter 6 || Class 10th (For Hindi Medium)

##### Triangles Part-2 || Chapter 6 || Class 10th (For Hindi Medium)

**Introduction to **Triangles

__What is Triangle?__

A triangle can be defined as a polygon which has three angles and three sides. The interior angles of a triangle sum up to 180 degrees and the exterior angles sum up to 360 degrees. Depending upon the angle and its length, a triangle can be categorized in the following types-

- Scalene Triangle – All the three sides of the triangle are of different measure
- Isosceles Triangle – Any two sides of the triangle are of equal length
- Equilateral Triangle – All the three sides of a triangle are equal and each angle measures 60 degrees
- Acute angled Triangle – All the angles are smaller than 90 degrees
- Right angle Triangle – Anyone of the three angles is equal to 90 degrees
- Obtuse-angled Triangle – One of the angles is greater than 90 degrees

**Similarity Criteria of Two Polygons Having the Same Number of Sides**

Any two polygons which have the same number of sides are similar if the following two criteria are met-

- Their corresponding angles are equal, and
- Their corresponding sides are in the same ratio (or proportion)

**Similarity Criteria of Triangles**

To find whether the given two triangles are similar or not, it has four criteria. They are:

**Side-Side- Side (SSS) Similarity Criterion –**When the corresponding sides of any two triangles are in the same ratio, then their corresponding angles will be equal and the triangle will be considered as similar triangles.**Angle Angle Angle (AAA) Similarity Criterion**– When the corresponding angles of any two triangles are equal, then their corresponding side will be in the same ratio and the triangles are considered to be similar.**Angle-Angle (AA) Similarity Criterion –**When two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are considered as similar.**Side-Angle-Side (SAS) Similarity Criterion –**When one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are said to be similar.

**Proof of Pythagoras Theorem**

**Statement: **As per **Pythagoras theorem, **“In a right-angled triangle, the sum of squares of two sides of a right triangle is equal to the square of the hypotenuse of the triangle.”

**Proof –**

Consider the right triangle, right-angled at B.

Construction-

Draw BD ⊥ AC

Now, △ADC ~ △ABC

So, AD/AB = AB/AC

or AD. AC = AB^{2} ……………(i)

Also, △BCD ~ △ ABC

So, CD/BC = BC/AC

or, CD. AC = BC^{2} ……………(ii)

Adding (i) and (ii),

AD. AC + CD. AC = AB^{2} + BC^{2}

AC(AD + DC) = AB^{2} + BC^{2}

AC(AC) = AB^{2} + BC^{2}

⇒ AC^{2} = AB^{2} + BC^{2}

Hence, proved.

Solved Example

**Question:**

In a right-angled triangle ABC, which is right-angled at A, where CM and BL are the medians of a triangle. Show that, 4(BL^{2} +CM^{2}) = 5 BC^{2}

**Solution:**

Given that,

Medians BL and CM, ∠A = 90°

From the triangle ABC, we can write it as:

BC^{2} = AB^{2} + AC^{2} (Using Pythagoras Theorem) …(1)

From the triangle, ABL,

BL^{2} = AL^{2} +AB^{2}

or we can write the above equation as:

BL^{2} =(AC/2)^{2} +AB^{2} (Where L is the midpoint of AC)

BL^{2} = (AC^{2}/4) + AB^{2}

4BL^{2} = AC^{2} + 4 AB^{2} ….(2)

From triangle CMA,

CM^{2} = AC^{2} + AM^{2}

CM^{2} = AC^{2} + (AB/2)^{2} (Where M is the midpoint of AB)

CM^{2} = AC^{2} + AB^{2}/4

4CM = 4 AC +AB ….(3)

Now, by adding (2) and (3), we get,

4(BL^{2}+ CM^{2}) = 5(AC^{2}+ AB^{2})

Using equation (1), we can write it as:

4(BL^{2}+ CM^{2}) = 5 BC^{2}

Hence, it is proved.

**Problems Related to Triangles**

- A girl having a height of 90 cm is walking away from a lamp-post’s base at a speed of 1.2 m/s. Calculate the length of that girl’s shadow after 4 seconds if the lamp is 3.6 m above the ground.
- S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Now, prove that triangle RPQ and triangle RTS are similar.
- E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD

at F. Show that triangles ABE and CFB are similar.