The Triangle and Its Properties - Free MCQ Practice Test with solutions,

In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Every triangle has exactly three medians, one from each vertex, and they all intersect each other at the triangle's centroid.

• A triangle has three altitudes.
• An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side.
• Each vertex of a triangle can have an altitude drawn to the opposite side.
• These altitudes can be inside or outside the triangle, depending on the type of triangle (acute, right, or obtuse)
• Therefore, option C: 3 is correct.

By Pythagoras property,
AB² = AC² + BC²
= 5² + 12² = 25 + 144 = 169 
or AB²= 169. So, AB = 13 or the length of AB is 13 cm.

In Triangle ΔPQR, the side opposite to vertex Q is the side that does not touch vertex Q.

The side opposite to Q would be side PR, as it does not include the vertex Q.

for example, 

The vertex opposite to side RT is S.

Step-by-step explanation:

In the given question,

We have a triangle RST which has sides RS, ST and TR.

We know that a triangle consists three vertices and three sides each of which is particularly opposite to any one of the side of the triangle and vice-versa.

Therefore, we can see here that in the triangle RST.

The side RS is opposite to the vertex T.

The side ST is opposite to the vertex R.

and,

The side TR is opposite to the vertex S.

Hence, the correct answer will be S as side RT is opposite to S.

• In triangle Δ PQR, PM is identified as the altitude.
• An altitude is a line segment from a vertex to the opposite side, forming a right angle (90°) with that side.
• In the provided diagram, PM is drawn from vertex P to side QR, intersecting at a right angle.
• This perpendicular relationship confirms PM as the altitude, as it meets the defining criteria of an altitude in a triangle.In Δ PQR, PM is altitude

An exterior angle of a triangle is equal to the sum of the opposite interior angles

x^o = 70^o + 50^o

x^o = 120^o .

So option C is the correct answer. 

An exterior angle of a triangle is equal to the sum of the opposite interior angles

So in the given figure 

110^o = 50^o + x 

110^o - 50^o = x 

60^o = x 

So option B is correct answer. 

An exterior angle of a triangle is equal to the sum of the opposite interior angles

So in the given figure 

80^o = 30^o + x 

it implies x = 50^o. 

The sum of the angles of a triangle is 180°.

As the ratio of the angles of the triangle is 4:5:9 they can be taken to be 4x, 5x 

⇒ 9x. 4x + 5x + 9x = 180°

⇒ 18x = 180°

⇒ x = 10°

This gives the angles of the triangle as 4x10=40°, 5x10=50° and 9x10=90° .

Check using Pythagoras theorem:
a²+b²=c²
we find 
3²+4²=5²
9 + 16 = 25

Since the sum of squares of the smaller two sides equals the square of the largest side, it forms a right angle

One angle of the triangle = 110°
Let the other two angles each be x
Sum of all the angles of a triangle = 180°
x + x + 110 = 180
2x = 180 - 110
2x = 70
x = 35°
The other two angles are 35°, 35°.

Let the angles be x
sum of the angles of a triangle=180
5x+3x+x=180
9x=180
x=20
angle I = 5x = 5*20 = 100
angle II = 3x = 3*20 = 60
angle III = x = 20

Third interior angle = 180 − (45 + 55) = 80°
Exterior angle = 180 − 80 = 100°