Introduction to Learn high school geometry problems

 Learning high school Geometry is a module of math which involves about the study of shapes, lines, angles, dimensions, etc. Through learning Geometry was the initial method of thoughts in which a small number of easy problems were assumed and then in high school we used to derive more complex ones. In high school geometry we are going to study how to solve geometry problems on rectangle, circle and triangles. It plays vital role in real time application like elevation, projection. Learning geometry provides many foundational skills and helps to build the thinking skills of logic, deductive reasoning, and analytical reasoning.

 

 

Learn high school geometry theorem based problems

 

We going to learn about high school geometry, which deals with some of the complicated solution as not like middle or lower school geometry. In high school geometry angles in a circle, when two points say A and B of a circle splits the circumference of the circle into two parts called the arcs of the circle. If the two parts are unequal, the smaller part is called the

minor arc and the larger one is called the major arc.

ACB is a major arc and ADB is a minor arc.

Theorem : The sum of the opposite angles of a cyclic quadrilateral is 180.

Given: ABCD is a cyclic quadrilateral in a circle with centre O.

To prove: m∠A + m∠C = 180°, m∠B + m∠D = 180°

Construction: Join B and D to the centre O

Proof: m∠BCD = ½ m∠BOD

(Angle at the centre = Twice angle at any point on the circumference)

m∠BAD = ½ reflex m∠BOD (same theorem)

m∠BCD + m ∠BAD = ½ m∠BOD + ½ reflex m∠BOD

= ½ (m∠BOD + reflex m∠BOD)

= ½ *360° = 180°

m∠A + m∠C = 180°

m∠A + m∠B + m∠C + m∠D = 360°

(The sum of angles of a quadrilateral is 360)

But m∠A + m∠C = 180° (proved) _ m∠B + m∠D = 180°

Hence the theorem is proved.

Therefore the opposite angles in a cyclic quadrilateral are supplementary.

Thus the theorem is explained.

 

 

Example for Learn high school geometry problems

 

Example for high school geomentry problems: In the figure O is the centre. m∠OAC = 35° and m∠OBC = 45° find m∠AOB

Solution : Join OC

OA = OB = OC; m∠OCA = m∠OAC = 35°

m∠OCB = m∠OBC = 45°;

m∠ACB = m∠OCA + m∠OCB

= 35° + 45° = 80°;

m∠AOB = 2m ∠ACB = 2 * 80° = 160°

Thus the high school geomentry problems are learn through the above solution.