Question 1: In the adjoining figure, is an obtuse triangle, obtuse at . If , prove that .

Answer:

Given is an obtuse triangle, is obtuse angle

In , we have

(by Pythagoras theorem)

In , we have

Hence proved.

Question 2: In the adjoining figure, of is acute angle and . Prove that

Answer:

In

Therefore (by Pythagoras theorem)

Similarly, in , we have

(by Pythagoras theorem)

Hence proved.

Question 3: Prove in any triangle, the sum of the square of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.

Answer:

Construction: Altitude,

The , is obtuse angle triangle, and is an acute angle triangle.

In is obtuse and .

Therefore … … … … … i)

Similarly, in is acute at and

Therefore

(since ) … … … … … ii)

Adding i) and ii) we get

… … … … … ii)

Putting in iii) we get

Hence proved.

Question 4: Prove that three times the sum of the square of the sides of a triangle is equal to 4 times the sum of square of medians of the triangle.

Answer:

Given and and are medians as shown in the diagram.

In , with as the median of , we have

… … … … … i)

Similarly with bisecting and bisecting , we get

and

Adding i), ii) and iii) we get

Hence proved.

Question 5: and are mid points of side and respectively in . is . Prove that (i) (ii) (iii)

Answer:

(i) In

(by Pythagoras theorem)

(multiply by 4 on both sides)

(since )

(ii) In ,

(since )

(iii) From i) and ii) we have

Adding, we get

(since we have )

Question 6: In the adjoining figure, is right angles at . and are two medians drawn from vertices and respectively. If is and , find the length of .

Answer:

In

Similarly in

Question 7: In . Let and and let be the length of the perpendicular from to . Prove that

(i)

(ii)

Answer:

Consider the figure

(i) , then

Area of

Also the area of

Therefore

(ii) In , therefore

(since )

Question 8: In an equilateral triangle with side , prove that (i) Altitude (ii) Area

Answer:

Consider adjoining figure.

(i) Draw

is the mid point since is equilateral with

Since is right angled at

(ii) Now Area of

Question 9: is a triangle in which and is any point on . Prove that

Answer:

Draw

In and , we have

,

is common

and (since )

Since and are right angles at

Question 10: From a point in the interior of , perpendiculars and are drawn to side and respectively. Prove:

i)

ii)

Answer:

Let be the point inside and let and

i) In and we have

and

Adding the above three

ii) In right triangle and we have

Similarly,

Adding i) , ii) and iii)

Hence proved.

Question 11: A point inside a rectangle is joined with each of the vertices and . Prove that

Answer:

Join to and

Draw a line through

In and we have

… … … … … i)

Similarly, in and we have

… … … … … ii)

Since and

Therefore from i) and ii) we get

Question 12: In is mid point of and . Prove:

i)

ii)

iii)

Answer:

Given

and

i) In is an obtuse angle

ii) In is an acute angle

iii) From i) and ii) we get

Question 13: In an equilateral is trisected at . Prove that .

Answer:

Draw

Join

In and , we have

and is common

Therefore we have

Since (equilateral triangle)

Therefore is an acute triangle

(since )

Question 14: In adjoining figure is a right angled triangle. and point and trisect . Prove

Answer:

Given

Let

and

In and we have

and

Now

Question 15: is a right angled triangle. . A circle is inscribed in it. The lengths of the two sides containing the right angle are and . Find the radius of the circle.

Answer:

In , we have

Now, Area of

Question 16: In adjoining figure, is the mid point of and . If and , prove that:

i)

ii)

iii)

Answer:

In … … … … … … i)

In … … … … … … ii)

In … … … … … … iii)

(ii) From i) and ii)

or

(i) from ii) and iii) we get

(iii) Adding (ii) and (iii) we get

Question 17: In the adjoining figure, . Prove:

i)

ii)

Answer:

In … … … … … … i)

In … … … … … … ii)

(i) We know,

or

(ii) From i and ii)

or

Question 18: is right angled. and . Prove that:

i)

ii)

iii)

iv)

Answer:

In … … … … … … i)

In … … … … … … ii)

In … … … … … … iii)

(ii)

(i)

(iii)

(iv) Dividing i) by ii) we get