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
(ii) In , therefore
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:
… … … … … … i)
… … … … … … ii)
… … … … … … 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