Congruent Triangles - Exercise 12.6

# Congruent Triangles - Exercise 12.6 | Extra Documents & Tests for Class 9 PDF Download

``` Page 1

Question:36
In ?ABC, if ?A = 40° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In the triangle ABC it is given that
Page 2

Question:36
In ?ABC, if ?A = 40° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In the triangle ABC it is given that
We have to find the longest and shortest side.
Here
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence will be the longest
Since is the shortest angle so that side in front of it will be the shortest.
And  is shortest side
Hence Is longest and  is shortest.
Question:37
In a ?ABC, if ?B = ?C = 45°, which is the longest side?
Solution:
In the triangle ABC it is given that
We have to find the longest side.
Here
(Since )
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence BC will be the longest side.
Question:38
In ? ABC, side AB is produced to D so that BD = BC. If ?B = 60° and ?A = 70°, prove that :
i
Solution:
It is given that
Page 3

Question:36
In ?ABC, if ?A = 40° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In the triangle ABC it is given that
We have to find the longest and shortest side.
Here
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence will be the longest
Since is the shortest angle so that side in front of it will be the shortest.
And  is shortest side
Hence Is longest and  is shortest.
Question:37
In a ?ABC, if ?B = ?C = 45°, which is the longest side?
Solution:
In the triangle ABC it is given that
We have to find the longest side.
Here
(Since )
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence BC will be the longest side.
Question:38
In ? ABC, side AB is produced to D so that BD = BC. If ?B = 60° and ?A = 70°, prove that :
i
Solution:
It is given that
, and
We have to prove that
1
2
1
Now
And since BD=BC, so , and

That is,
Now
And , so

Hence
1
Sideinfrontofgreateranglewillbelonger
And
2  Proved.
Question:39
Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
As we know that a triangle can only be formed if
The sum of two sides is greater than the third side.
Page 4

Question:36
In ?ABC, if ?A = 40° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In the triangle ABC it is given that
We have to find the longest and shortest side.
Here
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence will be the longest
Since is the shortest angle so that side in front of it will be the shortest.
And  is shortest side
Hence Is longest and  is shortest.
Question:37
In a ?ABC, if ?B = ?C = 45°, which is the longest side?
Solution:
In the triangle ABC it is given that
We have to find the longest side.
Here
(Since )
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence BC will be the longest side.
Question:38
In ? ABC, side AB is produced to D so that BD = BC. If ?B = 60° and ?A = 70°, prove that :
i
Solution:
It is given that
, and
We have to prove that
1
2
1
Now
And since BD=BC, so , and

That is,
Now
And , so

Hence
1
Sideinfrontofgreateranglewillbelonger
And
2  Proved.
Question:39
Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
As we know that a triangle can only be formed if
The sum of two sides is greater than the third side.
Here we have 2 cm, 3 cm and 7 cm as sides.
(Since 5 is less than 7)
Hence the sum of two sides is less than the third sides
So, the triangle will not exist.
Question:40
In ? ABC, ?B = 35°, ?C = 65° and the bisector of ?BAC meets BC in P. Arrange AP, BP and CP in descending order.
Solution:
It is given that
AP is the bisector of
We have to arrange ,  and  in descending order.
In  we have
(As AP is the bisector of )
So
Sidesinfrontorgreateranglewillbegreater              ........
1
In  we have
(As AP is the bisector of )
Since,
So                    ..........2
Hence
From
1 &
2 we have
Question:41
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Solution:
We have to prove that the perimeter of a triangle is greater than the sum of its altitude.
In
, ,
Page 5

Question:36
In ?ABC, if ?A = 40° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In the triangle ABC it is given that
We have to find the longest and shortest side.
Here
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence will be the longest
Since is the shortest angle so that side in front of it will be the shortest.
And  is shortest side
Hence Is longest and  is shortest.
Question:37
In a ?ABC, if ?B = ?C = 45°, which is the longest side?
Solution:
In the triangle ABC it is given that
We have to find the longest side.
Here
(Since )
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence BC will be the longest side.
Question:38
In ? ABC, side AB is produced to D so that BD = BC. If ?B = 60° and ?A = 70°, prove that :
i
Solution:
It is given that
, and
We have to prove that
1
2
1
Now
And since BD=BC, so , and

That is,
Now
And , so

Hence
1
Sideinfrontofgreateranglewillbelonger
And
2  Proved.
Question:39
Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
As we know that a triangle can only be formed if
The sum of two sides is greater than the third side.
Here we have 2 cm, 3 cm and 7 cm as sides.
(Since 5 is less than 7)
Hence the sum of two sides is less than the third sides
So, the triangle will not exist.
Question:40
In ? ABC, ?B = 35°, ?C = 65° and the bisector of ?BAC meets BC in P. Arrange AP, BP and CP in descending order.
Solution:
It is given that
AP is the bisector of
We have to arrange ,  and  in descending order.
In  we have
(As AP is the bisector of )
So
Sidesinfrontorgreateranglewillbegreater              ........
1
In  we have
(As AP is the bisector of )
Since,
So                    ..........2
Hence
From
1 &
2 we have
Question:41
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Solution:
We have to prove that the perimeter of a triangle is greater than the sum of its altitude.
In
, ,
We have to prove
Since
So  and
........
1
Now consider  then
, and
Again consider
, and
1,
2 and
3, we get
Hence the perimeter of a triangle is greater than the sum of all its altitude.
Question:42
In the given figure, prove that:
i
CD + DA + AB + BC > 2AC
ii
CD + DA + AB > BC

Solution:
1 We have to prove that
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## Extra Documents & Tests for Class 9

1 videos|228 docs|21 tests

## Extra Documents & Tests for Class 9

1 videos|228 docs|21 tests

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