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6. Angles Q and R of a  \small \Delta PQR  are  \small 25^{\circ}  and  \small 65^{\circ}. Write which of the following is true:

            (i)  \small PQ^2+QR^2=RP^2

            (ii) \small PQ^2+RP^2=QR^2

            (iii) \small RP^2+QR^2=PQ^2

        

As we know the sum of the angles of any triangle is always 180. So, Now. Since PQR is a right-angled triangle with right angle at P. So   Hence option (ii) is correct.

4.  Which of the following can be the sides of a right triangle?

          (i) \small 2.5\hspace{1mm} cm, \small 6.5\hspace{1mm} cm, 6 cm.

          (ii) 2 cm, 2 cm, 5 cm.

         (iii) \small 1.5\hspace{1mm} cm, 2cm, \small 2.5\hspace{1mm} cm.

          In the case of right-angled triangles, identify the right angles.

As we know,  In a Right-angled Triangle: By Pythagoras Theorem, (i) , , 6 cm. As we know the hypotenuse is the longest side of the triangle, So Hypotenuse = 6.5 cm Verifying the Pythagoras theorem,   Hence it is a right-angled triangle. The Right-angle lies on the opposite of the longest side (hypotenuse) So the right angle is at the place where 2.5 cm side and 6 cm side meet. (ii) 2 cm, 2...

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of  the ladder from the wall.    

            

Here. As we can see, The ladder with wall forms a right-angled triangle with 

the vertical height of the wall = perpendicular = 12 m

length of ladder = Hypotenuse = 15 m

Now, As we know

In a Right-angled Triangle: By Pythagoras Theorem,

(Hypotenus)^2=(Base)^2+(Perpendicular)^2

(15)^2=(Base)^2+(12)^2

(Base)^2=(15)^2-(12)^2

(Base)^2=225-144

(Base)^2=81

Base=9 m

Hence the distance of the foot of the ladder from the wall is 9 m.

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2. ABC is a triangle, right-angled at C. If \small AB = 25\hspace{1mm} cm and \small AC = 7\hspace{1mm} cm, find BC.
 

As we know,  In a Right-angled Triangle: By Pythagoras Theorem, As ABC is a right-angled triangle with  Base = AC= 7 cm. Perpendicular = BC Hypotenuse = AB = 25 cm So, By Pythagoras theorem, Hence, Length od BC is 24 cm.

1.  PQR is a triangle, right-angled at P. If  \small PQ = 10\hspace{1mm}cm  and  \small PR = 24\hspace{1mm}cm, find QR.
 

As we know,  In a Right-angled Triangle: By Pythagoras Theorem, As PQR is a right-angled triangle with  Base = PQ = 10 cm. Perpendicular = PR = 24 cm. Hypotenuse = QR So, By Pythagoras theorem, Hence, Length od QR is 26 cm.  

Q. Find the unknown length x in the following figures (Fig \small 6.29):

    
 

As we know in a Right-angled Triangle: By Pythagoras Theorem, So, using this theorem, i)  . ii) iii) iv) v) in this question as we can see from the figure, it is making the right angle with the half-length of x, so . vi)

5.  ABCD is quadrilateral. Is
    \small AB+BC+CD+DA< 2(AC+BD)            ?
        

Let the intersection point of the two diagonals be O. As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). So, In  : In  : In  : In  : Now, Adding all four equations we, get which can also be expressed as  Hence this is true.

4.  ABCD is a quadrilateral.
        Is  \small AB+BC+CD+DA> AC+BD ?

            

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). So, In  : In  : Adding (1) and (2) we get, Hence The given statement is True.  

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Let ABC be a triangle with AB = 12cm and BC = 15cm Now As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). AB + BC > CA 12 + 15 > CA CA < 27......(1) Also, in a similar way AB + CA  > BC  CA > BC -  AB CA > 15 - 12  CA > 3............(2) Hence from (1) and (2), we can say that the length of third side of the triangle must be...

3.  AM is a median of a triangle ABC.
            Is \small AB+BC+CA> 2AM ?
(Consider the sides of triangles \small \Delta ABM and \small \Delta AMC.)

        

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). So, In  : In  : Adding (1) and (2), we get As we can see M is the point in line BC So, we can say So our equation becomes . Hence it is a True statement.  

2. Take any point O in the interior of a triangle PQR. Is

            (i)  \small OP+OQ> PQ ?
            (ii) \small OQ+OR> QR ?
            (iii) \small OR+OP> RP?

        

i) As POQ is a triangle, the sum of any two sides will always be greater than the third side. so  Yes,   . ii) As ROQ is a triangle, the sum of any two sides will always be greater than the third side. so  Yes,   iii) As ROQ is a triangle, the sum of any two sides will always be greater than the third side. so  Yes,

1. Is it possible to have a triangle with the following sides?

            (i) 2 cm, 3 cm, 5 cm             (ii) 3 cm, 6 cm, 7 cm
            (iii) 6 cm, 3 cm, 2 cm

As we know, According to the Triangle inequality law, the sum of lengths of any two sides of a triangle would always greater than the length of the third side. So Verifying this inequality by taking all possible combinations, we have, (i) 2 cm, 3 cm, 5 cm 3 + 5 > 2 ----> True 2 + 5 >  3---->  True 2 + 3 > 5 ---->False Hence the triangle is not possible. (ii) 3 cm, 6 cm, 7 cm  3 + 6 > 7 ----->...

2.  Find angles x and y in each figure.

i) As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal. So, Now,  As we know the sum of internal angles of a triangle is 180. so, Hence, . ii)  As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal. AND the sum of internal angles of a triangle is 180. so, Also, . Hence . iii) As we...

1. Find angle x in each figure: 

        

As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal. and the sum of angles of the triangle is equal to . So, i)  ii)  iii)  iv)  v) vi) vii) viii) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle. So,  ix)As we know when two lines are intersecting, the opposite angles are equal. So

2. Find the values of the unknowns x and y in the following diagrams

        

  

i) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle. Now, As we know the sum of internal angles of a triangle is 180. so, Hence, . ii)  As we know when two lines are intersecting, the opposite angles are equal. So Now, As we know the sum of internal angles of a triangle is 180. so, Hence, . iii) As we know, the exterior angle is equal to...

1.  Find the value of the unknown x in the following diagrams

        

As we know that the sum of the internal angles of the triangle is equal to . So, i)  ii)  iii)  iv)  v)  vi) 

2. Find the value of the unknown interior angle \small x in the following figures:

        

As we know that the exterior angle is equal to the sum of the opposite internal angles. So,  i)  ii)  iii)  iv)  v)  vi) 

1. Find the value of the unknown exterior angle x in the following diagrams:

            

As we know that the exterior angle is equal to the sum of the opposite internal angles. So,  i)  ii)  iii)  iv)  v)  vi)   

3.  Is something wrong in this diagram (Fig \small 6.12)? Comment.
 

            

Yes, The measure of the exterior angle is given wrong. As we know that in a triangle, the exterior angle is equal to the sum of opposite interior angles. So, Exterior angle =                         =  Hence the exterior angle should be equal be  instead of .

1.  Exterior angles can be formed for a triangle in many ways. Three of them are shown here (Fig  \small 6.10)

There are three more ways of getting exterior angles. Try to produce those rough sketches.

In a triangle, there are a total of six exterior angles Exterior Angles in a Triangle  
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