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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.  

5.  A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
 

As we can see the tree makes a right angle with  Perpendicular = 5 m Base = 12 m As we know,  In a Right-angled Triangle: By Pythagoras Theorem, Here, The Hypotenuse of the triangle was also a part of the tree originally, So The Original height of the tree = height + hypotenuse                                                   = 5 m + 13 m                                                 ...

8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
 

As we know that the diagonals of the rhombus are perpendicular to each other and intersect at a point which is mid of both the diagonal. So. By Pythagoras Theorem we can say that Hence Side of the rhombus is 17 cm. So, The Perimeter of the rhombus = 4 x 17 cm                                                  = 68 cm. Hence, the perimeter of the rhombus is 68 cm.

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
 

As we can see in the rectangle,  By Pythagoras theorem, Now as given in the question, Diagonal = 41 cm. Length = 40 cm. So, Putting these value we get, Hence the width of the rectangle is 9 cm. So  The perimeter of the rectangle = 2 ( Length + Width )                                                  = 2 ( 40 cm + 9 cm )                                                  = 2 x 49 cm        ...
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