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D Devendra Khairwa
In the first part we have proved that   . Thus by  c.p.c.t. , we can conclude :                                                                      

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S seema garhwal
In,               ( given )               (common )       ( By AA rule)     ( corresponding sides of similar triangles )              

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S seema garhwal
Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal. To prove : PQRS is a parallelogram. Proof : In PQRS,     Since,         and       . So,PQRS is a parallelogram.

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D Devendra Khairwa
For finding the number of triangles we need to find the area of the figure. Consider the hexagonal structure :                       Area of hexagon  =   6     Area of 1 equilateral  Thus area of the equilateral triangle :                       or        or        So, the area of the hexagon is  :         And the area of an equilateral triangle having 1cm as its side is :        or   ...

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D Devendra Khairwa
The three main points form a triangle ABC. Now we have to find a point which is equidistant from all the three points. Thus we need to find the circumcenter of the .  We know that circumcenter is defined as the point as the intersection point of the perpendicular bisectors of the sides of the triangle. Hence the required point can be found out by drawing perpendicular bisectors of .

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S Sanket Gandhi
The required point is called in-centre of the triangle. This point is the intersection of the angle bisectors of the interior angles of a triangle. Hence the point can be found out in this case just by drawing angle bisectors of all the angles of the triangle.  

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D Devendra Khairwa
We know that circumcenter of a triangle is equidistant from all the vertices. Also, circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle. Thus, draw perpendicular bisectors of each side of the triangle ABC. And let them meet at a point, say O.  Hence O is the required point which is equidistant from all the vertices.  

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D Devendra Khairwa
Consider a right-angled triangle ABC with right angle at B. Then                          (Since ) Thus the side opposite to largest angle is also largest.         Hence the given statement is proved that all line segments are drawn from a given point, not on it, the perpendicular line segment is the shortest.

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S Sanket Gandhi
We are given that     . Thus                  Also,  PS bisects  , thus :                            Now, consider ,                                           (Exterior angle) Now, consider ,                            Thus from the above the result we can conclude that :                                  

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S Sanket Gandhi
                                            Consider  in the above figure :            (Given) Thus                            (as angle opposite to smaller side is smaller)       Now consider , We have :                 and                             Adding the above result we get,                         or                                                      Similarly, consider , we have   ...

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D Devendra Khairwa
In this question, we will use the property that sides opposite to larger angle are larger. We are given   and  .  Thus,                                   ..............(i) and                                     ...............(ii) Adding (i) and (ii), we get :                           or                                          Hence proved.

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D Devendra Khairwa
We are given that,                                                                                                                              ......................(i) Also,                                                (Linear pair of angles)          .....................(ii) and                                                  (Linear pair of angles)         ...

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S Sanket Gandhi
Consider a right-angled triangle ABC with right angle at A. We know that the sum of interior angles of a triangle is 180. So,                                     or                                        or                                                        Hence  and  are less than    (). Also, the side opposite to the largest angle is also the largest. Hence the side BC is largest is the...

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S Sanket Gandhi
Consider   and  , (i)          (Since it is given that AP is altitude.) (ii)                                         (Isosceles triangle) (iii)                                        (Common in both triangles) Thus by RHS axiom we can conclude that :                       Now, by c.p.c.t.we can say that :            

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D Devendra Khairwa
Using the given conditions, consider    and     , (i)                   (Right angle) (ii)                                  (Common in both the triangles) (iii)                                (Given that altitudes are of the same length. ) Thus by RHS axiom, we can say that  :                 Hence by c.p.c.t.,      And thus        (sides opposite to equal angles are also equal). Thus ABC is an...

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D Devendra Khairwa
(i)  From the figure we can say that :                                               or                                           or                                             Now, consider    and   , (a)                          (Given) (b)                             (Given) (c)                           (Prove above) Thus by SSS congruence rule, we can conclude that :                      ...

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D Devendra Khairwa
In the previous part of the question we have proved that     Thus by c.p.c.t., we can write :                Hence   bisects   .  

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S Sanket Gandhi
Consider    and    ,  (i)                 (Given) (ii)                (Common in both triangles) (iii)   Thus by RHS axiom we can conclude that :     Hence by c.p.c.t. we can say that  :               or        AD bisects BC.    

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S Sanket Gandhi
In the previous part we have proved that   . Thus by c.p.c.t. we can say that  :        Also,                                                          SInce BC is a straight line, thus  :            or                                                                                     or                                                                                        Hence it is clear...

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S Sanket Gandhi
In the first part, we have proved that . So, by c.p.c.t.   . Hence AP bisects . Now consider   and  , (i)                           (Common) (ii)                           (Isosceles triangle) (iii)                           (by c.p.c.t. from the part (b)) Thus by SSS congruency we have   :                                           Hence by c.p.c.t. we have :       or  AP bisects .
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