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Q : 1      ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$). AC is a diagonal. Show that :

(iii) PQRS is a parallelogram.

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.

4.  Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side $\small 1$ cm as you can. Count the number of triangles in each case. Which has more triangles?

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   ...

3.   In a huge park, people are concentrated at three points (see Fig.):
A : where there are different slides and swings for children,
B : near which a man-made lake is situated,
C : which is near to a large parking and exit.
Where should an icecream parlour be set up so that maximum number of persons can approach it?
(Hint : The parlour should be equidistant from A, B and C)

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 .

2.  In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

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.

1.  ABC is a triangle. Locate a point in the interior of $\small \Delta ABC$ which is equidistant from all the vertices of $\small \Delta ABC$.

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.

6.  Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

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.

5.  In Fig ,  $\small PR>PQ$  and PS bisects $\small \angle QPR$. Prove that  $\small \angle PSR>\angle PSQ$.

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 :

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig.). Show that $\small \angle A>\angle C$ and $\small \angle B>\angle D$.

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   ...

3.  In Fig., $\small \angle B <\angle A$  and $\small \angle C <\angle D$. Show that $\small AD .

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.

2. In Fig, sides AB and AC of $\small \Delta ABC$ are extended to points P and Q respectively. Also, $\small \angle PBC < \angle QCB$. Show that $\small AC> AB$.

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

1. Show that in a right angled triangle, the hypotenuse is the longest side.

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...

5. ABC is an isosceles triangle with $\small AB=AC$. Draw $\small AP\perp BC$ to show that $\small \angle B=\angle C$.

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 :

4.  BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

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...

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\small \Delta PQR$ (see Fig). Show that:
(i)  $\small \Delta ABM \cong \Delta PQN$
(ii) $\small \Delta ABC \cong \Delta PQR$

(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 :                      ...

2.(ii) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$. Show that

(ii) AD bisects $\small \angle A$.

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

2.(i)  AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$. Show that

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.

1.(iv) $\small \Delta ABC$ and  $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that

(iv) AP is the perpendicular bisector of BC.

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...

1.(iii)  $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that

(iii) AP bisects $\small \angle A$  as well as $\small \angle D$.

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 .

1.(ii)  ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A  and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that

(ii) $\small \Delta ABP \cong \Delta ACP$

Consider   and   , (i)    is common side in both the triangles. (ii)                         (This is obtained from the c.p.c.t. as proved in the previous part.) (iii)                                       (Isosceles triangles) Thus by SAS axiom, we can conclude that :

1.(i) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that

(i) $\small \Delta ABD\cong \Delta ACD$

Consider   and    , (i)               (Common) (ii)               (Isosceles triangle) (iii)             (Isosceles triangle) Thus by SSS congruency we can conclude that :
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