4 Question. xercise 4. Fill in the blanks using the correct word given in brackets. (i) ll circles are....(congruent, similar) (ii) ll squares are....(similar, congruent) (iii) ll... triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are... (b) their corresponding sides are....(equal, proportional). Solution (i) ll circles are similar because they have similar shape but not same size. (ii) ll squares are similar because they have similar shape but not same size. (iii) ll equilateral triangles are similar because they have similar shape but not same size. (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal. (b) their corresponding sides are proportional. Question. Give two different examples of pair of (i) similar figures. (ii) non-similar figures. Solution (i) (a) air of equilateral triangle are similar figures. (b) air of squares are similar figures. (ii) (a) triangle and a quadrilateral form a pair of non-similar figures. (b) square and a trapezium form a pair of non-similar figures. Question 3. not S State whether the following quadrilaterals are similar or.5 cm R 3 cm.5 cm.5 cm 3 cm 3 cm.5 cm Q Solution The two quadrilaterals, in figure are not similar because their corresponding angles are not equal. It is clear from the figure that, is 90 but is not 90. 3 cm
4 xercise 4. Question. In figures, (i) and (ii),. Find in figure (i) and in figure (ii)..5 cm cm 7. cm.8 cm 3 cm 5.4 cm Solution (i) In figure (i), (Given) (y basic proportionality theorem).5 3 (Q.5 cm, 3 cm and cm, given) 3 cm 5. (ii) In figure (ii), (Given) (y basic proportionality theorem).8 7. 5.4 (Q.8 cm, 5. 4 cm and 7. cm, given).8 7. 4. cm 54. Question. and F are points on the sides Q and R respectively of a QR, for each of the following cases, state whether F QR (i) 3. 9 cm, Q 3 cm, F 3. 6 cm and FR. 4 cm. (ii) 4 cm, Q 4. 5 cm, F 8 cm and RF 9 cm. (iii) Q. 8 cm, R. 5 6 cm, 0. 8 cm and F 0. 3 6 cm. Solution (i) In figure, Q 39. 3.3, F FR 36. 4 3..5 F Q FR (i) F is not parallel QR because converse of basic proportionality theorem is not satisfied. Q (ii) 3 cm 3.9 cm 3.6 cm F.4 cm R
(ii) In figure, and Q 4 4.5 40 45 F FR 8 9 8 9 4 cm 8 cm 4.5 cm F 9 cm F satisfied. (iii) In figure, F Q FR QR because converse of basic proportionality of theorem is Q Q 0.8 0.8 Q.8 0.8.0 R 9 55.8 cm 0.8 cm 0.36 cm.56 cm F and Q F F FR R F 0. 36 0. 36 9 56. 036.. 0 55 F Q FR F QR because converse of basic proportionality theorem is satisfied. Question 3. In figure, if LM and LN, prove that M N. R M L N Solution In, LM (Given) M M L (i) L (asic proportionality theorem)
In, LN (Given) N L N L (ii) (asic proportionality theorem) From qs. (i) and (ii), we get M N M N M N M N M N + + (dding both sides by ) M N M + M N + N M N M N M N M + M N + N Hence proved. Question 4. In figure, and F. rove that F F. F Solution In, (Given) (i) (asic proportionality theorem) In, F (Given) F F (ii) (asic proportionality theorem) From qs. (i) and (ii), we get Hence proved. F F Question 5. In figure, OQ and F OR. Show that F QR. F O Q R
Solution In figure, theorem, In QO, we have, and in OR, From qs. (i) and (ii), OQ and F OR, then by basic proportionality (i) Q O F (ii) FR O F Q FR Now, in QR, we have proved that F F Q FR F QR Q R (y converse of basic proportionality theorem) Hence proved. Question 6. In figure, and are points on O, OQ and OR, respectively such that Q and R. Show that QR. O Q R Solution In figure, Q (Given) O O Q (i) (asic proportionality theorem) lso, in figure, R (Given) From qs. (i) and (ii), we get O O (ii) R (asic proportionality theorem) O O Q R QR (onverse of basic proportionality theorem)
Question 7. Using theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (recall that you have proved it in lass IX). Solution In, is the mid-point of. i.e., (i) s straight line l. Line l is drawn through and it meets at. y basic proportionality theorem, [From q. (i)] is the mid-point of. Hence proved. Question 8. Using theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (recall that you done it in lass IX). Solution In, and are mid-points of side and, respectively. l and (See in figure) (y converse of basic proportionality theorem) Question 9. is a trapezium in which and its diagonals intersect each other at the point O. Show that O O. O O Solution O F
We draw, OF (lso ) In, O O O (asic proportionality theorem) (i) In, O O O (asic proportionality theorem) O O (ii) From qs. (i) and (ii), we get O O i.e., Hence proved. O O O O O O Question 0. The diagonals of a quadrilateral intersect each other at the point O such O O. Show that is a trapezium. O O Solution In figure, Through O, we draw O O meets at. In, O O O O O (ii) O O O O (Given) O O O O (Given) (i) From qs. (i) and (ii), we get O O O (y converse of basic proportionality theorem) Now, we have O and O Quadrilateral is a trapezium. Hence proved. O
4 xercise 4.3 Question. State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form (i) 60 60 80 40 Q 80 40.5 cm R (ii) 3 6 5.5 cm Q 4 R (iii) L.7 3 4 6 M 5 F (iv) M 70 3 5 6 70 N L Q 0 R
(v) 80.5 5 80 3 6 F (vi) 70 80 80 30 F Q R Solution (i) Yes. In and QR, 60, Q 80 and R 40 Here, corresponding angles are equal. Therefore, ~ QR (y similarity criterion) (ii) Yes. In and QR,.5, QR 4 R 5 and Q 3 6 Here, all corresponding sides are equal in proportional. Therefore, ~ QR (y SSS similarity criterion) (iii) No. In LM and F M 4 i.e.,, L F 3 6 L M F LM and F 7. 5 LM F Here, all corresponding sides are not equal in proportional. Thus, the two triangles are not similar. (iv) Yes. In LMN and QR MN 3 M Q 70, and ML Q 6 QR 5 0 MN ML i.e., Q OR Here, corresponding two adjacent sides are in proportional and one angle is equal. Therefore, MNL ~ QR (y SS similarity criterion)
(v) No. In, is given but the included side is not given. (vi) Yes. 70, 80 and F 30 (QIn F, + + F 80 ) Q 80, R 30, then 70 (QIn QR, Q + + R 80 ) Here,, Q, F R Therefore, F ~ QR (y similarity criterion) Question. In figure, O ~ O, O 5 and O 70. Find O, O and O. 70 O 5 Solution O + 5 80 (QO is a straight line) O 80 5 55 O + O + O 80 (Sum of three angles of O) O + 70 + 55 80 O + 5 80 O 80 5 55 Now, we are given that, O ~ O. (Similar triangle) O O O O O 55 i.e., O 55 Hence, we have O 55, O 55 and O 55. Question 3. iagonals and of a trapezium with intersect each other the point O. Using a similarity criterion for two triangles, show that O O Solution at O. O. O raw is a trapezium and and are diagonals intersect 4 3 O
In figure, (Given) 3, 4 (lternate interior angles) lso, O O (Vertically opposite angles) O ~ O (Similar triangle) O O O O Hence proved. Question 4. In figure, QS ~ TQR. (Ratios of the corresponding sides of the similar triangles) O O (Taking reciprocals) O O QR QS QT and, show that R T Q S R Solution In figure, (Given) Q R We are given that, Now, in QS and TQR, we have and (Sides opposite to equal angles of QR) QR QT QS R QR QT QS Q (QQ R proved) QS Q QR QT (Taking reciprocals) (i) QS TQR (ach ) QS Q QR QT [y q. (i)] Therefore, by SS similarity criterion, we have QS ~ TQR.
Question 5. S and T are points on sides R and QR of QR such that RTS. Show that RQ ~ RTS. Solution raw a RQ such that S and T are points on R and QR and joining them. In figure, we have RQ and RTS in which RQ RTS (Given) RQ SRT (ach R) Then, by similarity criterion, we have RQ ~ RTS Note If any two corresponding angles of the triangles are equal, then their third corresponding angles are also equal by. Question 6. In figure, if, show that ~. S R T Q Solution In figure, (Given) and (T) and (ach ) Now, in and, we have (roved) i.e., and also, (ach ) ~ (y SS similarity criterion) Hence proved.
Question 7. In figure, altitudes and of intersect each other at the point. Show that (i) ~ (ii) ~ (iii) ~ (iv) ~ Solution (i) In figure, (ach 90 ) and (Vertically opposite angles) ~ (y similarity criterion) (ii) In figure, (ach 90 ) and (ach ) ~ (y similarity criterion) (iii) In figure, (ach 90 ) and (ommon angle) ~ (y similarity criterion) (iv) In figure, (ach 90 ) and (ommon angle) ~ (y similarity criterion) Question 8. is a point on the side produced of a parallelogram and intersects at F. Show that ~ F. Solution joining. raw a parallelogram and produce a line to and F In parallelogram, (i) Now, for and F, we have F [From q. (i)] F (lternate angles as F) ~ F ( similarity)
Question 9. In figure, and M are two right triangles, right angled at and M, respectively. rove that (i) ~ M (ii) M M Solution (i) In figure, we have M (ach 90 ) ecause the and M are right angled at and M, respectively. lso, M (ommon angle ) ~ M (y similarity criterion) (ii) s ~ M, M (Ratio of the corresponding sides of similar triangles) M Hence proved. Question 0. and GH are respectively the bisectors of and GF such that and H lie on sides and F of and FG, respectively. If ~ FG. Show that (i) (ii) ~ HG (iii) ~ HGF GH FG Solution raw two and FG along that draw two bisectors and GH of and GF. H F G Since, ~ FG (i) In and FGH GFH (i) Q ~ FG GF GFH
From qs. (i) and (ii), we get (ii) In and HG, FGH (ii) Q ~ FG FG FG (Halves of equals are equal FGH ~ FGH (Q similarity criterion) GH FG (Qorresponding sides of two similar triangles are proportional) HG (iii) Q ~ FG FG HG HG (iv) Q ~ FG FG FG (Halves of equals are equal) HG From qs. (iii) and (iv), we get ~ HG (Q similarity criterion) (iii) In and HGF, HFG (v) Q ~ FG GF GFH HFG HGF (vi) Q ~ FG, FG FG ( Halves equals are equal) HGF From qs. (v) and (vi), we get ~ HGF (Q similarity criterion)
Question. In figure, is a point on side produced of an isosceles with. If and F, prove that ~ F. F Solution In figure, we are given that is isosceles and (i) For and F, F [From q. (i)] and F [ach 90 ] ~ F ( similarity criterion) Question. Sides and and median of a are respectively proportional to sides Q and QR and median M of QR. Show that ~ QR. Q M R Solution Given in and QR, and M are their medians, respectively. (i) Q R M To prove ~ QR onstruction roduce to such that and produce M to N such that M MN. Join,, QN and RN. 3 4 Q M R N
roof Quadrilaterals and QNR are parallelograms because their diagonals bisect each other at and M, respectively. and QN R QN R QN Q [y q. (i)] i.e., Q QN (ii) From q. (i), Q M M N i.e., From qs. (ii) and (iii), we have (Qiagonals are bisect each other) (iii) Q N Q QN ~ QN N (iv) Similarly, we can prove On adding qs. (iv) and (v), we have ~ RN 3 4 (v) + 3 + 4 ~ QR (SS similarity criterion) Question 3. is point on the side of a such that. Show that. Solution raw a such that is a point on and join. For and, we have (Given) and (ommon ) ~ ( similarity criterion)
. Question 4. Sides and and median of a are respectively proportional to sides Q and R and median M of another QR. Show that ~ QR. Solution Given, in and QR, lso, and M are their medians, respectively. (i) Q R M To prove ~ QR onstruction roduce to such that and produce M to N such that M MN. Join,, QN and RN. 3 4 Q M R roof Quadrilaterals and QNR are parallelograms because their diagonals bisect each other at and M, respectively. and QN R QN R QN Q [y q. (i)] i.e., Q QN (ii) From q. (i), Q M M N i.e., N (Qiagonals are bisect each other) (iii) Q N
From qs. (ii) and (iii), we have Q Q N QN ~ QN N (iv) Similarly, we can prove that On adding qs. (iv) and (v), we have ~ RN 3 4 (v) + 3 + 4 ~ QR (SS similarity criterion) Question 5. vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 8 m long. Find the height of the tower. Solution In figure (i), is a pole and behind it a Sun is risen which casts a shadow of length 4 cm and makes a angle θ to the horizontal and in figure ii, M is a height of the tower and behind a Sun risen which casts a shadow of length, NM 8 cm. In and NM, N θ and MN 90 ~ MN ( similarity criterion) M M MN MN 6 h 6 8 h 4 m 4 8 4 Sun Sun 6 cm h cm θ 4 cm (i) N θ 8 cm (ii) M
Question 6. If and M are medians of and QR, respectively, where ~ QR, prove that. Q M Solution raw two and QR taking and M points on and QR such that and M are the medians of the and QR. Q M R ~ QR (Given) ;, Q, R Q QR R (i) Now, and QM RM QR (ii) From q. (i), Thus, we have (Q is mid-point of and M is mid-point of QR) Q QR Q QM Q QM Q QM [y q. (ii)] and QM (Q Q) ~ QM (by SS similarity criterion) Q M Hence proved.
4 xercise 4.4 Question. Let ~ F and their areas be, 64 cm and cm, respectively. If F 5. 4 cm, find. Solution ~ F (Given) ar( ) ar( F ) F (Using property of area of similar triangles) 64 F F 8 F 8 8 F 8 5 4 Question. iagonals of a trapezium with intersect each other at the point O. If. Find the ratio of the area of O and O. Solution ar( O) ar( O) (Using property of area of similar triangles) ( ) (Q ) 4 4 Question 3. In figure, and are two triangles on the same base. If intersects at O, show that ar ( ) O ar( ) O. O O
Solution M O L raw L and M (See figure) In OL and OM LO MO 90 and OL OM (Vertically opposite angle) OL ~ OM ( similarity criterion) L O M O (i) Now, ( ) ( L) ar( ) L O ar( ) ( ) ( M) M O [y q. (i)] Hence, ar( ) O ar( ) O Question 4. If the areas of two similar triangles are equal, prove that they are congruent. Solution Let ~ QR and ar( ) ar( QR) (Given) Q R i.e., ar( ) ar( QR) Q QR R (Using property of area of similar triangles) Q, QR and R (SSS proportionality criterion) QR.
Question 5., and F are respectively the mid-point of sides, and of. Find the ratio of the areas of F and. Solution raw a taking mid-points, and Fon, and and join them. F Here, F, and F (i) (Q, and Fare mid-points of sides, and, respectively) F F (SSS proportionality criterion) F ~ ar( F ) ar( ) Hence, the required ratio is : 4. [From q. (i)] 4 (Using property of area of similar triangle) ar( F ) [ Qar( ) ar( )] ar( ) 4 Question 6. rove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians. Solution Q R M In figure, is a median of and M is a median of QR. Here, is mid-point of and M is mid-point of QR. Now, we have,
In figure, is a median of and M is a median of QR. Here, is mid-point of and M is mid-point of QR. Now, we have, ~ QR Q (i) (orresponding angles are equal) lso, Q QR In and QM, and Q QM (Ratio of corresponding sides are equal) (Q is mid-point of and M is mid-point of QR) (ii) Q QM QM [y q. (i)] Q QM [y q. (ii)] ~ QM (SS similarity criterion) Now, (iii) Q M ar( ) ar( QR) Q ar( ) ar( QR) M (Using property of area of similar triangles) [From q. (iii)] Question 7. rove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. Solution a a Q a a a a
raw is a square having sides of length a Then, the diagonal, a We construct equilateral and Q. ~ Q (quilateral triangles are similar) Hence proved. ar( ) ar( Q) (Using property of area of similar triangles) a ar( ) ar ( Q) ( a ) Question 8. and are two equilateral triangles such that is the mid-point of. Ratio of the area of and is (a) : (b) : (c) 4 : (d) : 4 Solution (c) Here, a (Say) (Q is an equilateral) a (Q is mid-point of ) Now, ~ (Qoth the triangles are equilateral) i.e., The ratio is 4 :. ar( ) ar( ) (Using property of area of similar to triangles) a 4 a Question 9. Sides of two similar triangles are in the ratio 4 : 9. reas of these triangles are in the ratio (a) : 3 (b) 4 : 9 (c) 8 : 6 (d) 6 : 8 Solution (d) reas of two similar triangles are in the ratio of the square of their corresponding sides 4 6 9 8
4 xercise 4.5 Question. Sides of some triangles are given below. etermine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 4 cm, 5 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 00 cm (iv) 3 cm, cm, 5 cm Solution We know that, in right triangle, sum of squares of two smaller sides is equal to the square of the third (large) side. (i) Here, ( 7) + ( 4) 49 + 576 65 ( 5) Therefore, given sides 7 cm, 4 cm and 5 cm make a right triangle and length of its hypotenuse is 5 cm. (ii) Here, () 3 + () 6 9 + 36 45 and ( 8 ) 6 4. oth values are not equal. Therefore, given sides 3 cm, 8 cm and 6 cm does not make a right triangle. (iii) Here, ( 50) + ( 80) 500 + 6400 8900 and ( 00) 0000. oth values are not equal. Therefore, given sides 50 cm, 80 cm and 00 cm does not make a right triangle. (iv) Here, ( ) + ( 5) 44 + 5 69 ( 3) Therefore, given sides 3 cm, cm and 5 cm make a right triangle and length of its hypotenuse is 3 cm. Question. QR is a triangle right rangled at and M is a point on QR such that M QR. Show that M QM MR. Solution In QR and MQ, 4 Similarly, 3 + + 4 (ach 90 ) R M 3 4 Q
and MR MQ (ach 90 ) QM ~ RM ( criterion) ar( QM) M ar( RM) RM (Using property of area of similar triangles) ( QM) ( M) M ( RM) ( M) RM (rea of a triangle ase Height) QM RM M RM M QM RM or M QM MR Hence proved. Question 3. In figure, is a triangle right angled at and. Show that (i) (ii) (iii) Solution s proved in above question, ~ ~ (i) ~ Then, ar( ) ar( (Using property of area of similar triangles) ( ) ( ) ( ) ( ) (rea of triangle ase Height)
(ii) ~ ar( ) ar( ) (Using property of area of similar triangles) ( ) ( ) ( ) ( ) (rea of triangle ase Height) (iii) ~ ar( ) ar( ) (Using property of area of similar triangles) ( ) ( ) ( ) ( ) (rea of triangle ase Height) Hence proved. Question 4.. is an isosceles triangle right angled at. rove that Solution raw is an isosceles triangle right angled at. and (i) y ythagoras theorem, we have + + [Q by q. (i)] Hence proved. Question 5. is an isosceles triangle with. If, prove that is a right triangle. Solution raw an isosceles with.
In, we are given that (i) and (ii) Now, + + [y q. (i)] [y q. (ii)] i.e., + Hence, by the converse of the ythagoras theorem, we have is right angled at. Question 6. is an equilateral triangle of side a. Find each of its altitudes. Solution raw equilateral, each side is a. lso, draw. Where is an altitude. In and (ommon) and 90 (RHS congruency) a (Qin an equilateral triangle altitude is the perpendicular bisector of ). Now, from by ythagoras theorem, we get + ( a) + a 3a 3a Question 7. rove that the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals. Solution raw is a rhombus in which a (Say) Its diagonal and are right angled bisector of each other at O. O In O, O 90,
O and O In O, use ythagoras theorem, we have O + O + + 4 or 4 + + + + + ( Q ) Hence proved. Question 8. In figure, O is a point in the interior of a, O, O and OF. Show that F O (i) O + O + O O O OF F + + (ii) F + + + + F Solution In, from point O join lines O, O and O. (i) In right angled OF, F O OF + F (y ythagoras theorem) O OF F (i) Similarly, in O, O O (ii) and in O, O O (iii) On adding qs. (i), (ii) and (iii), we get O + O + O O O OF F + + (ii) From part q. (i), we get O + O + O O O OF F + + Similarly, O + O + O O O OF F + + From qs. (iv) and (v), we have F + + + + F O (iv) (v)
Question 9. ladder 0 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall. Solution Let be the position of the window and be the length of the ladder. Then, 8 m (Height of window) 0 m (Length of ladder) 0 m 8m Let x m be the distance of the foot of the ladder from the base of the wall. Using ythagoras theorem in, we get + x + () 8 + ( 0) x 00 64 36 x m x 6, i.e., 6 m Question 0. guy wire attached to a vertical pole of height 8 m is 4 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? Solution Let be the vertical pole of height 8 m. guy wire is of length 4 m. ole 4 m Wire 8 m stake x m Let x m be the distance of the stake from the base of the pole. Using ythagoras theorem in, we get i.e., + x + ( 8) ( 4) x ( 4) ( 8) 576 34 5 x 5 m x 6 7m (QWe take positive sign because cannot be negative)
Question. n aeroplane leaves an airport and flies due North at a speed of 000 kmh. t the same time, another aeroplane leaves the same airport and flies due West at a speed of 00 kmh. How far apart will be two planes after h? Solution The first plane travels distance in the direction of North in h at speed of 000 km/h. 000 500 km (lane I) 3 km N North W (lane II) The second plane travels distance in the direction of West in h at a speed of 00 km/h. West 00 3 800 km In right angled, + (y ythagoras theorem) ( 800) + ( 500) 340000 + 50000 5490000 5490000 m 300 6 m Question. Two poles of heights 6 m and m stand on a plane ground. If the distance between the foot of the poles is m, find the distance between their tops. Solution Let and be the two poles of heights m and 6 m. Then, 6 5 cm S
x m 5m 6m m m m Let distance between tops of two poles x m Using ythagoras theorem in, we get i.e., + x ( ) + () 5 69 x 3 Hence, distance between their tops 3 m Question 3. and are points on the sides and, respectively of a right angled at. rove that + +. Solution raw a right at. Take and points on the sides and and join, and. In right angled, and in right angled, On adding qs. (i) and (ii), we get + (i) (y ythagoras theorem) + (ii) + ( + ) + ( + ) ( + ) + ( + ) (QIn, + and In, + ) + (y ythagoras theorem) + + Hence proved.
Question 4. The perpendicular from on side of a intersects at such that 3 (see in figure). rove that +. Solution Given, 3 4 and 3 4 (i) In, + (ii) In, + (y ythagoras theorem) (iii) On subtracting q. (iii) from q. (ii), we get 3 4 4 9 6 6 + Hence proved. Question 5. In an equilateral, is a point on side such that. rove that 9 7. 3 Solution raw is an equilateral triangle, is a point on side such that. raw a line is perpendicular to. 3
a (Say) (y property of equilateral triangle) a 3 3 a 3 3 Q a (QIn an equilateral triangle altitude is perpendicular bisector of.) a a a 3 6 Using ythagoras theorem in, + + (QRight, ) a a + a 6 a a + a 4 36 ( 36 9+ ) a 36 8 a 36 7 9 9 7 Hence proved.
Question 6. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. Solution raw is an equilateral triangle of side a. (Say) a x and Let x Now, a (In an equilateral triangle altitude is a perpendicular bisector of ) In right angled, + a x + a a x + a 4 4a 4x + a 3a 4x Hence proved. Question 7. Tick the correct answer and justify : In, 6 3, cm and 6 cm. The angle is : (a) 0 (b) 60 (c) 90 (d) 45 Solution (c) Given 6 cm and 6 3 cm and cm Now, + ( 6 3) + () 6 is right angled at 90 lso, < is less than cannot be more than 45 30 08 + 36 44 ( ) ( ) 90 30 60.
4 Question. QS Q. SR R xercise 4.6 (Optional)* In figure, S is the bisector of QR of QR, prove that Q S R Solution Given, in figure, S is the bisector of QR of QR. Now, draw RT S to meet Q produced int. roof Q RT S and transversal R intersects them (lternate interior angle) (i) Q RT S and transversal QT intersects them 4 T 3 Q S R 3 4 (orresponding angle) (ii) ut 3 (Given) 4 [From qs. (i) and (ii)] T R (iii) (QSides opposite to equal angles of a triangle are equal) Now, in QRT, S RT (y construction) QS Q SR T (y basic proportionally theorem) QS SR Q [From q. (iii)] R
Question. In figure, is a point on hypotenuse of, such that, M and N. rove that (i) M N M (ii) N M N N M Solution Given that, is a point on hypotenuse of, M and N. Now, join NM. Let and NM intersect at O. roof (i) In M and NM, M NM (ach equal to 90 ) M MN Let M Then, M 90 OM 90 ( 90 ) MN ( Q M + M + M 80 ) MO ~ NM ( similarity criterion) M M N M (orresponding sides of the similar triangles are proportional) M M N M O N
(ii) In NM and N, Let N NM N (ach equal to 90 ) NM N Then, N 90 ON 90 ( 90 ) NO ( Q N + N + N 80 ) NM ~ N ( similarity criterion) N M N N N M N N N M N Question 3. In figure, is a triangle in which > 90 and produced. rove that +. Solution Given that, in figure, is a triangle in which > 90 and roof produced. In right, Q 90 + (y ythagoras theorem) + ( + ) [ Q + ] ( + ) + + + + [ Q( a + b) a + b + ab] (QIn right with 90, + ) (y ythagoras theorem)
Question 4. In figure, is a triangle in which < 90 and. rove that +. Solution Given that, in figure, is a triangle in which < 90 and. roof In right, 90 + (y ythagoras theorem) + ( ) (Q + ) + + [ Q( a b) a + b ab] ( + ) + + { In right with 90, + } (y ythagoras theorem) Question 5. In figure, is a median of a and M. rove that (i) + M + (ii) M + (iii) + + Solution Given that, in figure, is a median of a and M. roof (i) In right M, Q M 90 M + M (y ythagoras theorem) M + ( M + ) ( QM M + ) ( M + M ) + + M + + M [ Q( a + b) a + b + ab] [QIn right M with M 90, M + M (y ythagoras theorem)] M
+ + M [Q ( is a median of )] + M + (i) (ii) In right M, Q M 90 M + M (y ythagoras theorem) M + ( M) [ Q M + M] M + + M M ( M + M ) + M + M [ Q( a b) a + b ab] [Q In right M with M 90, M + M (y ythagoras theorem] + M M + (iii) On adding qs. (i) and (ii), we get + + ( ). (Q, is a median of ) (ii) Question 6. rove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Solution Given that, is a parallelogram whose diagonals are and. M N Now, draw M and N (produced). roof In right M and N,
M (Opposite sides of a parallelogram) N (oth are altitudes of the same parallelogram to the same base) M ~ N (RHS congruence criterion) M N (T) (i) In right N, Q N 90 N + N (y ythagoras theorem) In right M, M 90 N + ( + N) ( QN + N) N + + N + N [ Q( a + b) a + b + ab] ( N + N ) + + N + + N (ii) (QIn right N with N 90 ) N + N (y ythagoras theorem) M + M ( QM M) M + ( M) [ Q( a b) a + b ab] M + + M M ( M + M ) + M + M [Q In right triangle M with M 90, M + M (y ythagoras theorem)] [Q + N (iii), opposite sides of parallelogram and M N from q. (i)] Now, on adding qs. (iii) and (ii), we get + ( + ) + ( + ) + + + Question 7. In figure, two chords and intersect each other at the point. rove that (i) ~ (ii)
Solution Given that, in figure, two chords and intersects each other at the point. roof (i) and (Vertically opposite angles) (ngles in the same segment) ~ ( similarity criterion) (ii) ~ [roved in (i)] (Qorresponding sides of two similar triangles are proportional) Question 8. In figure, two chords and of a circle intersect each other at the point (when produced) outside the circle. rove that (i) ~ (ii) Solution Given that, in figure, two chords and of a circle intersect each other at the point (when produced) out the circle. roof (i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angle. Therefore, (i) and (ii) In view of qs. (i) and (ii), we get ~ (Similar) (Q similarity criterion) (ii) ~ [roved in (i)] (Qorresponding sides of the similar triangles are proportional)
Question 9. In figure, is a point on side of such that. rove that is the bisector of. Solution Given that, is a point on side of such that. Now, from produce cut off. Join. roof (Given) [Q (by construction)] In, (y converse of basic proportionality theorem) (orresponding angle) (i) and (lternate interior angle) (ii) Q (y construction) (iii) (ngles opposite equal sides of a triangle are equal) Using qs. (i), (ii) and (iii), we get i.e., is the bisector of.
Question 0. Nazima is fly fishing in a stream. The tip of her fishing rod is.8 m above the surface of the water and the.8 m fly at the end of the string rests on the water 3.6 m away and.4 m from a point directly under the tip of the rod. ssuming that her string (from the tip of her rod to the fly) is taut, how much.4 m. m string does she have out? If she pulls in the string at the rate of 5 cms, what will be the horizontal distance of the fly from her after s? Solution Length of the string that she has out (.8) + (.4) 3.4 + 5.76 3m (Using ythagoras theorem).8 m.4 m Hence, she has 3 m string out. Length of the string pulled in s 5 60 cm 0.6 m.4 m.8 m Length of remaining string left out 3.0 0.6.4m (y ythagoras theorem) (.4) (.8) 5.76 3.4.5.5.59m (pprox.) Hence, the horizontal distance of the fly from Nazima after s. +.59.79 m (pprox.)