## NCERT Class 10 Chapter 6: Triangles## Exercise 6.1
- All circles are _ (congruent, similar)
- All squares are _ (similar, congruent)
- All triangles are similar _ (isosceles, equilateral)
- Two polygons of the same number of sides are similar, if (a) their corresponding angles are _ and (b) their corresponding sides are _ (equal, proportional)
- All circles are similar.
- All squares are similar.
- All equilateral triangles are similar.
- Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.
- similar figures.
- non-similar figures.
- A pair of two squares with different length of its sides are similar.
A pair of two equilateral triangles with different lengths of sides are similar. - A pair of two isosceles triangles with different angles are non-similar.
A pair of square and a rhombus whose angles are not 90° are non-similar.
In the quadrilateral ABCD, all the angles are 90°. However, in the quadrilateral PQRS, the angles are not equal to 90°. Hence, the given quadrilaterals are not similar. ## Exercise 6.2
- PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
- PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
- PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
I. EF ‖ QR, if the basic proportionality theorem is satisfied, i.e. Since, LHS ≠ RHS, the theorem isn't satisfied. Hence, EF is not parallel to QR. II. EF ‖ QR, if the basic proportionality theorem is satisfied, i.e. Since, LHS = RHS, the theorem is satisfied. Hence, EF is parallel to QR. III.EF ‖ QR, if the basic proportionality theorem is satisfied, i.e. Since, LHS = RHS, the theorem is satisfied. Hence, EF is parallel to QR.
It is given that LM is parallel to CB in ∆ABC, therefore:
Let the triangle be ABC with DE as the line drawn through D as midpoint of side AB and E as a point on line AC. To Prove : AE = EC Proof :
Let the triangle be ABC, where DE is the line formed by joining the D which is the midpoint of AB and E which is the midpoint of AC.
Therefore, by using the converse basic proportionality theorem we can conclude that EO ‖ AB. It is also known that EO ‖ CD, which implies that AB ‖ CD. Since two of the sides of the given quadrilateral are parallel, so it satisfies the condition for being a trapezium. Hence, ABCD is a trapezium. ## Exercise 6.3
Hence, by AAA criterion ∆ABC ~ ∆PQR.
Hence, by SSS criterion ∆ABC ~ ∆PQR.
The ratios obtained are not same. Hence, ∆MPL is not similar to ∆DEF.
Hence, by SAS criterion ∆LMN ~ ∆PQR.
Hence, ∆ABC is not similar to ∆DEF.
Hence, by AAA criterion ∆DEF ~ ∆PQR.
∠DOC and ∠BOC form a linear pair. Therefore Hence ∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.
In ∆AOB and ∆COD, ∠BAO = ∠OCD (Alternate interior angles) ∠ABO = ∠ODC (Alternate interior angles) ∠AOB = ∠COD (Vertically opposite angles) Hence, by the AAA similarity criterion, ∆AOB ~ ∆COD. Since, ∆AOB ~ ∆COD Therefore, the corresponding sides will be proportional Hence, proved that OB/OD = OA/OC.
In ∆PQR, ∠PQR = ∠PRQ Therefore, it is an isosceles triangle which implies that PQ = PR. Using QR/QS = QT/PQ and ∠Q = ∠Q in ∆PQS and ∆TQR, we can conclude that the triangles are similar (by SAS similarity criterion). Hence, proved that ∆PQS ~ ∆TQR.
In ∆ RPQ and ∆ RTS, ∠RTS = ∠QPS (Given) ∠R = ∠R (Common Angle) Hence, by the AA similarity criterion ∆ RPQ ~ ∆ RTS.
In ∆ ABE and ∆ ACD AB = AC (Using c.p.c.t) ... Equation (I) AD = AE (Using c.p.c.t) ... Equation (II)
- ∆AEP ~ ∆ CDP
- ∆ABD ~ ∆ CBE
- ∆AEP ~ ∆ADB
- ∆ PDC ~ ∆ BEC
I. In ΔAEP and ΔCDP, ∠AEP = ∠CDP (Right Angles) ∠APE = ∠CPD (Vertically Opposite Angles) Therefore, by AA similarity criterion ΔAEP ~ ΔCDP. II. In ΔABD and ΔCBE, ∠ADB = ∠CEB (Right Angles) ∠ABD = ∠CBE& (Common Angle) Therefore, by AA similarity criterion ΔABD ~ ΔCBE. III. In ΔAEP and ΔABD, ∠AEP = ∠ADB (Right Angles) ∠PAE = ∠DAB (Common Angle) Therefore, by AA similarity criterion ΔAEP ~ ΔABD. IV. In ΔPDC and ΔBEC, ∠PDC = ∠BEC (Right Angles) ∠PCD = ∠BCE (Common Angle) Therefore, by AA similarity criterion ΔPDC ~ ΔBEC.
In ∆ABE and ∆CFB, ∠BAE = ∠BCF (Opposite angles of a parallelogram) ∠ABE = ∠CFB (Alternate interior angles) Therefore, by AA similarity criterion ΔABE ~ ΔCFB.
- ∆ ABC ~ ∆AMP
- CA/PA = BC/MP
I. In ΔABC and ΔAMP, ∠ABC = ∠AMP (Right Angles) ∠BAC = ∠MAP (Common Angle) Therefore, by AA similarity criterion ΔABC ~ ΔAMP. II. Since, ∆ABC is similar to ∆AMP, their corresponding sides will have an equal ratio. Hence,
- CD/GH = AC/FG
- ∆ DCB ~ ∆ HGE
- ∆ DCA ~ ∆ HGF
It is given that ∆ABC ~ ∆ FEG, therefore ∠A = ∠F ∠B = ∠E ∠ACB = ∠FGE Divide both sides by 2 ∠ACB/2 = ∠FGE/2 ∠ACD = ∠FGH (Bisected Angles) I. Now, in ∆ACD and ∆FGH: ∠A = ∠F (As shown above) ∠ACD = ∠FGH (As shown above) Therefore, by the AA similarity criterion ∆ACD ~ ∆FGH. Which implies that II. Similarly, ∠DCB = ∠HGE In ∆DCB and ∆HGE: ∠B = ∠E (As Shown above) ∠DCB = ∠HGE (As shown above) Therefore, by the AA similarity criterion ∆DCB ~ ∆HGE. III. In ∆DCA and ∆HGF ∠A = ∠F (As shown above) ∠DCA = ∠HGF (As shown above) Therefore, by the AA similarity criterion ∆DCA ~ ∆HGF.
In an isosceles triangles, angles opposite to equal sides are also equal. Therefore, ∠ABD = ∠ACD = ∠ECF In ∆ ABD and ∆ ECF: ∠ADB = ∠EFC (Right Angles) ∠ABD = ∠ECF (As Shown above) Therefore, by the AA similarity criterion ∆ ABD ~ ∆ ECF.
It is given that Therefore, by the SSS similarity criterion, ∆ABD ~ ∆PQM. This implies that ∠ABD = ∠PQM (Corresponding angles of similar triangles) ∠ABC = ∠PQR In ∆ABC and ∆PQR AB/PQ = BC/QR (Given) ∠ABC = ∠PQR (As shown above) Hence, by the SAS similarity criterion ∆ABC ~ ∆PQR.
In ∆ADC and ∆BAC: ∠ADC = ∠BAC (Given) ∠ACD = ∠BCA (Common Angles) Therefore, by the AA similarity criterion ∆ADC ~ ∆BAC. This implies that CA/CB = CD/CA (Corresponding sides of similar triangles)
First, we will produce AD and PM lines to exterior points E and N respectively such that AD = DE and PM = MN, then join E and N with C and R respectively. In ∆ABD and ∆CDE: . ∠ADB = ∠ CDE (Vertically opposite angles) AD = DE (By construction) BD = DC (D is the mid-point of BC) Therefore, by the SAS congruency criterion ∆ABD ≅ ∆CDE. This implies that AB = CE (By c.p.c.t.) In ∆PQM and ∆MNR: . ∠PMQ = ∠NMR (Vertically opposite angles) PM = MN (By construction) QR = MR (M is the mid-point of QR) Therefore, by the SAS congruency criterion ∆PMQ ≅ ∆NMR. This implies that PQ = RN (By c.p.c.t.) It is given that Therefore, by the SSS similarity criterion ∆ACE ~ ∆PRN. This implies that ∠CAE = ∠RPN (Corresponding angles of similar triangles) ∠CAD = ∠RPM Similarly, ∠BAD = ∠QPM. Add both the above equations ∠CAD + ∠BAD = ∠RPM + ∠QPM ∠BAC = ∠QPR In ∆ABC and ∆PQR: ∠BAC = ∠QPR (As shown above) AB/PQ = AC/PR (Given) Hence, by the SAS similarity criterion ∆ABC ~ ∆PQR.
Let AB be the pole whose shadow is BC and let the tower be PQ with QR as the shadow. If we connect the top of pole and tower to the end of their respective shadows, we can obtain triangles ABC and PQR. Since, the shadows BC and QR are being cast at the same time, the sun's elevation will be the same. Therefore, ∠ACB = ∠PRQ In ∆ABC and ∆PQR: ∠ACB = ∠PRQ (As shown above) ∠ABC = ∠PQR (Right Angles) Therefore, by the AA similarity criterion ∆ABC ~ ∆PQR. This implies that AB/PQ = BC/QR (Corresponding sides of similar triangles) AB = 6 m, BC = 4 m QR = 28 m 6/PQ = 4/28 6/PQ = 1/7
Hence, the height of the tower is 42 m.
It is given that ∆ ABC ~ ∆ PQR, which implies that ∠ABC = ∠PQR (Corresponding Angles of similar triangles) ∠ABD = ∠PQM In ∆ABD and ∆PQM: AB/PQ = BD/QM (As shown above) ∠ABD = ∠PQM (As shown above) Therefore, by the SAS similarity criterion ∆ABD ~ ∆PQM. This implies that
Hence, proved that AB/PQ = AD/PM. ## Exercise 6.4
Since ∆ ABC ~ ∆ DEF, therefore Hence, BC = 11.2 cm
In ∆AOB and ∆COD: ∠AOB = ∠COD (Vertically opposite angles) ∠ABO = ∠CDO (Alternate interior angles) Therefore, by the AA similarity criterion ∆AOB ~ ∆COD. This implies that Hence, the ratio of the areas of triangles AOB and COD is 4 : 1.
We need to construct two lines AP and DM where P and M are points on BC such that AP and DM are perpendicular to BC. Area of triangle = ½ × Base × Altitude Area of ∆ABC = ½ × BC × AP Area of ∆DBC = ½ × BC × DM Ratio of Areas of ∆ABC and ∆DBC = (½ × BC × AP)/ (½ × BC × DM)
In ∆AOP and ∆DOM: ∠AOP = ∠DOM (Vertically opposite angles) ∠APO = ∠DMO& (Right Angles) Therefore, by the AA similarity criterion ∆AOP ~ ∆DOM. This implies that AO/DO = PA/MD (Corresponding sides of similar triangles)
From Equation (I) and Equation (II),
Let there be two similar triangles ABC and PQR where Area of ABC = Area of PQR. AB/PQ = 1
Similarly,
Therefore, by the SSS congruency criteria ∆ABC ≅ ∆PQR. Hence, proved that two similar triangles with equal areas are congruent.
Since D and F are mid-points of sides AB and AC, therefore, by the mid-point theorem DF ‖ BC. In ∆ABC and ∆ADF: ∠BAC = ∠DAF (Common Angles) ∠ABC = ∠ADF (Corresponding Angles) Therefore, by the AA similarity criterion ∆ABC ~ ∆ADF. This implies that From the above equations, we can conclude that Ar (∆ADF) = �(Ar (∆ABC)) Ar (∆BED) = �(Ar (∆ABC)) Ar (∆CEF) = �(Ar (∆ABC)) Now, Hence, the ratio of the areas of triangles DEF and ABC is 1 : 4.
Let there be two similar triangles ABC and PQR, where AD and PM are the respective medians of the triangles. To Prove : ar (∆ABC)/ ar (∆PQR) = AD Proof : It is given that ∆ABC ~ ∆PQR, which implies that ∠ABC = ∠PQR (Corresponding angles of similar triangles) ∠ABD = ∠PQM Now, in ∆ABD and ∆PQM: ∠ABD = ∠PQM (As shown above) AB/PQ = BD/QM (As shown above) Therefore, by the SAS similarity criterion ∆ABD ~ ∆PQM. This implies that Hence, proved that ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Let there be a square ABCD with equilateral triangles ABD and ACE described on its one side and one diagonal. To Prove : ar (∆ABD) = ½ (ar (∆ACE)) Proof : By using the Pythagoras theorem in ABC, we get: AC AB = BC as ABCD is a square AC It is known that ABD and ACE are both equilateral triangles. Therefore, by the AAA similarity criterion ∆ABD ~ ∆ACE. This implies that Hence, proved 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. ## Tick the correct answer and justify:
D is the mid-point of BC, therefore BD = DC = ½(BC) ... Equation (I) AB = BC (Sides of an equilateral triangle) ... Equation (II) Since, ∆ABC and ∆BDE are both equilateral triangles, therefore by the AAA similarity criterion ∆ABC ~ ∆BDE. This implies that Hence, the correct answer is (C) 4 : 1.
Let there be two similar triangles ABC and DEF. Then, according to question Since, it is given that the triangles ABC and DEF are similar. Therefore, Hence, the correct answer is (D) 16 : 81. ## Exercise 6.5
- 7 cm, 24 cm, 25 cm
- 3 cm, 8 cm, 6 cm
- 50 cm, 80 cm, 100 cm
- 13 cm, 12 cm, 5 cm
I. Squares of the smaller two sides = 7 Square of the longest side = 25 Therefore, the triangle satisfies the Pythagoras theorem and is a right angled triangle. The length of hypotenuse of this triangle is 25 cm.
Square of the longest side = 8 Therefore, the triangle does not satisfy the Pythagoras theorem and is not a right angled triangle. III. Squares of the smaller two sides = 50 Square of the longest side = 100 Therefore, the triangle does not satisfy the Pythagoras theorem and is not a right angled triangle. IV. Squares of the smaller two sides = 5 Square of the longest side = 13 Therefore, the triangle satisfies the Pythagoras theorem and is a right angled triangle. The length of hypotenuse of this triangle is 13 cm.
By using Pythagoras Theorem in ∆PQM, we get PQ PM By using Pythagoras Theorem in ∆PRM, we get PR PM Add Equation (I) and Equation (II) 2PM 2PM 2PM 2PM 2PM 2PM 2PM PM Hence, proved that PM
- AB
^{2}= BC . BD - AC
^{2}= BC . DC - AD
^{2}= BD . CD
I. In ∆ABD and ∆CBA: ∠BAD = ∠BCA (Right Angles) ∠ABD = ∠ABC (Common Angles) Therefore, by the AA similarity criterion ∆ABD ~ ∆CBA. This implies that AB/CB = BD/AB (Corresponding Sides of similar triangles) AB II. Since, we have proven that ∆ABD ~ ∆CBA, therefore ∠CAB = ∠BDA (Corresponding angles of similar triangles) ∠CAB = ∠CDA In ∆CBA and ∆CAD: ∠ACB = ∠ACD (Right Angles) ∠CAB = ∠CDA (As shown above) Therefore, by the AA similarity criterion ∆CBA ~ ∆CAD. This implies that AC/BC = DC/AC (Corresponding Sides of similar triangles) AC III. In ∆DCA and ∆DAB: ∠DCA = ∠DAB (Right Angles) ∠CDA = ∠ADB (Common Angles) Therefore, by the AA similarity criterion ∆DCA ~ ∆DAB. This implies that CD/AD = AD/BD (Corresponding Sides of similar triangles) AD
It is given that ∆ABC is an isosceles triangle. This implies that the sides opposite to the equal angles are equal. Since, ∠ACB is given to be the right angle. Therefore, ∠ABC and ∠BAC are the equal angles with sides AC = BC. By using Pythagoras Theorem in ABC, we get AB AB AB Hence, proved that AB
It is given that AB AB AB The above equation satisfies the Pythagoras Theorem. Hence, ∆ABC is a right angles triangle.
Let us draw the altitude AD in the triangle ABC. In ∆ADB and ∆ADC: ∠ADB = ∠ADC (Right Angles) AD = AD (Common Side) AB = AC (Sides of equilateral triangle ABC) Therefore, by the RHS congruency criteria ∆ADB ≅ ∆ADC. This implies that BD = CD (By c.p.c.t.) Therefore, BC = 2BD BD = BC/2 BD = a By applying Pythagoras theorem in ∆ADB, we get AB AD AD AD AD = ?3a Hence, the length of each of the altitude of the triangle ABC is ?3a.
Let there be a rhombus ABCD whose diagonals bisect each other at O. To Prove : AB Proof : We know that the diagonals of a rhombus bisect each other at right angles. Therefore, AOB, AOD, BOC, COD are all right angles triangles. This also implies that AO = CO and BO = DO as O is the point of bisection. By applying Pythagoras Theorem to ∆AOB, we get AB Similarly, BC CD AD Adding all of the above obtained equations, we get AB AB AB Since, AO = CO and BO = DO. Therefore, AB AB AB O is the midpoint of AC and BD, so 2AO = AC and 2BO = BD. Therefore, >AB Hence, proved that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
- OA
^{2}+ OB^{2}+ OC^{2}- OD^{2}- OE^{2}- OF^{2}= AF^{2}+ BD^{2}+ CE^{2}, - AF
^{2}+ BD^{2}+ CE^{2}= AE^{2}+ CD^{2}+ BF^{2}.
Let us join OB, OA and OC. I. By applying Pythagoras Theorem in ∆AOF, we get OA By applying Pythagoras Theorem in ∆BOD, we get OB By applying Pythagoras Theorem in ∆COE, we get OC Adding the above obtained equations, we get OA
Hence, proved. II. OA (OA By applying Pythagoras Theorem in ∆AOE, we get OA OA Similarly, OC OB Hence, (OA
Let AB be the ladder which reaches the window at A with the base of the wall at C. The wall will be perpendicular to the ground, therefore ABC will form a right triangle which is right angled at C. AB = 10 m , AC = 8m By applying Pythagoras Theorem in ABC, we get AB 10 100 = 64 + BC 36 = BC BC = 6 m Hence, the distance between foot of the ladder and the base of the wall is 6 m.
Let the wire be AB and the pole be AC. BC is the distance between the stake and base of the pole. ∆ACB is a right triangle which is right angled at C. By Applying Pythagoras Theorem in ABC, we get AB 24 576 - 324 = BC 252 = BC BC = 6?7 m Hence, the stake needs to be driven 6?7 m away from the base of the pole to make the wire taut.
Let A be the current position of the plane flying to north and B be the current position of the plane flying to the west. Let O be the starting point for the two planes. AOB will form a right triangle, right angled at O. Speed of plane going north = 1000 km/h = 1200 × 3/2 = 1800 km By Applying the Pythagoras theorem in AOB, we get AB AB AB AB AB = 300?61 km
Let AB be the taller pole and CD be the smaller pole. The distance between their tops is AC and the distance between their feet is BD. We will draw an imaginary line from C to O which is a point on the pole AB such that OC ‖ BD. The poles stand perpendicular to the ground, so ∠BDC is a right angle. ∠AOC = ∠BDC (Corresponding Angles) Therefore, ∆AOC is a right triangle. OCBD forms a rectangle as OC ‖ BD and AB ‖ CD. This implies that OC = BD = 12 m OB = CD = 6 m Now, AB = AO + OB AO = AB - OB = 11 - 6 = 5 m By Applying Pythagoras Theorem in AOC, we get AC AC AC AC AC = 13 m Hence, the distance between the top of the two poles is 13 m.
By Applying Pythagoras Theorem in ABC, we get AB By Applying Pythagoras Theorem in AEC, we get AE By Applying Pythagoras Theorem in BDC, we get BD By Applying Pythagoras Theorem in DEC, we get DE Therefore, AE AE From Equation (I) and Equation (II), we get
Hence, proved.
BC = BD + CD BC = 3CD + CD BC/4 = CD By applying Pythagoras Theorem in ADB, we get AB By applying Pythagoras Theorem in ADC, we get AC Subtract Equation (II) from Equation (I): AB AB AB AB AB AB 2(AB 2AB
Hence, proved.
Let us construct altitude AE in ∆ABC. Since AE is the altitude of an equilateral triangle, therefore it will also act as a median. BE = CE = BC/2 By applying Pythagoras Theorem in ABE, we get AB AE AE By applying Pythagoras Theorem in ADE, we get AD AD AD AD AD AD AD AD 9AD
Hence, proved.
Let there be an equilateral triangle ABC with altitude AD and length of each side a. To prove : 3a Proof : We know that the altitude of an equilateral triangle also acts as a median. Therefore, BD = CD = BC/2 By applying Pythagoras Theorem in ABD, we get AB AB a 3a
Hence, proved that three times the square of one side is equal to four times the square of one of its altitudes.
AB AC BC It can be observed that AB This implies that the triangle satisfies the Pythagoras Theorem with AC as the hypotenuse. Therefore, angle B is 90°. Hence, correct options is ## Exercise 6.6 (Optional)
Let us produce the line QP to an external point T such that TR is parallel to PS. ∠QPS = ∠SPR as PS is the angle bisector for ∠QPR. ∠QPS = ∠QTR (Corresponding Angles) ∠SPR = ∠PRT (Alternate Interior Angles) Therefore, ∠PRT = ∠QTR. We know that if two angles of a triangle are equal then the triangle is isosceles and sides opposite to the equal angles are also equal. Therefore, PT = PR By using the basic proportionality theorem in QRT, we get QS/SR = PQ/PT
Hence, proved.
- DM
^{2}= DN . MC - DN
^{2}= DM . AN
It can be concluded from the figure that DBMN is a rectangle as its opposite sides are parallel and the angles are 90° each. This implies that DN = BM and DM = BN. (i). ∠CDM = ∠CDB - ∠BDM ∠CDM = 90° - ∠BDM ∠CDM + ∠BDM = 90° Also, in ∆CDM: ∠CDM + ∠CMD + ∠MCD = 180° ∠CDM + 90° + ∠MCD = 180° ∠CDM + ∠MCD = 90° Therefore, we can say that ∠CDM + ∠MCD = ∠CDM + ∠BDM (Both sides are equal to 90°) ∠MCD = ∠BDM In ∆CDM and ∆MDB: ∠MCD = ∠BDM (As shown above) ∠CMD = ∠BMD (Right Angles) Therefore, by the AA similarity criterion ∆CDM ~ ∆MDB. This implies that DM/BM = MC/DM DM We have already shown that BM = DN.
(ii). ∠ADB = ∠ADN + ∠BDN ∠ADN + ∠BDN = 90° Also, in ∆ADN: ∠AND + ∠ADN + ∠DAN = 180° 90° + ∠ADN + ∠DAN = 180° ∠ADN + ∠DAN = 90° Therefore, we can say that ∠ADN + ∠DAN = ∠ADN + ∠BDN (Both sides equal to 90°) ∠DAN = ∠BDN In ∆ADN and ∆NDB: ∠DAN = ∠BDN (As shown above) ∠CMD = ∠BMD (Right Angles) Therefore, by the AA similarity criterion ∆ADN ~ ∆NDB. This implies that DN/BN = NA/DN DN We have already shown that BN = DM.
Hence, proved.
By applying Pythagoras Theorem in ABD, we get AB By applying Pythagoras Theorem in ACD, we get AC AC AC We already know that AD
Hence, proved.
By applying Pythagoras Theorem in ABD, we get AB By applying Pythagoras Theorem in ACD, we get AC AC AC We already know that AD
Hence, proved.
- AC
^{2}= AD^{2}+ BC . DM + (BC/2)^{2} - AB
^{2}= AD^{2}- BC . DM + (BC/2)^{2} - AC
^{2}+ AB^{2}= 2 AD^{2}+ BC^{2}/2
It is given that AD is the median of triangle ABC, therefore
By applying Pythagoras Theorem in AMD, we get AD (i). By applying Pythagoras Theorem in ACM, we get AC AC AC By Using Equation (I) AC We already know that AM AC
(ii). By applying Pythagoras Theorem in ABM, we get AB We know that AD AM Now, AB AB AB By Using Equation (I) AB
(iii). By applying Pythagoras Theorem in ABM, we get AB By applying Pythagoras Theorem in AMC, we get AC Adding the above two equations, we get AB AB AB AB AB We know that AM AB By Using Equation (I) AB AB
Hence, proved.
Let ABCD be a parallelogram with AB ‖ CD and altitude AE. We will produce BA to a point F such that DF ? BF. To Prove : AC Proof : By applying Pythagoras Theorem in ADF, we get AD By applying Pythagoras Theorem in BDF, we get BD BD BD We know that AF
By applying Pythagoras Theorem in ADE, we get AD By applying Pythagoras Theorem in ACE, we get AC AC AC We know that AE
Add Equation (I) and Equation (II) AC Since ABCD is a parallelogram, therefore AB = CD and BC = AD (Opposite sides of a parallelogram) Also, it can be observed that AEDF forms a rectangle, which implies that AF = DE; (Opposite sides of a rectangle) Therefore, AC
Hence, proved that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
- ∆APC ~ ∆ DPB
- AP . PB = CP . DP
AP = BP = CP = DP as they are radii of the given circle. Therefore, AP/DP = 1 = CP/BP. (i). In ∆APC and ∆DPB: AP/DP = CP/BP (As shown above) ∠APC = ∠DPB (Vertically opposite angles) Therefore, by the SAS similarity criterion ∆APC ~ ∆DPB. (ii). We have already shown that AP/DP = CP/BP. This implies that AP.BP = CP.DP AP.PB = PC.DP Hence, proved.
- ∆ PAC ~ ∆ PDB
- PA . PB = PC . PD
(i). Since ABCD is a quadrilateral which is contained in a circle, therefore, ABCD forms a cyclic quadrilateral. This implies that, ∠PAC = ∠BDC (Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle) In ∆PAC and ∆PDB: ∠PAC = ∠BDC (As shown above) ∠APC = ∠BPD (Common Angle) Therefore, by the AA similarity criterion ∆PAC ~ ∆PDB.
This implies that AP/DP = PC/PB (Corresponding sides of similar triangles) AP.PB = PC.DP Hence, proved.
Let us produce BA to an external point P such that AP = AC. BC/ CD = AB/ AC (Given) BC/CD = AB/AP (AP = AC) Therefore, by using the converse of basic proportionality theorem, we can say that AD ‖ CP. ∠BAD = ∠APC (Corresponding Angles) ∠CAD = ∠ACP (Alternate Interior Angles) ∆ACP is an isosceles triangle as AP = AC. This implies that ∠APC = ∠ACP (Angles opposite to equal sides in an isosceles triangle) Therefore, ∠BAD = ∠CAD AD divides the ∠BAC in two equal parts. Hence, AD is the bisector of ∠BAC.
The situation can be represented in form of a triangle ABC, where A is the tip of fishing rod, C is the end of the string and BC is the surface of water. By applying Pythagoras theorem in ABC, we get AC AC AC AC AC = 3 Therefore, Nazima has 3 m of her fishing rod string out. It is given that she is pulling the string by 5 cm per second. Therefore, Length of string pulled by Nazima in 1 second = 5 cm Length of string pulled by Nazima in 12 seconds = 12 × 5 = 60 cm = 0.6 m Let the new position of the fly be D which is a point on the water surface BC. Length of string remaining out after 12 seconds = AD = 3 - 0.6 m = 2.4 m By applying Pythagoras Theorem in ABD, we get AD (2.4) BD BD BD BD = 1.59 m (Approx.) Horizontal distance of the fly = 1.2 + BD = 1.2 + 1.59 = 2.79 m Hence, the fly is at a horizontal distance of 2.79 m (approx. value) from Nazima. Next TopicClass 10 Maths Chapter 7 |