SOLUTIONS
1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm 2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that $PM^2 = QM × MR$. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR . Let ∠MPR = $Ø$ In ∆MPR, ∠MRP = $180^0 - 90^0 -Ø$ => $90^0 - Ø$ Similarily, in ∆PQM, ∠MPQ = $(90 - Ø)$ Therefore , in ∆PQM, => $180^0 = ∠Q + ∠MPQ+∠PMQ $ => $180^0 = ∠Q + (90^0 - Ø) + 90^0$ => ∠Q = Ø . Therefore by AAA similarity,∆PRM ~ ∆PQM, Since the triangles are similar, therefore , ${PR}/{QR} ={MR}/{PM}={PM}/{QM}$ i.e $PM^2 = MR × QM$ Hence proved3. In Fig., ABD is a triangle right angled at A and AC ⊥BD. Let ∠BCA = θ, Therefore in ∆ABC, ∠ABC = (90-θ) In ∆ACD, ∠CAD = (90-θ) and ∠ADC = θ (i) In ∆ABD and ∆ABC, ∠A =∠C ( 90 degree) ∠ABD = ∠ABC ∠ADB = ∠BAC Therefore by AAA similarity, ∆ABD~∆ABC Hence, ${AB}/{CB}={BD}/{AB} = {AD}/{AC}$ Therefore, $AB^2 = BC × BD$ Hence Proved (ii) In ∆ABC and ∆ACD, ∠C = ∠C (90 degree and common angle) ∠ABC = ∠CAD ∠BAC = ∠ADC Therefore by AAA similarity,∆ABC ~ ∆ACD Hence, ${AB}/{AD}={BC}/{AC}={CA}/{CD}$ => $AC^2=BC × CD$ Hence Proved (iii) In ∆ABD and ∆ACD ∠D = ∠D ( common angle) ∠BAD =∠ACD (90 degree) ∠ABD = ∠CAD Therefore by AAA similarity, ∆ABD ~ ∆ACD Hence ${AB}/{AC}={BD}/{AD}={AD}/{CD}$ Hence $AD^2 = BD × CD$ Hence Proved
Show that (i) $AB^2 = BC ×BD$ (ii) $AC^2 = BC × DC $ (iii) $ AD^2 = BD × CD$4. ABC is an isosceles triangle right angled at C. Prove that $AB^2 = 2AC^2$. Let ABC be a isosceles triangle right angles at C. Therefore AC = BC Draw a line CD from vertex C to the side AB such that CD⊥AB Let ∠B = θ So , in ∆BCD , ∠BCD = $180^0-90^0-θ$ = $90-θ$ and in ∆ABC , ∠BAC = $180^0-90^0-θ$ = $90-θ$ SO, in ∆ABC and ∆BCD , ∠B = ∠B (θ, common) ∠ACB = ∠BDC ($90^0$ degree) and ∠BAC = ∠BCD Therefore by AAA similarity, ∆ABC ~ ∆BCD hence, ${AB}/{BC}={BC}/{BD}={AC}/{CD}$ ..........(1) Also in ∆ACD,∠ACD = $θ$ ∠CAD = $90-θ$ SO, in ∆ABC and ∆ACD , ∠ABC = ∠ACD ∠C = ∠D (right angle) ∠A = ∠BCD Therefore by AAA similarity,∆ABC ~ ∆ACD ${AB}/{AC}={BC}/{CD}={CA}/{AD}$ => $AB × AD = AC^2$ ................(2) Since $AD= AB - DB$ therefore eqn (2) becomes, $AB×(AB-DB)=AC^2$ $AB×AB - AB×BD = AC^2$ $AB^2 - AB×BD = AC^2$ Since AB×BD = $BC^2$ (from eqn (1)) $AB^2 - BC^2 = AC^2$ $AB^2 = AC^2 + BC^2$ Since BC = AC , Therefore $AB^2 = AC^2 +AC^2$ $AB^2 = 2AC^2$ Hence Proved.5. ABC is an isosceles triangle with AC = BC. If $AB^2 = 2 AC^2$, prove that ABC is a right triangle. Given $AB^2 = 2 AC^2$ $AB^2 = AC^2 +AC^2$ Since AC=BC Therefore ,$AB^2 = AC^2 +BC^2$ These sides satisfy the pythagoras theorem, Hence the triangle ABC is a right angles triangle.6. ABC is an equilateral triangle of side $2a$. Find each of its altitudes. Let ABC be any equilateral triangle with sides $2a$.Perpendicular AD is drawn from A to BC. So, AD bisects the BC. Hence BD=DC= $a$ In triangle ADB, $AB^2=AD^2+BD^2$ $(2a)^2= AD^2+a^2$ $4a^2= AD^2+a^2$ $3a^2= AD^2$ $AD=√3a$ Altitutes AD = $√3a$7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Let ABCD be a rhombus , and diagonals meeting at O. TO prove, $AB^2+BC^2+CD^2+DA^2 = AC^2+BD^2$ SO, in ∆AOB, $AB^2=AO^2+BO^2$ ........(1) ∆AOD , $AD^2=AO^2+DO^2$ .......(2) ∆COD , $CD^2=CO^2+DO^2$ ........(3) ∆BOC , $BC^2=CO^2+BO^2$ .........(4) Adding (1), (2),(3) and (4), we get $AB^2+AD^2+CD^2+BC^2=AO^2+BO^2+AO^2+DO^2+CO^2+DO^2+CO^2+BO^2$ $AB^2+AD^2+CD^2+BC^2=2AO^2+2BO^2+2CO^2+2DO^2$ Since $AO=CO$ and $BO=DO$ $AB^2+AD^2+CD^2+BC^2=4AO^2+4BO^2$ $AB^2+AD^2+CD^2+BC^2=4(AO^2+BO^2)$ $AB^2+AD^2+CD^2+BC^2=4(({AC}/2)^2+({BD}/2)^2)$ $AB^2+AD^2+CD^2+BC^2={AC}^2+{BD}^2$ Hence Proved8. In Fig. , O is a point in the interior of a triangle ABC, OD ⊥BC, OE ⊥AC and OF ⊥AB. Show that (i) $OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 =AF^2 + BD^2 + CE^2$ (ii) $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$. Join OA,OB and OC. (i) In ∆AOF , $OA^2=OF^2+AF^2$ ...........(1) In ∆BOF , $OB^2=OD^2+BD^2$ ...........(2) In ∆COE , $OC^2=CE^2+OE^2$ ...........(3) Adding (1) ,(2) and (3), we get $OA^2 + OB^2+ OC^2= OF^2+AF^2+ OD^2+BD^2+CE^2+OE^2$ i.e $OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 =AF^2 + BD^2 + CE^2$ .......(4) Hence Proved (ii) From eqn(4), $AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2$ $AF^2 + BD^2 + CE^2 = (OA^2 – OE^2)+ (OB^2 – OF^2) + (OC^2 – OD^2) $ Hence Proved9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall. Distance of the ladder from the base of the wall = $√(10^2-8^2)$ =$√(100-64)$ =$√36$ =$6$ $m$10. A guy wire attached to a vertical pole of height 18 m is 24 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? By pythagoras theorem, disatnce of the pole from the ase of wire = $√(24^2-18^2)$ =$√(576-324)$ =$√252$ =$6√7 $ $m$11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after $1{1/2}$ hours? Distance travelled by each planes in $1{1/2}$ hours = 1500 km and 1800 km. Distance between the two planes = $√(1500^2+1800^2)$ = $100 √(225+324) $ = $100 √(549) $ = $300 √61$ $km$12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. Distance between the tip of two poles = $√(5^2+12^2)$ = $√(25+144)$ = $√169$ = $13$ $m$13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $AE^2 + BD^2 = AB^2 + DE^2$.
In ∆ACE, $AC^2+CE^2=AE^2$ ..........(1) In ∆BCD , $BC^2+CD^2=DB^2$ ..........(2) From (1) and (2) $AC^2+CE^2+BC^2+CD^2 =AE^2 + DB^2$ .....(3) In ∆CDE, $DE^2=CD^2+CE^2$ .....(4) In ∆ABC, $AB^2=AC^2+BC^2$ ......(5) From equation (3), (4) and (5) $DE^2+AB^2 =AE^2+DB^2$14. The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD . Prove that $2 AB^2 = 2 AC^2 + BC^2$. In ∆ACD, $AC^2=AD^2+DC^2$ => $AC^2-DC^2=AD^2$ .........(1) In ∆ABD , $AB^2 = AD^2 + BD^2$ => $AB^2 - BD^2 = AD^2 $ .......(2) From equation (1) and (2) $AC^2-DC^2 = AB^2 - BD^2$ .........(3) Given , 3DC = DB , $DC={BC}/4$ and $BD = {3BC}/4$ .....(4) Therefore from eqn (3) and (4) $AC^2-({BC}/4)^2 = AB^2 - ({3BC}/4)^2$ $AC^2-({BC}^2/16) = AB^2 - ({9BC^2}/16)$ $16AC^2-({BC}^2) = 16AB^2 - ({9BC^2})$ $16AC^2 = 16AB^2 - {8BC^2}$ $2AC^2 = 2AB^2 - {BC^2}$ $2 AB^2 = 2 AC^2 + BC^2$. Hence Proved15. In an equilateral triangle ABC, D is a point on side BC such that $BD = 1/3 BC$. Prove that $9AD^2 = 7 AB^2$. ∆ABC is a equilateral traingle with sides $a$ unit. Draw an altitute AE from A to BC. Therefore , BE =EC = $a/2$ In ∆AEB , $AB^2 = AE^2 + BE^2$ => $a^2 = AE^2 +a^2/4$ => $AE^2 = {3a^2}/4$ => $AE = {√3a}/2$ Given that BD = ${1/3}$BC $BD=a/3$ DE = BE - BD = $a/2 - a/3$ = $a/6$ In ∆AED, $AD^2 = AE^2+ DE^2$ $AD^2 = ({√3a}/2)^2 + (a/6)^2$ = ${3a^2}/4 + a^2/36 $ = ${28a^2}/{36}$ = $7/9{a}^2$ $AD^2$ = $7/9{AB}^2$ $9AD^2$ = $7{AB}^2$ Hence proved16. 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. Let triangle ABC be a equilateral triangle with side $a$ . Altitute AE is drawn from A to BC. Therefore $BE = EC = BC/2 = a/2$ In ∆ABE, $AB^2 = AE^2+EB^2$ $a^2 = AE^2 + (a/2)^2$ = $AE^2+a^2/4$ $AE^2 = a^2-a^2/4$ $AE^2 = 3/4a^2$ $4AE^2 = 3a^2$17. Tick the correct answer and justify : In ∆ABC,AB = 6√3 cm,AC = 12 cm and BC = 6 cm. The angle B is : (A) 120° (B) 60° (C) 90° (D) 45° Given, AB = 6√3 cm , AC = 12 cm and BC = 6 cm Therefore $AB^2 = 108$ , $AC^2 = 144$ and $BC^2 = 36$ Therefore $AB^2+BC^2 = 108 + 36$ = 144 = $AC^2$ The sides are satisfying the Pythagoras theroem. Hence , these are the sides of right angles triangle. Therefore angle B = 90° Ans: C
