Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 3 Revision Question Bank Triangles 1. In the figure, DE AC and DC AP. Prove that BE BC EC CP . Solution: In BPA, we have DC AP [given] Therefore, by basic proportionality theorem, we have BC BD CP DA ..(i) In BCA , we have DE AC [given] Therefore, by basic proportionality theorem, we have BE BD EC DA ..(ii) From Eqs. (i) and (ii) BC BE BE BC or CP EC EC CP Hence proved. 2. In figure, if EDC~ ESA BEC = 115° and EDC=70°. Find DEC, DCE, EAB, AEB and EBA. Solution: Since, BD is a line and EC is a ray on it. DEC + BEC = 180° [linear pair axiom] DEC +115 0 =180 0 [ BEC =115 0 ,given] DEC = 180° – 115° = 65° But AEB = DEC [vertically opposite angles] AEB=65° In CDE, we have CDE + DEC+ DCE = 180° [by property of sum of the angles of a triangle is 180 0 ]
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Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes
1. 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. [4] Solution: Given A ABC in which AB = BC = CA and AD BC.
To prove 3AB2 =4AD2 Proof In ADB and ADC, AB = AC [given]
B = C [each 60°] and ADB = ADC [each 90°]
ADB ADC [by SAA congruence rule]
1BD DC BC
2 ..(i)
In right ADB, by Pythagoras theorem, AB2 =AD2+ BD2
To prove ABC DEF Proof Since, ABC - DEF and ar( ABC) = ar( DEF)
B E
and C F ..(i)
Now, 2 2
2 2
ar ABC ar ABCBC BC
ar DEF EF ar ABC EF
[ ar ABC ar DEF , given]
2
2 2
2
BC1 EF BC
EF
EF = BC [taking positive square root] Now, in ABC and DEF, we get B = E [from Eq. (i)] C = F [from Eq. (i)] and BC = EF [proved above]
ABC DEF
[by ASA congruence rule] Hence proved.
3. If the sides AB and AC and median AD of a ABC are proportional to sides PQ and PR and median PM of another PQR. Then, prove that ABC PQR. [4] Solution: Sol : Given Two ABC and PQR in which AD and PM are medians such that AB AC AD
PQ PR PM
To prove ABC PQR Construction Produce AD to E and PM to N such that AD = DE and PM = MN. Join CE and NR.
Proof In ADB and EDC, AD = ED [by construction] BD = CD [since, D is mid-point of BC] and ADB = CDE [vertically opposite angles]
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Hence proved. 7. Gopal is walking away from the base of a lamp-post at a speed of 2 m/s and his height
(MN) is 0.9 m. Suppose the lamp-post 4.5 m above from the ground as shown in adjoining figure.
(i) Find the distance of Gopal from the base of lamp-post in 5s. (ii) Is the PQR and MNR similar? (iii) Find the length of his shadow after 5 s. [4] Solution: (i) Let N be the position of Gopal after 5 s, then distance of Gopal From the base of lamp-post, QN = 2 × 5=10m (ii) In PQR and MNR, LQ = LN [since, each angle is 90° because lamp-post as well as Gopal is standing vertically]
R = R [common angle] PQR MNR [by AA similarity criterion]
(iii) Let length of his shadow after 5 s in NR = x Since, PQR – MNR
Construction Join BE, CD and draw EF BA and DG CA. Proof Since EF is perpendicular to AB. Therefore, EF is the height of triangles ADE and DBE.
Now, Area 1
ADE2
(base × height) = 1
2(AD. EF)
and, Area DBE = 1
2 (base × height) =
1
2 (DB. EF)
1AD.EFArea ADE AD2
1Area ABE DBDB.EF2
..(i)
Similarly, we have
1AE. DGArea ADE AE2
1Area DEC ECEC. DG2
..(ii)
But, DAC are on the same base DE and between the same parallels DE and BC.
Area DBE = Area DEC
1 1
Area DBE Area DEC [Taking reciprocals of both sides]
Area ADE Area ADE
Area DBE Area DEC [Multiplying both sides by Area ( ADE)]
AD AE
DB EC [Using (i) and (ii)]
10. In the given figure, ABCD is a quadrilateral P, Q, R and S are the points of trisection of the sides AB, BC, CD and DA, respectively. Prove that PQRS is a parallelogram. [4]
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Solution. Given A quadrilateral ABCD in which P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respectively and are adjacent to A and C. To prove PQRS is a parallelogram i.e., PQ SR and QR PS.
Construction Join AC. Proof Since, P, Q, R and S are the points of trisection of AB, BC, CD and DA respectively.
BP = 2PA, BQ = 2 QC, DR = 2 RC and, DS = 2 SA In ADC, we have DS 2SA DR 2RC
2 and , 2SA SA RC RC
DS DR
SA RC
S and R divide the sides DA and DC respectively in the same ratio. SR AC ..(i)
[By the converse of Thale’s Theorem]
In ABC, we have BP 2PA BQ 2QC
2 and 2PA PA QC QC
BP BQ
PA QC
P and Q divide the sides BA and BC respectively in the same ratio. PQ AC ..(ii) [By the converse of Thale’s Theorem]
From equations (i) and (ii), we have
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