Problem 1 :
In the adjacent figure, triangle ABC is right angled at C and DE ⊥ AB . Prove that ΔABC ∼ ΔADE and hence find the lengths of AE and DE.
In triangle ABC,
AB2 = BC2 + AC2
AB2 = 122 + 52
AB2 = 144 + 25 = 169
AB = 13
In triangles ABC and EAD
<BCA = <AED
<CAB = <DAE
So, the triangles ABC and EAD are similar.
BC/EA = CA/ED = AB/AD
12/EA = (2+3)/ED = 13/3
12/EA = 5/ED = 13/3
12/EA = 13/3
EA = 36/13
5/ED = 13/3
ED = 5(3)/13
ED = 15/13
Problem 2 :
In the adjacent figure, ΔACB ∼ ΔAPQ . If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.
Solution :
Since the triangles ACB and APQ are similar.
BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm
BC/PQ (Angles opposite to <BAC and <PAQ) --(1)
AC/PA (Angles opposite to <ABC and <AQP) --(2)
AB/AQ (Angles opposite to <ACB and <APQ) --(3)
(1) = (2) = (3)
BC/PQ = AC/PA = AB/AQ
8/4 = AC/2.8 = 6.5/AQ
8/4 = AC/2.8
AC = 8(2.8)/4
AC = 5.6 cm
8/4 = 6.5/AQ
AQ = 6.5(4)/8
AQ = 3.25 cm
Problem 3 :
If figure OPRQ is a square and <MLN = 90° . Prove that (i) ΔLOP ∼ ΔQMO (ii) ΔLOP ∼ ΔRPN (iii) ΔQMO ∼ ΔRPN (iv) QR2 = MQ × RN
Solution :
(i) ΔLOP ∼ ΔQMO
In triangles ΔLOP and ΔQMO
<OMQ = <LOP (Corresponding angles) A
<MQO = <OLP (90) A
OQ = OP (side)
By ASA, the above triangles are similar.
(ii) ΔLOP ∼ ΔRPN
<LPO = <PNR (corresponding angles) A
<OLP = <PRN (90) A
OP = PR (side)
By ASA, the above triangles are similar.
(iii) ΔQMO ∼ ΔRPN
<OQM = <PRN (A)
OQ = PR (Side)
Since the triangles ΔLOP and ΔQMO are similar,
OP/MO = OL/OQ = LP/MQ
OP/PN = OL/RN = LP/PR
MQ = PR ----(1)
OQ = RN ----(2)
MQ = RN
Hence the ΔQMO ∼ ΔRPN.
(iv) QR2 = MQ × RN
Let us join OR,
OR2 = OQ2 + QR2
OP2 + PR2 - OQ2 = QR2
OP2 = QR2
OP = MQ = RN
OP (OP) = QR2
MQ x RN = QR2
Hence proved.
Problem 4 :
If ΔABC ∼ ΔDEF such that area of ΔABC is 9cm2 and the area of ΔDEF is 16cm2 and BC = 2.1 cm. Find the length of EF.
Solution :
Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
Area of ΔABC / Area of ΔDEF = BC2/EF2
(9/16) = (2.1)2/EF2
EF2 = 4.41 (16/9)
EF = 2.8 cm
Problem 5 :
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.
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