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.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Mar 29, 24 12:11 AM
Mar 28, 24 02:01 AM
Mar 26, 24 11:25 PM