Mathematics Secondary 2. Chapter 1 Congruence and Similarity. Page 1

Chapter 1 Congruence and Similarity.
 
1.  ∆PQR is similar to ∆ABC. Given that PQ = 5 cm, QR = 4 cm and AB = 8 cm, calculate the length of BC.

^PQ/_QR = ^AB/_BC
⁵/₄ = ⁸/_BC
BC = ^{8 x 4}/₅
BC = ³²/₅=6²/₅

2.  ABCD is a trapezium where AB is parallel to DC. Name a pair of similar triangles. If
     AM = 5 cm, MC = 7 cm, BM = 6 cm, AB = 8 cm, DC = x cm and MD = y cm, find the
    values of x and y.

^DC/_AB = ^MC/_AM
^x/₈ = ⁷/₅
x = ^{8 x 7}/₅
x = ⁵⁶/₅=11¹/₅
^MD/_BM = ^MC/_AM
^y/₆ = ⁷/₅
y = ^{6 x 7}/₅
y = ⁴²/₅=8²/₅

3. State, with reasons, whether ∆ABC is similar to ∆PQR and find the value of x.

 

angle of P = 45, angle of C= 35, so ∆ABC is similar to ∆PQR
^QR/_BC = ^QP/_BA
^x/₇ = ⁶/₄
x = ^{6 x 7}/₄
x = ⁴²/₄=10¹/₂

4.  ∆OAB is similar to ∆OPQ.

(a) Explain clearly why AB is parallel to QP.

(b) If OA = 5 cm, OB = 6 cm, OQ = 8 cm, QP = 5.5 cm, OP = x cm and AB = y cm, find the values of x and y.

(a). The angle of Q = angle of B and they form alternate angles, AB // QP.
(b). ^OP/_OA = ^OQ/_OB
x = ^{(OQ x OA)}/_OB
x = ^{8 x 5}/₆=⁴⁰/₆
= 6²/₃
^AB/_QP = ^OB/_OQ
^y/_5.5 = ⁶/₈
y = ^{6 x 5.5}/₈
y = ³³/₈=4¹/₈

5. In Figure, ∆ABC is similar to ∆PQR. Given that AB = 8 cm, AC = 10 cm, PQ = 12 cm, QR = 9cm, BC = x cm and PR = y cm, calculate the values of x and y.

^BC/_QR = ^AB/_QP
^x/₉ = ⁸/₁₂
x = ^{8 x 9}/₁₂
x = ⁷²/₁₂=6
^RP/_CA = ^QP/_AB
^y/₁₀ = ¹²/₈
y = ^{12 x 10}/₈
y = 15

Need more practise, go to number 6