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Active Maths 2 (Strands 1–5): Ch 19 Solutions
Chapter 19 Exercise 19.1
1
Q. 1. (i) An axiom is a statement that is accepted but cannot be proven, e.g. x + 0 = x. (ii) A statement that can be proven logically: for example, Pythagoras’ Theorem. (iii) The logical steps used to prove a theorem, e.g. when proving two triangles are
similar, we show they have equal angles. (iv) A statement derived (found) from a theorem, e.g. it can be derived from Pythagoras’
Theorem that |A| + |B| > |C|, where A, B and C are the sides of a right-angled triangle and C is the hypotenuse.
(v) A theorem in reverse order, e.g. the angles in a quadrilateral sum to 360°. The converse is: a polygon whose angles sum to 360° is a quadrilateral.
Q. 2. (i) Statement: a = b Converse: b = a (true) (ii) Statement: when you stick two equal squares together you get a rectangle Converse: A rectangle is made up of two equal squares (false)
Square Square
Rectangle
Q. 3. Statement Converse Converse is True/False
If the largest angle in a triangle is obtuse, then the two smaller angles are acute.
If the two smaller angles in a triangle are acute, then the largest angle is obtuse.
False
If two triangles are congruent, then they are similar.
If two triangles are similar then they are congruent
False
If a triangle is equilateral, then it is also isosceles.
If a triangle is isosceles then it is also equilateral
False
If a number is divisible by 6, then it is also divisible by 3.
If a number is divisible by 1, then it is divisible by 3
False
Q. 4. Given: Triangles ABC and BDE with parallel lines l and m as shown below |AC| = |DE|.
B
C
m
I
ED
A1
3 4
2
To prove: ΔABC ≡ ΔDBE Construction: Label angles 1, 2, 3 and 4 Proof: |∠1| = |∠4| … alternate |AC| = |DE| … given |∠2| = |∠3| … alternate ⇒ Δ ABC ≡ ΔDBE …… ASA Q.E.D.
2 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 5. (i) Given: Circle with chords [PQ] and [ST] as shown below.
12
3
4
ST
R
Q
P
To prove: Δ PQR ≡ Δ SRT
Construction: Label angles 1, 2, 3, 4
Proof: |∠1| = |∠2| … vertically opposite
|RQ| = |SR| … radii
|∠4| = |∠3| … alternate
⇒ ΔPQR ≡ ΔSRT (ASA)
(ii) |PT| = |QS| … diameters
[|PR| = |RT| = |RQ| = |RS|]
Note: R is the centre of the circle.
Q. 6. (i) Given: Triangles DEF and LMO
|∠E| = 90°
|∠D| = 50°
|∠M| = 90°
|∠O| = 50°
|DF| = |LO|
To prove: ΔDEF = ΔLMO
Proof: |DF| = |∠LO| … given
|∠D| = |∠O| … given
|∠F| = 180° – (90° + 50°)
= 40° … angle is a Δ
|∠L| = 180° – (90° + 50°)
= 40° … angle is a Δ
⇒ |∠F| = |∠L|
⇒ ΔDEF ≡ ΔLMO (ASA)
Q.E.D.
(ii) Given: Triangles ABC and BCD
|∠ABC| = |∠CBD| = 30°
|AB| = |BC| = |BD|
To prove: ΔABC ≡ ΔCBD
Proof: |AB| = |BD| … given
|∠ABC| = |∠CBD| = 30°
… given
BC is a common side
⇒ ΔABC ≡ ΔCBD (SAS)
(iii) Yes, by SSS
(iv) Yes, by ASA
Q. 7. (i) Yes |∠BAC| = |∠DAE|
… common angle
|∠ABC| = |∠ADE|
.… corresponding
|∠ACB| = |∠AED|
.… corresponding.
(ii) Yes |∠PSQ| = |∠RSQ|
.… both 90°
|∠PQS| = |∠QRS| … because triangle
PQR is similar to ΔPSQ and therefore
|∠PQS| = |∠PRQ|
Exercise 19.2
Q. 1. Given: Square ABCD as shown.
B 12
3
C
DA
To prove: |∠ACB| = 45°
Construction: Label angles 1, 2 and 3
3Active Maths 2 (Strands 1–5): Ch 19 Solutions
Proof: |∠2| = 90° … square
|AB| = |BC| … square
⇒ |∠1| = |∠3| … isosceles Δ
⇒ 90° + |∠1| + |∠3|
= 180° … angle is a Δ
⇒ 2|∠1| = 90°
⇒ |∠1| = 45°
Q.E.D
Q. 2. Given: Diagram as shown
Y Z
X
W
To prove: |∠YZW| = 90°
Proof: |XY| = |XZ| = |XW|
So we can draw a circle around Δ WYZ with X as the centre with |XY|, |XZ| and |WX| as radii
⇒ |∠YZW| = 90° … angle in a semicircle.
Q. 3. Given: Diagram as shown.
B
C
DA1
2
3
To prove: |∠ABC| > |∠CDB|
Construction: Label angles 1, 2, 3
Proof: |∠1| = |∠2| + |∠3|
… exterior angle
⇒ |∠1| > |∠3|
Q.E.D.
Q. 4. Given: Triangle PQS as shown.
1
Q S
R
P
AA+B
B
To prove: Δ PQR is isosceles Construction: Label angle 1 Proof: |∠1| = A + B … exterior angle ⇒ Δ PQR is isosceles. Q.E.D.
Q. 5. Given: Triangle BCD with extension A and E as shown |∠ABC| = |∠CDE|.
B
C
EDA1 4 23
To prove: |∠3| = |∠4| Construction: Label angles 1, 2, 3 and 4 Proof: |∠3| = 180° – |∠1| … straight
angle |∠4| = 180° – |∠2| … straight
angle but |∠1| = |∠2| … given ⇒ |∠4| = 180° – |∠1| ⇒ |∠4| = |∠3| Q.E.D.
Q. 6. Given: Diagram as shown.
BC
E
DA
12
4
3
To prove: |∠BCE| = |∠CED| + |∠ABC| Construction: Label angles 1, 2, 3, 4 Proof: |∠1| = |∠2| + |∠4| …
exterior angle |∠4| = |∠3| … alternate ⇒ |∠1| = |∠2| + |∠3|
Q.E.D.
4 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 7. Given: Triangle PQR as shown.
P
X
Q R
1
2 3
To prove: |∠PXQ| >|PRQ|
Construction: Label angles 1, 2 and 3
Proof: |∠1| = |∠2| + |∠3|
… exterior angle
⇒ |∠1| > |∠3|
Q.E.D.
Q. 8. Given: Isosceles Δ ABC as shown.
DC
B
A
2
3 4 1
To prove: |∠DCB| = 90° + |∠ABC|
________ 2
Construction: Label angles 1, 2, 3, 4
Proof: |∠3| = |∠4| … isosceles Δ
⇒ |∠4| = 180° – |∠1|
… straight angle
|∠2| = 180° – (|∠3| + |∠4|)
… angles in a Δ
⇒ |∠2| = 180° – (180° – |∠1| + 180° – |∠1|)
⇒ |∠2| = 2 |∠1| – 180°
⇒ 2|∠1| = 180° + |∠2|
⇒ |∠1| = 90° + |∠2|
_____ 2
Q.E.D.
Q. 9. Given: Diagram as shown.
DCBE
A
7
51 4 6 2
8
3
To prove: Δ ABE ≡ Δ ACD
Construction: Label angles 1, 2, 3, 4, 5, 6, 7, 8
Proof: |AE| = |AD| … given S |AB| = |AC| … given S |∠1| = |∠2| … isosceles Δ
|∠3| = |∠4| … isosceles Δ
⇒ |∠5| = 180° – |∠3| = 180° – |∠4| = |∠6| ⇒ |∠7| = |∠8|
… 2 equal angles in a Δ implies the 3rd angles are equal A
⇒ Δ ABE ≡ Δ ACD (SAS)
Q.E.D.
Q. 10. (i) |AM| = |AM| (common)
|BM| = |MC| (given)
|AB| = |AC| (given)
∴ ΔABM ≡ ΔAMC (side, side, side)
(ii) ∠AMB in ΔAMB corresponds to ∠AMC in ΔAMC, Therefore |∠AMB| = |∠AMC|
(iii) |∠AMB| = |∠AMC| = 180° (Straight line)
but |∠AMB| = |∠AMC|
⇒|∠AMB| = |∠AMC| = 90°
Therefore AM ⊥ BC
5Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 11. Given: Δ XYZ as shown.
X
ZY
2
4 6 3
5
1
To prove: |∠1| + |∠2| + |∠3| = 360°
Construction: Label angels 4, 5 and 6
Proof: |∠1| = |∠5| + |∠6| … exterior angle |∠2| = |∠4| + |∠6| … exterior angle |∠3| = |∠4| + |∠5| … exterior angle ⇒ |∠1| + |∠2| + |∠3| = 2 ( |∠4| + |∠5| + |∠6| ) but |∠4| + |∠5| + |∠6| = 180° … angle in a Δ
⇒ |∠1| + |∠2| + |∠3| = 2(180°) = 360° Q.E.D.
Q. 12. (i) Given: Δ PQR as shown below |PQ| = |PR|
X
Y
RQ
P
1 23
4
5
6
To prove: Δ XYP is isosceles Construction: Label angles 1–6 Proof: |∠1| = |∠2| … isosceles Δ
|∠3| + |∠2| + 90° = 180° … angles in a Δ
⇒ |∠3| = 90° – |∠2| ⇒ |∠3| = 90° – |∠1| |∠4| = |∠3| … vertically
opposite
⇒ |∠4| = 90° – |∠1| |∠5| + |∠1| + 90° = 180° … angles in a Δ
⇒ |∠5| = 90° – |∠1| ⇒ |∠4| = |∠5| ⇒ Δ XYP is isosceles Q.E.D.
(ii) But Δ PXY may be equilateral
|∠6| = |∠1| + |∠2| … exterior angle ⇒ |∠6| = 2|∠1| If Δ PXY is equilateral |∠6| = |∠4| = |∠5| ⇒ 2|∠1| = 90° – |∠1| ⇒ 3|∠1| = 90°
⇒ |∠1| = 30° So when |∠PQR| = 30°
Δ XYP is equilateral
Q. 13. Given: Diagram as shown.
A C D
B
1 3
245
To prove: |∠ABC| = 2|∠CDB|
Construction: Label angles 1, 2, 3, 4, 5, 6
Proof: |∠1| = |∠4| … isosceles Δ
|∠2| = |∠3| … isosceles Δ
|∠1| + |∠2| + |∠3| + |∠5|
= 180° … angles in a Δ
⇒ |∠1| + 2|∠2| + |∠5| = 180°
|∠1| + |∠4| + |∠5| = 180°
… angles in a Δ
⇒ 2|∠1| + |∠5| = 180°
|∠1| + 2|∠2| + |∠5|
= 2|∠1| + |∠5|
⇒ |∠1| = 2|∠2|
Q.E.D.
6 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 14. Given: Triangles PQS and SQR as shown below
P Q R
S
21
3 4 5 6
To prove: |∠PSQ| + |∠QSR| = 90°
Construction: Label angles 1–6 as shown
Proof: |∠1| = |∠3| … isosceles Δ
|∠5| = |∠1| + |∠3|
… exterior angle
⇒ |∠5| = 2|∠1|
|∠2| = |∠6| … isosceles Δ
|∠4| = |∠2| + |∠6|
… exterior angle
⇒ |∠4| = 2|∠2|
|∠4| + |∠5| = 180°
… straight angle
⇒ 2|∠2| + 2|∠1| = 180°
⇒ 2(|∠1| + |∠2|) = 180°
⇒ |∠1| + |∠2| = 90°
Q.E.D.
Q. 15. (a) Given:
A
B C
D
Ezw
T
X Y
To prove: [AE bisects DAC.
Proof: |Y| = |W| Alternate
|Y| = |X| isosceles
|X| = |Z| Corresponding
⇒ |W| = |Z|
Therefore [AE bisects ∠DAC
Q.E.D.
(b) No, the result in (a) would not still apply. Angle Y would not be equal to angle X.
Exercise 19.3
Q. 1. Given: Parallelogram ABCD as shown.
X
CD
A B
Y 24
6
5
1
3
To prove: Δ ADY similar to Δ BXY
Construction: Label angles 1, 2, 3, 4, 5, 6
Proof: |∠1| = |∠5| … alternate |∠2| = |∠4| … vertically
opposite |∠3| = |∠6| … alternate ⇒ Δ ADY and Δ BXY are similar
Q.E.D.
Q. 2. Given: Rhombus PQRS as shown.
P S
RQ
21
34
To prove: [PR] bisects |∠QPS| Construction: Label angles 1– 4 Proof: |∠1| = |∠3| … alternate |∠3| = |∠2| … isosceles Δ
⇒ |∠1| = |∠2|
⇒ [PR] bisects |∠QPS|
Q.E.D.
7Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 3. Given: Parallelogram ABCD and ABEC as shown.
A
D C E
B
To prove: Δ ADC ≡ BCE
Construction: Δ ADC ≡ Δ BCE
Proof: |AD| = |BC| … parallelogram
|AC| = |BE| … parallelogram
|DC| = |AB| … parallelogram
|AB| = |CE| … parallelogram
⇒ |DC| = |CE|
⇒ Δ ADC ≡ Δ BCE (SSS)
Q.E.D.
Q. 4. Given: Rectangle DEFG as shown
D G
E F21
To prove: |DF| = |EG|
Construction: Label angles 1 and 2
Proof: |DE| = |GF| … opposite sides
|∠1| = |∠2| … right angles
EF is a common side in Δ DEF and Δ EFG
⇒ Δ DEF ≡ Δ EFG
⇒ |DF| = |EG|
… Corresponding sides
Q.E.D.
Q. 5. Given: Parallelogram LMOP as shown
L M
OP
1 2
3 4
To prove: |∠1| + |∠2| = 180°
Construction: Label angles 3 and 4
Proof: |∠3| = |∠2| … opposite angles
|∠4| = |∠1| … opposite angles
|∠1| + |∠2| + |∠3| + |∠4|
= 360° … quadrilateral
⇒ 2(|∠1| + |∠2|) = 360°
⇒ |∠1| + |∠2| = 180°
Q.E.D.
Q. 6. Given: Parallelogram ABCD as shown. |AD| = 2|AB|. M is the midpoint of [BC]
A D
B M C
21
3 47
56
To prove: |∠AMD| = 90°
Construction: Label angles 1, 2, 3, 4, 5, 6, 7
Proof: |AD| = |BC| … opposite sides
⇒ |AD| = 2|AB|
⇒ 1 __ 2
|AD| = |AB|
⇒ 1 __ 2 |BC| = |AB|
⇒ |BM| = |AB|
⇒ |∠1| = |∠3|
|∠2| = |∠3| … alternate
⇒ |∠1| = |∠2|
Similarly |CM| = |CD|
⇒ |∠4| = |∠6|
|∠5| = |∠4| … alternate
⇒ |∠5| = |∠6|
|∠1| + |∠2| + |∠5| + |∠6|
= 180°
⇒ 2(|∠2| + |∠5|) = 180°
⇒ |∠2| + |∠5| = 90°
8 Active Maths 2 (Strands 1–5): Ch 19 Solutions
|∠2| + |∠5| = |∠7| = 180° … angles in a Δ
⇒ 90° + |∠7| = 180° ⇒ |∠7| = 90°
Q.E.D.
Q. 7. (a) Given: Parallelogram PQRS as shown.
X
Y
Q R
SP
2
3
1 4
5
6
To prove: |SY| = |QX|
Construction: Label angles 1, 2, 3, 4, 5, 6
Proof: |∠1| = |∠6| … alternate |PQ| = |SR| … opposite sides |∠3| = |∠5| … given ⇒ Δ RSY ≡ Δ PXQ (ASA)
(b) ⇒ |SY| = |QX|
… corresponding sides
Q.E.D.
Q. 8. Given: Parallelogram ABCD. M is the midpoint of [AC].
D X A
M
C Y B
2
3
4
1
To prove: |MX| = |MY|
Construction: Label angles 1, 2, 3, 4 Proof: |∠1| = |∠4| … alternate |CM| = |AM| … given |∠2| = |∠3| … vertically
opposite ⇒ Δ CYM ≡ Δ AXM (ASA)
⇒ |MX| = |MY|
… corresponding sides
Q.E.D.
Q. 9. (i) Given: Parallelogram PQRS. T is the midpoint of [SR].
R
U
S T
P Q
213 4
To prove: Δ PST ≡ Δ RTU
Construction: Label angles 1, 2, 3, 4
Proof: |∠1| = |∠4|
… alternate
|ST| = |TR| … given
|∠2| = |∠3|
… vertically opposite
⇒ Δ PST ≡ Δ RTU (ASA)
Q.E.D.
(ii) |RU| = |PS| … corresponding sides
Q.E.D.
(iii) |QU| = |QR| + |RU|
|QR| = |PS| … opposite sides
|RU| = |PS| … part (ii)
⇒ |QU| = 2|PS|
Q.E.D.
Q. 10. (i) Given: Square DEFG. P, Q, R, S are the midpoints of the sides.
D P E
QSO
G R F14
572
3
68
9Active Maths 2 (Strands 1–5): Ch 19 Solutions
To prove: Δ DOS ≡ Δ GRO
Construction: Label angles 1, 2, 3, 4, 5, 6
Proof: |∠1| = |∠2| … both right angles |DO| = |GO| … diagonal bisect |GR| = |SO| ⇒ Δ DOS ≡ Δ GRO (RHS) Q.E.D.
(ii) Given: Same as part (i)
To prove: Δ DOG ≡ Δ DOE
Construction: Label angles 7, 8
Proof: |∠7| = |∠8|
… diagonals meet at night angles
|DG| = |DE| … square
|GO| = |EO|
… diagonals bisected
⇒ Δ DOG ≡ Δ DOE (RHS)
Q.E.D.
(iii) It can be shown that
Δ ROF ≡ Δ QOF ≡ Δ QOE ≡ Δ POE ≡ etc.
Using a similar proof on part (i)
⇒ Total area = 8 × area Δ ROF
= 8 × 8
= 64 cm2
Q. 11. Given: Parallelogram ABCD as shown, [AX] ⊥ [BC] and [CY] ⊥ [AD]
B X C
P
A Y D
Q
13
5 42
6
To prove: Δ PAB ≡ Δ CDQ
Construction: Label angles 1, 2, 3, 4, 5, 6
Proof: |∠1| = |∠2|
… alternate
As AD || BC and AX ⊥ BC and CY ⊥ AD
⇒ AX || CY
⇒ |∠3| = |∠4|
… alternate
⇒ |∠5| = |∠6|
… 2 equal angles implies the 3rd angles are equal
|AB| = |CD|
… opposite side
⇒ Δ PAB ≡ Δ CDQ (ASA)
Q.E.D.
∴ |BP| = |DQ|
… corresponding sides
Q.E.D.
Exercise 19.4
Q. 1. Consider Δs BEF and DEA.
|∠FBE| = |∠ADE| … alternate
|∠BFE| = |∠DAE| .… alternate
|∠BEF| = |∠DEA|
.… vertically opposite
∴ ΔBEF is similar to ΔDEA.
∴ |BE|
_____ |DE|
= |EF|
_____ |EA|
|DE|.|EF| = |BE|.|AE|.
Q. 2. RQ || TS
∴ |∠PQR| = |∠PST|
.… corresponding
|∠PRQ| = |∠PTS|
.… corresponding
|∠TPS| = |∠RPQ|
… common angle.
∴ ΔRPQ is similar to ΔTPS
∴ |RQ|
_____ |PR|
= |TS|
_____ |PT|
====
10 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 3. (i)
C
1
2 3
4
DA
B
E
Label angles 1 → 4
|∠1| + |∠2| = 90° as ΔABC is right-angled.
|∠2| + |∠3| = 90° … straight angle
∴ |∠1| + |∠2| = |∠2| + |∠3|
∴ |∠1| = |∠3|
|∠3| + |∠4| = 90° as ΔCDE is right-angled
|∠2| + |∠3| = 90° .… straight angle
∴ |∠3| + |∠4| = |∠2| + |∠3|
∴ |∠4| = |∠2|
Also |∠BAC| = |∠EDC|
… both 90°
∴ ΔBAC is similar to ΔCDE.
(ii) Redrawing Δs
C
1
2
3
4D EA
B
C
We have |AC|
_____ |DE|
= |BC|
_____ |CE|
… similar triangles theorem.
Q. 4. We note that |AC| . |FG| = |AG| . |BC|
could also be written as |AC|
____ |BC| = |AG|
____ |FG| .
BD || CE
⇒ |AC|
____ |BC| = |AE|
____ |DE|
DF || EG
⇒ |AE|
____ |DE| = |AG|
_____ |FG|
⇒ |AC| ____
|BC| =
|AG| _____ |FG|
∴ |AC| . |FG| = |AG| . |BC| (cross-multiply)
Q. 5. (i) Consider ΔABE and ΔFCE
|∠ABE| = |∠FCE| .… alternate |∠BAE| = |∠CFE| .… alternate |∠BEA| = |∠CEF| … vertically
opposite ∴ ΔABC is similar to ΔFCE
∴ |AE|
_____ |FE|
= |BE|
_____ |CE|
(ii) From (i) ΔABE is similar to ΔFCE
∴ |AB|
_____ |FC|
= |BE|
_____ |CE|
|AB|.|CE| = |BE|.|FC|
But |AB| = |DC|
… opposite sides of a parallelogram
∴ |DC|.|CE| = |BE|.|FC|
Q. 6.
S
Q
T
R
Δ SRT is similar to Δ QRS
as |∠QRS| = |∠SRT| … common angle and |∠RQS| = |∠RST| … both 90° ⇒ |∠RSQ| = |∠RTS| Also |∠RQS| = |∠SQT| … both 90°
∴ Δ RQS is similar to Δ SQT.
∴ |RQ|
_____ |QS|
= |QS|
_____ |QT|
⇒ x __ y = y _____
|QT|
|QT|x = y2
|QT| = y2
__ x
11Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 7. Δs ADF and AEC
|AD|
_____ |DE|
= |AF|
_____ |FC|
… 1
Δs AEF and ABC
|AE|
_____ |EB|
= |AF|
_____ |FC|
… 2
1 = 2
|AD|
_____ |DE|
= |AE|
_____ |EB|
Q. 8. (i) |∠PTQ| = |∠TQS| ..… alternate
|∠TQP| = |∠QST| … both 90°
∴ Δ QTS is similar to Δ QTP.
(ii) Δ QTS is similar to Δ QTP
∴ |QT|
_____ |QS|
= |PT|
_____ |QT|
|QT|2 = |PT|.|QS|
Q. 9. In Δ ABC and Δ BDC
|∠ABC| = |∠BDC| … both 90° |∠ACB| = |∠BCD| … common angle. ∴ Δ ABC is similar to Δ BDC. ⇒ |∠CAB| = |∠CBD| also |∠ADB| = |∠BDC| … both 90° ∴ Δ ADB is similar to Δ BDC
and also Δ ABC is similar to Δ ADB
⇒ |AB|
_____ |AC|
= |AD|
_____ |AB|
⇒ |AB|2 = |AD|.|AC|
Q. 10. (a) In Δ ADB and Δ APC
|∠ADB| = |∠ACP|
... angles at the circle subtended by the same are
|∠DAB| = |∠CAP|
... as |AD| bisects ∠BAC
∴ Δ ABD is similar to Δ APC
(b) As Δ ABD is similar to Δ APC
|AC|
_____ |AD|
= |PC|
_____ |BD|
|AC|.|BD| = |AD|.|PC|
Q. 11. (i) In Δ PQT and Δ RWQ
|∠PQT| = |∠RWQ|
..… alternate angles
|∠TPQ| = |∠QRW|
… opposite angles in a parallelogram.
∴ ΔPQT is similar to ΔRWQ.
(ii)
RQ
P Sx2x
3x
T
W
Let |TS| = x, ⇒ |PT| = 2x
⇒ |QR| = 3x
From (i) |PQ|
______ |RW|
= |PT|
_____ |QR|
⇒ |PQ|
______ |RW|
= 2x ___ 3x = 2 __ 3
⇒ |PQ| = 2 __ 3 |RW|
But |PQ| = |RS|
∴ |RS| = 2 __ 3 |RW|
⇒ |RS| : |RW|
2 : 1
Exercise 19.5
Q. 1. Given: Square ABCD and DBFE.
A Bx
x xy
y y
y
x CD
E
F
To prove: Area DBEF = 2 × (area ABCD)
Construction: Mark sides of ABCD as x and DBEF as y.
12 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Proof: Using Pythagoras’ Theorem x2 + x2 = y2
⇒ 2x2 = y2
⇒ 2(area of ABCD) = area of BDFE
Q.E.D.
Q. 2. In Δ PSQ , |PQ|2 = |PS|2 + |QS|2
In Δ RSQ , |QR|2 = |SR|2 + |QS|2
But |PQ| = |QR| ∴ |PQ|2 = |QR|2
∴ |PS|2 + |QS|2 = |SR|2 + |QS|2
|PS|2 = |SR|2
Q. 3. Given: Triangle ABC as shown.
A
B D C
To prove: |AB|2 + |DC|2 = |BD|2 + |AC|2
Proof: |AB|2 = |BD|2 + |AD|2 … Pythagoras’ Theorem ⇒ |AD|2 = |AB|2 – |BD|2
|AC|2 = |AD|2 + |CD|2 … Pythagoras’ Theorem ⇒ |AD|2 = |AC|2 – |CD|2
⇒ |AC|2 – |CD|2 = |AB|2 – |BD|2
⇒ |AB|2 + |DC|2 = |BD|2 + |AC|2
Q.E.D.
Q. 4. Given: Triangles ABC and APQ as shown.
C
A P B
Q
To prove: |BQ|2 + |CP|2
= |PQ|2 + |BC|2
Construction: Draw PC and BQ.
Proof: |AQ|2 + |AP|2 = |PQ|2
… Pythagoras
⇒ |AC|2 + |AB|2 = |BC|2
… Pythagoras
⇒ |AQ|2 + |AP|2 + |AC|2 + |AB|2
= |PQ|2 + |BC|2
⇒ (|AQ|2 + |AB|2) + (|AP|2 + |AC|2)
= |PQ|2 + |BC|2
⇒ |BQ|2 + |CP|2 = |PQ|2 + |BC|2
… Pythagoras
Q.E.D.
Q. 5. Given: Circle with centre O, tangents [AB] and [AC] as shown.
1
2
AO
B
C
To prove: |AB| = |AC| Construction: Draw the line AO.
Label angles 1 and 2 Proof: |∠1| = |∠2| … right angles AO is a common side |OB| = |OC| … radii ⇒ Δ AOB ≡ Δ AOC (RHS) ⇒ |AB| = |AC| … corresponding sides
Q.E.D.
Q. 6. In Δ ABC
|AC|2 = |BC|2 + |AB|2
In semicircle Q
Area = 1 __ 2 pr2
= 1 __ 2 p ( |AC| _____ 2 ) 2
= |AC|2 p
_______ 4
13Active Maths 2 (Strands 1–5): Ch 19 Solutions
In semicircle R we similarly have
Area = |BC|2p
_______ 4
and in semicircle P, Area = |AB|2p
_______ 4
So is |AC|2p
_______ 4 = |BC|2p
_______ 4 + |AB|2p
_______ 4 ?
Is |AC|2 p = |BC|2 p + |AB|2 p ?
Is |AC|2 = |BC|2 + |AB|2 True
Q. 7. By Pythagoras’ Theorem
|AB|2 = |AC|2 + |BC|2
But |AC| = |BC|
so |AB|2 = 2|BC|2
|AB| = √__
2 |BC|
Q. 8.
RP S
Q
Construction: QS, the perpendicular bisector of PR
In Δ QSP, |QS|2 = |PQ|2 − |PS|2
But Δ QPR is equilateral so |PQ| = |PR| and also |PS| = 1 __ 2 |PR|
∴ |QS|2 = |PR|2 − ( 1 __ 2 |PR| ) 2 = |PR|2 − 1 __ 4 |PR|2
|QS|2 = 3 __ 4 |PR|2
∴ |QS| = √__
3 __ 4 |PR|
= √__
3 ___ 2 |PR|
Q. 9. Q R
P N
M
S
T
Construction: [MN] through T such that [MN] || [QP] and [MN] ⊥ [PS] Join T to Q, R, P and S
By Pythagoras’ Theorem: |TM|2 = |QT|2 – |QM|2 = |TR|2 – |MR|2
|QT|2 – |QM|2 = |TR|2 – |MR|2
|QT|2 – |TR|2 = |QM|2 – |MR|2
Similarly |TN|2 = |PT|2 – |PN|2 =|TS|2 – |NS|2
So |PT|2 – |TS|2 = |PN|2 – |NS|2
But |QM|2 – |MR|2 = |PN|2 – [NS]2
as |QM| = |PN| and |MR| = |NS| ∴|QT|2 – |TR|2 = |PT|2 – |TS|2
|QT|2 + |TS|2 = |PT|2 + |TR|2
14 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 10. By Pythagoras’ Theorem
|EC|2 = |EF|2 + |FC|2
Now |EF| = |AB|
∴ |EF|2 = |AB|2
And by Pythagoras’ Theorem |FC|2 = |AF|2 + |AC|2
∴ |EC|2 = |AB|2 + |AF|2 + |AC|2
Exercise 19.6
Q. 1. Given: Triangle EDF inside circle as shown.
D
O
FE1 23
4
56
To prove: Δ DEO ≡ Δ EFO
Construction: Label angles 1, 2, 3, 4, 5, 6
Proof: |∠5| = |∠6| = 90°
… given + straight angle
|OD| = |OF| … radii
EO is a common side
⇒ Δ DEO ≡ Δ EFO (SAS)
Q.E.D.
Q. 2. Given: Diagram as shown.
O3
4
2
1
To prove: |∠1| = |∠2|
Construction: Label angle 3. Draw extension from ∠3 as shown and label angle 4
Proof: |∠1| + |∠3| = 180° … opposite angles in cyclic
quadrilateral |∠2| = |∠4| … alternate angle |∠3| + |∠4| = 180° … straight
angle ⇒ |∠3| + |∠2| = 180° ⇒ |∠1| + |∠3| = |∠3| + |∠2| ⇒ |∠1| = |∠2|
Q.E.D.
Q. 3. Given: Diameter [BD]. Centre O.
B
A
O
D
13
2
To prove: |∠1| + |∠2| = 90°
Construction: Mark equal radii. Label angle 3
Proof: |∠1| + |∠3| = 90°
… angle in a semi circle
|∠2| = |∠3| … isosceles Δ
⇒ |∠1| + |∠2| = 90°
Q.E.D.
Q. 4. Given: Chords [PQ] and [RS] of a circle as shown.
P R
QS
X1
2
3 45
6
To prove: Δ PXS and Δ RXQ are similar
15Active Maths 2 (Strands 1–5): Ch 19 Solutions
Construction: Mark angles 1, 2, 3, 4, 5, 6
Proof: |∠1| = |∠6| … angles standing on the same arc
|∠2| = |∠5|
… angles standing on the same arc
|∠3| = |∠4|
… vertically opposite
⇒ Δ PXS and Δ RXQ are similar
Q.E.D.
Q. 5. Given: Cyclic quadrilateral ABCD as shown.
A D E
B
C
1
2 3
To prove: |∠CDE| = |∠CBA|
Construction: Label angles 1, 2, 3
Proof: |∠1| + |∠2| = 180°
… opposite
|∠2| + |∠3| = 180°
… straight angle
⇒ |∠1| + |∠2| = |∠2| + |∠3|
⇒ |∠1| = |∠3|
Q.E.D.
Q. 6. (i) Given: Diameter [AB] and [PQ] of circle C as shown.
A Q
P B
1 2
4 3
To prove: |∠APQ| ≡ |∠PQB|
Construction: Construct the quadrilateral AQBP and label angles 1, 2, 3, 4
Proof: |∠1| = 90°
… angle in a semi circle
|∠2| = 90°
… angle in a semi circle
also |∠3| = |∠4| = 90°
… angle in a semi circle
⇒ AQBP is a rectangle
⇒ |∠APQ| = |∠PQB|
… alternate angles
Q.E.D.
(ii) AP || BQ
This holds and AQBP is a rectangle
Q.E.D.
Q. 7. Given: Diagram as shown
O
CB
A
To Prove: |∠ABO| = |∠CBO|
Construction: Draw [OC] and [OA]
Proof: |OC| = |OA| … radii |CB| = |AB| … given [OB] is a common side ⇒ Δ OCB ≡ Δ OAB (SSS) ⇒ |∠ABO| = |∠CBO| … corresponding angle
16 Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 8. (i) Given: Diagram as shown
A B1
23
4C
O
D
To Prove: |∠ABO| = 2q
Construction: Label angles 1, 2
Proof: |∠2| = q … isosceles Δ
|∠1| = |∠2| + q
… exterior angle
⇒ |∠1| = 2q
Q.E.D.
(ii) To prove: |∠AOD| = 4q
Construction: Label angles 3, 4
Proof: |∠3| = |∠1| + |∠4|
… exterior angle
but |∠1| = |∠4| = 2q
… isosceles Δ (radii)
⇒ |∠3| = 4q
Q.E.D.
Q. 9. Given: O is the centre of a circle. M is the midpoint of [AB], as shown.
A BM
O
P
1 2
3
To prove: |∠AOM| = |∠APB|
Construction: Label angles 1, 2, 3
Proof: |AM| = |MB| … given
|AO| = |OB| … radii
OM is a common side
⇒ Δ AOM ≡ Δ BOM (SSS)
⇒ |∠1| = |∠2|
… corresponding angles
|∠1| + |∠2| = 2|∠3|
… angle at the centre
⇒ 2|∠1| = 2|∠3|
⇒ |∠1| = |∠3|
Q.E.D.
Q. 10. (i) Given: Diameter [AB] of circle with centre C. |AB| = |BQ| as shown.
AC B
P
Q
12
To prove: |∠BPQ| = 90°
Construction: Join B to P. Label angles 1, 2.
Proof: |∠1| = 90°
… angles in a semi circle
|∠2| = 180° – 90° = 90°
… straight angle
Q.E.D.
(ii) To prove: |AP| = |PQ| Proof: |∠1| = |∠2| … right angles |AB| = |BQ| … given PB is a common side ⇒ Δ APB ≡ Δ QPB
⇒ |AP| = |PQ| … corresponding sides ⇒ P is the midpoint of
[AQ]
17Active Maths 2 (Strands 1–5): Ch 19 Solutions
Q. 11. Given: Quadrilateral ABCD as shown.
AB
CD
1 2
34
To prove: ABCD is a cyclic quadrilateral
Construction: Label angles 1, 2, 3, 4
Proof: |∠1| + |∠2| + |∠3| + |∠4|
= 360° … quadrilateral
|∠2| = |∠4| = 90° … given
⇒ |∠1| + 90°+ |∠3| + 90°
= 360°
⇒ |∠1| + |∠3| = 180°
So opposite sides sum to 180°
⇒ ABCD is a cyclic quadrilateral
Q.E.D.
Q. 12. (i) Given: Diagram as shown.
CA B
P
Q
1
2
435
To prove: PC || QB
Construction: Label angles 1, 2
Proof: |∠1| = |∠2| = 90°
… angles in a semicircle
⇒ PC || QB
Q.E.D.
(ii) To prove: |AP| = |PQ|
Proof: Label angles 3, 4, 5
|∠1| = |∠2| … above
|∠3| = |∠4|
… corresponding
|∠5| is common to Δ APC and Δ AQB
⇒ Δ APC and Δ AQB are similar
|AP|
_____ |PQ|
= |AC|
_____ |CB|
but |AC| = |CB| … radii
⇒ |AP|
_____ |PQ|
= 1
⇒ |AP| = |PQ|
Q.E.D.