Exam practice

$\angle BDC = \angle BAC = 30°$ (same arc BC). In △BCD: x = 180 − 70 − 30 = 80.

$\angle APB = 90°$ (angle in a semicircle), so $\angle PAB = 180° - 90° - 35° = 55°$. x = ∠PQB = ∠PAB (same arc PB) = 55.

$\angle OAB = \angle OBA = 25°$ and $\angle OAC = \angle OCA = 35°$ (radii, isosceles triangles), so $\angle BAC = 60°$. x = ∠BOC = 2∠BAC = 120.

$\angle BOC = \angle BOD + \angle DOC = 90° + 60° = 150°$. x = ∠BAC = ½ ∠BOC (same arc BC) = 75.

OA = OC, so $\angle OCA = \angle OAC = 50°$, that is $\angle DCA = 50°$. x = ∠DBA = ∠DCA (same arc DA) = 50.

Reflex $\angle AOC = 360° - 130° = 230°$, and ∠ABC stands on the major arc AC, so x = ½ × 230 = 115.

$\angle ACB = 90°$ (angle in a semicircle), so $\angle CBA = 180° - 90° - 50° = 40°$. In the right triangle CDB: x = 180 − 90 − 40 = 50.

$\angle BOC = 180° - 130° = 50°$ (AOB is a straight line). x = ∠CDB = ½ ∠COB (same arc CB) = 25.

$\angle BDC = \angle BAC = 40°$ (same arc BC). In △DBC: x = ∠DCB = 180 − 80 − 40 = 60.

OA = OC, so $\angle OCA = \angle OAC = 50°$, that is $\angle BCA = 50°$. x = ∠BDA = ∠BCA (same arc AB) = 50.

$\angle OAB = 25°$ and $\angle OAC = 30°$ (isosceles triangles with radii as sides), so $\angle BAC = 55°$. Obtuse $\angle BOC = 2 \times 55° =$ 110∘°.

QR is a diameter, so $\angle QPR = 90°$ and $\angle PRQ = 180° - 90° - 40° = 50°$. x = ∠PSQ = ∠PRQ (same arc PQ) = 50∘°.

• i.∠AFC = ∠CED = 68∘° (an exterior angle of a cyclic quadrilateral equals the interior opposite angle).
• ii.The angle AOC on the arc AEC = 2∠AFC = 136∘° (the central angle on the same arc). The reflex angle $AOC = 360° - 136° =$ 224∘°.

ABCD is cyclic, so $\angle ABC = 180° - 100° = 80°$. $\angle ACB = 90°$ (angle in a semicircle), so $\angle BAC = 180° - 90° - 80° =$ 10∘°.

Add arc AF to each of the equal arcs. Which angles stand on arc AFE and arc FAB?

Proved.

Join BC.

Proved.

Proved.

Proved.

Join DG. Which triangle has ∠EHG as an exterior angle?

Join DG.

Proved.

Join MS and look at the triangle MSX.

Join MS.

Proved. (Arcs in degrees: arc MR is the far one, between the lines, and arc NS the near one.)

Join AR. Show that $\angle ARB = 90°$.

Join AR. Let ∠QPA = ∠APS = a and ∠QRB = ∠BRS = b.

Proved.

Show that the exterior angle QPN of PQRS equals the interior opposite angle QRS.

Proved.

Proved.

They are equal: they stand on the same base AB between the same parallels AB and EN.

△AMN has base AM = 2AB and the same height, so its area is ½ × 2AB × h = AB × h: area of △AMN = area of parallelogram ABCD.

Equal (same base, same parallels).

35 cm², the same.

Yes. $\angle AED = \angle BFC = 90°$ (corners of the rectangle), hypotenuse AD = BC (opposite sides of the parallelogram), and AE = BF (opposite sides of the rectangle): congruent by R.H.S.

$AC = \sqrt{25^2 - 7^2} = \sqrt{576} =$ 24 cm (Pythagoras, BC the hypotenuse).

Area of △ABC = ½ × 7 × 24 = 84 cm². The parallelogram on the same base between the same parallels is twice the triangle: 168 cm².

To prove: area of △ABC = ½ area of parallelogram BCED.

Construction: through C draw $CF \parallel BA$, meeting DE at F.

Proved.

The parallelogram stands on half the base, AD = ½ AC, between the same parallels, so its area is AD × h = ½ AC × h, the triangle’s.

They are equal in area. Both are twice △DPC: with ABCD it stands on DC between $AB \parallel DC$; with PQRD it stands on DP between $DP \parallel RQ$.

Area of ABCD = 8 × 7 = 56 cm², so area of PQRD = 56 cm².

Proved.

The four triangles fill the parallelogram, and XAB and XCD make half of it, so area of △AXD + area of △BCX = ½ area of ABCD too. (Or directly: a line through X parallel to AD does for them what MN did.)

△PBE has twice the base, BE = 2BC, between the same parallels, so its area is ½ × 2BC × h, the parallelogram’s.

The height to BC is $AB \sin 60° = 5 \times 0.866 \approx$ 4.33 cm (you should measure about 4.3 cm). Area of $ABCD = 4 \times 4.33 \approx 17.32$ cm², so area of $\triangle PBE \approx 17.32$ cm² (½ × 8 × 4.33).

The same construction as Question 2c:

BF is the height between the parallels: $BF = BC \sin 60° = 5 \times 0.866 \approx$ 4.33 cm ($\angle ABC = 120°$, so BC makes 60∘° with AB produced). Area of $ABCD = 7 \times 4.33 \approx$ 30.31 cm².

∠BOC = 2∠BDC.

Draw two circles of radius 3 cm or more, each with a central angle BOC and an inscribed angle BDC on the same arc, and measure:

Conclusion: the central angle is twice the inscribed angle on the same arc.

$$\begin{aligned} 7x &= 2(3x + 5) \\ 7x &= 6x + 10 \\ x &= 10 \end{aligned}$$
x = 10 (the angles are 70∘° and 35∘° ✓).

$\angle ABC + \angle ADC = 180°$ (opposite angles of a cyclic quadrilateral).

Proved. (Here ∠AOC is the angle on the arc ABC, the one ∠ADC stands on.)

$\angle ADC = 180° - 120° = 60°$, so $\angle AOC = 2 \times 60° =$ 120∘°.

∠PSQ and ∠PRQ.

∠PRQ = ∠PSQ = 60∘° (same arc PQ).

Proved.

90∘°: the angle in a semicircle (SH is a diameter).

Proved.

$\angle SIU = \angle SHU = 90° - 50° =$ 40∘°.

90∘°: PR passes through the center, so it is a diameter, and ∠PQR is an angle in a semicircle.

OS = OR (radii), so $\angle ORS = \angle OSR = 63°$, that is $\angle PRS = 63°$. ∠POS = 2∠PRS (on the same arc PS) = 126∘°.

Proved. (So $y = 63°$ and $x = 27°$.)

OQ and OR are radii of the same circle, so OQ = OR: △QOR is isosceles.