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Question 1 of 25

The figure below shows a trapezium. Which one of the following options is correct?

$Question Image of The figure below shows a trapezium. Which one of the following options is correct? .$

$\angle$a = $\angle$c

$\angle$a = $\angle$b

$\angle$a + $\angle$b = $\angle$c + $\angle$d

$\angle$a +$\angle$d = $\angle$c + $\angle$b

None of the above

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$\angle$a +$\angle$d = $\angle$c + $\angle$b

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The figure shown below, a trapezium ABCD was folded as shown. AB is parallel to CD. AE is parallel to BF. $\angle$AEF = 65$^\circ$. Find $\angle$FAE.

$Question Image of The figure shown below, a trapezium ABCD was folded as shown. AB is parallel to CD. AE is parallel to BF. $\angle$AEF = 65$^\circ$. Find $\angle$FAE. .$

20$^\circ$

30$^\circ$

40$^\circ$

50$^\circ$

None of the above

Sorry. Please check the correct answer below.

$\angle$XDA $\rightarrow$ 65$^\circ$
$\angle$CDX $\rightarrow$ 65$^\circ$
$\angle$DXB $\rightarrow$ $180^\circ – 65^\circ = 115^\circ$
$\angle$DAX = $\angle$FAE $\rightarrow$ $150^\circ – 65^\circ = 50^\circ$

The figure below shows a trapezium PQRS. What is the area of the shaded part?

$Question Image$

43.5

45.5

47.5

49.5

None of the above

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Area of Shaded Part = $\frac{1}{2} \times 9 \times 11 = 49.5$

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49.5

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Area of Shaded Part = $\frac{1}{2} \times 9 \times 11 = 49.5$

The figure below shows a trapezium ABCD. Given that AB//DC, $\angle$BAD = 105$^\circ$, and $\angle$BCD = 127$^\circ$, Find $\angle$ABC.

$Question Image$

51$^\circ$

53$^\circ$

55$^\circ$

57$^\circ$

None of the above

Sorry. Please check the correct answer below.

$\angle$ABC = $180^\circ - 127^\circ = 53^\circ$

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53$^\circ$

You are Right

$\angle$ABC = $180^\circ - 127^\circ = 53^\circ$

In the figure below, ABCD is a trapezium and PQCR is a square. The size of $\angle$DCR is $\frac{5}{4}$ of $\angle$DCQ. Find $\angle$DSQ?

$Question Image$

120$^\circ$

140$^\circ$

160$^\circ$

180$^\circ$

None of the above

Sorry. Please check the correct answer below.

$180 – 110 = 70$
$5 + 4 = 9$
$90 \div 9 = 10$
$10 \times 5 = 50$
$90 – 50 = 40$
$180 – 70 – 40 = 70$
$180 \times 2 = 360$
$\angle$DSQ = $360 – 70 – 79 – 40 = 160$

View Correct Answer

160$^\circ$

You are Right

$180 – 110 = 70$
$5 + 4 = 9$
$90 \div 9 = 10$
$10 \times 5 = 50$
$90 – 50 = 40$
$180 – 70 – 40 = 70$
$180 \times 2 = 360$
$\angle$DSQ = $360 – 70 – 79 – 40 = 160$

ABCD is a trapezium and EBD is an equilateral triangle. Find $\angle$ABE.

$Question Image$

18$^\circ$

20$^\circ$

22$^\circ$

24$^\circ$

None of the above

Sorry. Please check the correct answer below.

$\angle$ABE = $180^\circ - 110^\circ - 50^\circ = 20^\circ$

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20$^\circ$

You are Right

$\angle$ABE = $180^\circ - 110^\circ - 50^\circ = 20^\circ$

What is the area of the trapezium below?

$Question Image$

110cm$^2$

112cm$^2$

114cm$^2$

116cm$^2$

None of the above

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112cm$^2$

You are Right

In the figure, ABCD and EFGH are identical rhombuses and BKMN is a trapezium. Which one of the following statements is true?

$Question Image$

HE // GF // KM

AB // DC //KM

AD // BC // NM

HG // EF // NM

None of the above

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AD // BC // NM

You are Right

In the figure below, not drawn to scale, WXYZ is a trapezium. AY, GW and YC are straight line. Find $\angle$YWX

$Question Image$

68$^\circ$

70$^\circ$

72$^\circ$

74$^\circ$

None of the above

Sorry. Please check the correct answer below.

$180^\circ - 78^\circ = 102^\circ$
$180^\circ - 102^\circ – 42^\circ= 36^\circ$
$(180^\circ - 36^\circ) ÷ 2= 72^\circ$

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72$^\circ$

You are Right

$180^\circ - 78^\circ = 102^\circ$
$180^\circ - 102^\circ – 42^\circ= 36^\circ$
$(180^\circ - 36^\circ) ÷ 2= 72^\circ$

In the figure below, WXYZ is a trapezium. WZ is parallel to XY. $\angle$XWY = 52$^\circ$,$\angle$WYZ = 66 $^\circ$ and $\angle$WZY = 55$^\circ$. Find $\angle$WXY.

$Question Image$

56$^\circ$

58$^\circ$

60$^\circ$

62$^\circ$

None of the above

Sorry. Please check the correct answer below.

$180^\circ - 70^\circ - 70^\circ = 32^\circ$
$90^\circ - 32^\circ = 58^\circ$

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58$^\circ$

You are Right

$180^\circ - 70^\circ - 70^\circ = 32^\circ$
$90^\circ - 32^\circ = 58^\circ$

The figure is not drawn to scale. ABCD is a trapezium. Find $\angle$ADC.

$Question Image$

79$^\circ$

73$^\circ$

76$^\circ$

80$^\circ$

None of the above

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73$^\circ$

You are Right

In the figure below, not drawn to scale, ABCD is a trapezium. Find $\angle$ADB.

$Question Image$

10$^\circ$

15$^\circ$

20$^\circ$

25$^\circ$

None of the above

Sorry. Please check the correct answer below.

$180^\circ - 110^\circ = 70^\circ$
$180^\circ - 90^\circ - 45^\circ = 45^\circ$
$70^\circ - 45^\circ = 25^\circ$

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25$^\circ$

You are Right

$180^\circ - 110^\circ = 70^\circ$
$180^\circ - 90^\circ - 45^\circ = 45^\circ$
$70^\circ - 45^\circ = 25^\circ$

In the figure below, WXYZ is trapezium. Given that $\angle$WXY = 132$^\circ$ and $\angle$ZYW = 37$^\circ$, Find $\angle$XYW

$Question Image$

11$^\circ$

12$^\circ$

13$^\circ$

14$^\circ$

None of the above

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11$^\circ$

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In the figure below, not drawn to scale, PQST is a trapezium. PR // TS. Find $\angle$QRT.

$Question Image$

22$^\circ$

24$^\circ$

26$^\circ$

28$^\circ$

None of the above

Sorry. Please check the correct answer below.

$180^\circ - 104^\circ = 76^\circ$
$\angle$QST = $180^\circ - 102^\circ = 78^\circ$
$\angle$STR = $\angle$QRT = $180^\circ - 76^\circ - 78^\circ = 26^\circ$

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26$^\circ$

You are Right

$180^\circ - 104^\circ = 76^\circ$
$\angle$QST = $180^\circ - 102^\circ = 78^\circ$
$\angle$STR = $\angle$QRT = $180^\circ - 76^\circ - 78^\circ = 26^\circ$

In the figure, ABCD is a trapezium and BCD is an isosceles triangle. DB = DC, BAF = 52$^\circ$ and $\angle$AFE = 110$^\circ$. Find $\angle$BDC.

$Question Image$

None of the above

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View Correct Answer

You are Right

In the figure below, EBCD is a trapezium. ED is parallel to BC. $\angle$FAB = 74$^\circ$ and $\angle$EBC = 113$^\circ$. Find the sum of $\angle$w, $\angle$x, $\angle$y,and $\angle$z.

$Question Image$

213$^\circ$

215$^\circ$

217$^\circ$

219$^\circ$

None of the above

Sorry. Please check the correct answer below.

$180^\circ - 74^\circ = 106^\circ$ $\angle$y + $\angle$z
$180^\circ - 113^\circ = 67^\circ$
$180^\circ – 67^\circ = 113^\circ$ $\angle$x + $\angle$w
$113^\circ + 106^\circ = 219^\circ$

View Correct Answer

219$^\circ$

You are Right

$180^\circ - 74^\circ = 106^\circ$ $\angle$y + $\angle$z
$180^\circ - 113^\circ = 67^\circ$
$180^\circ – 67^\circ = 113^\circ$ $\angle$x + $\angle$w
$113^\circ + 106^\circ = 219^\circ$

The figure below shows a trapezium ABCD. AD is parallel to BE and BE = BC. $\angle$BEC = 70$^\circ$
(a) Find $\angle$ADC
(b) Find $\angle$ABC.

$Question Image$

(a) 50$^\circ$
(b) 100$^\circ$

(a) 60$^\circ$
(b) 105$^\circ$

(a) 70$^\circ$
(b) 110$^\circ$

(a) 80$^\circ$
(b) 115$^\circ$

None of the above

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(a) 70$^\circ$
(b) 110$^\circ$

You are Right

ABCD is a trapezium. AB is parallel to DC. Which one of the following statements is true?

$Question Image$

$\angle$DAB = $\angle$ADC

$\angle$DAB + $\angle$ADC = 180$^\circ$

$\angle$DAB + $\angle$ABC = 180$^\circ$

$\angle$DAB = $\angle$DCB

None of the above

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$\angle$DAB + $\angle$ADC = 180$^\circ$

You are Right

In the figure, YZ is a straight line while ABCD and PQRS are overlapping trapeziums. Given that $\angle$YAB = 50$^\circ$, $\angle$QDW = 135$^\circ$,$\angle$DQR = 60$^\circ$,
(a) Find $\angle$DXS.
(b) Find $\angle$QRW.

$Question Image$

(a) 50$^\circ$
(b) 75$^\circ$

(a) 60$^\circ$
(b) 80$^\circ$

(a) 70$^\circ$
(b) 85$^\circ$

(a) 80$^\circ$
(b) 90$^\circ$

None of the above

Sorry. Please check the correct answer below.

(a) $180^\circ - 50^\circ - 90^\circ = 40^\circ$
$180^\circ - 40^\circ - 90^\circ = 50^\circ$
(b) $180^\circ - 50^\circ = 130^\circ$
$360^\circ - 130^\circ - 135^\circ = 95^\circ$
$180^\circ - 45^\circ - 60^\circ = 75^\circ$

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(a) 50$^\circ$
(b) 75$^\circ$

You are Right

Name a trapezium in the figure below.

$Question Image$

SOP

OPQR

RQPS

SOT

None of the above

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View Correct Answer

RQPS

You are Right

The line PQ and QR are part of an incomplete trapezium PQRS. Complete drawing the trapezium such that $\angle$QRS = 120$^\circ$. Measure the line PS.

$Question Image$

6.2cm

6.4cm

9.2cm

9.4cm

None of the above

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View Correct Answer

9.4cm

You are Right

The figure below is not drawn to scale. ABCD is a trapezium and $\angle$ BDC = $\angle$ ADB. Find $\angle$ CAD.

$Question Image$

120$^\circ$

125$^\circ$

130$^\circ$

135$^\circ$

None of the above

Sorry. Please check the correct answer below.

120$^\circ$ $\rightarrow$ $360 – 280 = 80$
$80 ÷ 2 = 40$
$\angle$ CAD = $180^\circ - 40^\circ - 20^\circ = 120^\circ$

View Correct Answer

120$^\circ$

You are Right

120$^\circ$ $\rightarrow$ $360 – 280 = 80$
$80 ÷ 2 = 40$
$\angle$ CAD = $180^\circ - 40^\circ - 20^\circ = 120^\circ$

The figure below is not drawn to scale. ACDE is a trapezium. Given the AE = AB, $\angle$EAB =78$^\circ $ and $\angle$ACD is 90 $^\circ $ .
(a) Find $\angle$CDE.
(b) Find $\angle$ABE.
(c) Find $\angle$DEB.

$Question Image$

(a) 102$^\circ$
(b) 51$^\circ$
(c) 51$^\circ$

(a) 104$^\circ$
(b) 53$^\circ$
(c) 52$^\circ$

(a) 106$^\circ$
(b) 55$^\circ$
(c) 53$^\circ$

(a) 108$^\circ$
(b) 57$^\circ$
(c) 54$^\circ$

None of the above

Sorry. Please check the correct answer below.

(a) $180^\circ - 99^\circ = 81^\circ$
(b) $180^\circ - 78^\circ = 102^\circ$
$102^\circ ÷ 2 = 51^\circ$
(c) $\angle$DEB = $\angle$AEB = 51$^\circ$

View Correct Answer

(a) 102$^\circ$
(b) 51$^\circ$
(c) 51$^\circ$

You are Right

(a) $180^\circ - 99^\circ = 81^\circ$
(b) $180^\circ - 78^\circ = 102^\circ$
$102^\circ ÷ 2 = 51^\circ$
(c) $\angle$DEB = $\angle$AEB = 51$^\circ$

In the figure, ABCD is a trapezium where AD is parallel to BC. Find $\angle$BDC.

$Question Image$

30$^\circ$

35$^\circ$

40$^\circ$

45$^\circ$

None of the above

Sorry. Please check the correct answer below.

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35$^\circ$

You are Right

The figure below shows a trapezium ABCD with AB // DC, $\angle$ DAB = 68 $^\circ$. And $\angle$ ABC is an isosceles triangle with AB = AD. Find $\angle$BCD.

$Question Image$

25$^\circ$

30$^\circ$

35$^\circ$

40$^\circ$

None of the above

Sorry. Please check the correct answer below.

30$^\circ$ $\rightarrow$ $\angle$ $ABD = (180^\circ-68^\circ) ÷ 2 = 56^\circ$
$\angle$ ABD = $\angle$ BDC
$\angle$ $BCD = 180^\circ – (94^\circ+56^\circ) = 30^\circ$

View Correct Answer

30$^\circ$

You are Right

30$^\circ$ $\rightarrow$ $\angle$ $ABD = (180^\circ-68^\circ) ÷ 2 = 56^\circ$
$\angle$ ABD = $\angle$ BDC
$\angle$ $BCD = 180^\circ – (94^\circ+56^\circ) = 30^\circ$

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