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Question 1 of 31
In the figure below, ABC is a triangle. CF and CE are straight line. AD = DB, $\angle$ ADE = 130 $^\circ$. and $\angle$ ABF = 155$^\circ$. Find $\angle$ X.
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$180 – 130 = 50$
$180 – 155 – 25$
$180 – 130 – 25 = 25$
$(180 - 50) ÷ 2 = 65^\circ$
You are Right
$180 – 130 = 50$
$180 – 155 – 25$
$180 – 130 – 25 = 25$
$(180 - 50) ÷ 2 = 65^\circ$
The figure below is made up of three triangle. ABC is an equilateral triangle, BCD is a right-angled triangle and DEF is an isosceles triangle. $\angle$EFD =68$^\circ$. Find the sum of $\angle$CDE and $\angle$ABD.
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194$^\circ$ $\rightarrow$ $180 – 68 – 68 = 44$
$180 – 90 = 90$
$\angle$CDE + $\angle$ABD $\rightarrow$ $90 + 44 + 60 = 194$
The figure below is not down to scale. ABCD is a trapezium. Find the area of triangle ABC.
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60cm$^2$ $\rightarrow$ $\frac{1}{2} \times 12 \times 10cm = 60cm$
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60cm$^2$ $\rightarrow$ $\frac{1}{2} \times 12 \times 10cm = 60cm$
In the diagram below, KL and KJ are straight line, MKN is an isosceles triangle and JL is parallel to MN. Find
(a) $\angle$KJL
(b) $\angle$JNL.
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A square piece of paper, ABCD, is shaded on one side as shown in figure1. It is then folded at its comer B to form an isosceles triangle as shown in figure 2. The perimeter and area of the remaining shaded region in figure 2 is 72cm and 180cm$^2$ respectively. Find the area of the isosceles triangle.
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A. of ABCD $\rightarrow$ $18 \times 18 = 324cm^2$
A. of BEFG $\rightarrow$ $324 – 180 = 144cm^2$
A. of Triangle $\rightarrow$ $144 ÷ 2 = 72cm^2$
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A. of ABCD $\rightarrow$ $18 \times 18 = 324cm^2$
A. of BEFG $\rightarrow$ $324 – 180 = 144cm^2$
A. of Triangle $\rightarrow$ $144 ÷ 2 = 72cm^2$
The figure below is made up of a square and a triangle. Square QPSR has an area of 36cm$^2$. Find the area of triangle QRT.
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$\frac{1}{2} \times 6 \times 11 = 33$
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$\frac{1}{2} \times 6 \times 11 = 33$
The figure below shows an equilateral triangular piece of paper folded along line CX. $\angle$ACB is 22 $^\circ$. Find $\angle$ BCX.
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The figure below shows a triangle pond in a park with pavements on two of its sides. The total width of the pavements is 4m. The length of the park is 59m and its breadth is 37m. What is the total area of the two pavements?
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$59 – 4 = 55$
$55 \times 37 = 2035$
$59 \times 37 – 2035 = 148$
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$59 – 4 = 55$
$55 \times 37 = 2035$
$59 \times 37 – 2035 = 148$
The figure is made up of a semicircle and an equilateral triangle. The diameter of the semicircle is 10cm. What is the perimeter of the figure?
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Semi Circle $\rightarrow$ $\frac{1}{2} \times \pi \times 15 = 5\pi$cm
Triangle $\rightarrow$ $10 \times 2 = 20cm$
Total $\rightarrow$ $5\pi$cm + $20cm = (5\pi +20)$cm
You are Right
Semi Circle $\rightarrow$ $\frac{1}{2} \times \pi \times 15 = 5\pi$cm
Triangle $\rightarrow$ $10 \times 2 = 20cm$
Total $\rightarrow$ $5\pi$cm + $20cm = (5\pi +20)$cm
The figure below shown three overlapping triangle. ABC is an isosceles triangle and AB // FK. $\angle$ ABC = 106$^\circ$, $\angle$ CDH = 18$^\circ$, $\angle$ KFH = 52$^\circ$. and $\angle$ GJH = 40$^\circ$. Find
(a) $\angle$ FHD.
(b) $\angle$FKG.
(b) 11$^\circ$
(b) 13$^\circ$
(b) 15$^\circ$
(b) 17$^\circ$
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(a) $\angle$BEF = $\angle$CBA = $(180^\circ – 106^\circ) ÷ 2 = 37^\circ$
$\angle$BEF = $\angle$DEC = 37$^\circ$
$\angle$CDE = $180^\circ – 37^\circ - 106^\circ = 37^\circ$
$37 + 18 = 55^\circ$
$\angle$FHD = $180^\circ - 52^\circ - 55^\circ = 73^\circ$
(b) $\angle$JGH = $180^\circ - 73^\circ - 40^\circ = 67^\circ$
$\angle$ FGK = 113%
$\angle$FKG = $180^\circ - 113^\circ - 52^\circ = 15^\circ$
(b) 15$^\circ$
You are Right
(a) $\angle$BEF = $\angle$CBA = $(180^\circ – 106^\circ) ÷ 2 = 37^\circ$
$\angle$BEF = $\angle$DEC = 37$^\circ$
$\angle$CDE = $180^\circ – 37^\circ - 106^\circ = 37^\circ$
$37 + 18 = 55^\circ$
$\angle$FHD = $180^\circ - 52^\circ - 55^\circ = 73^\circ$
(b) $\angle$JGH = $180^\circ - 73^\circ - 40^\circ = 67^\circ$
$\angle$ FGK = 113%
$\angle$FKG = $180^\circ - 113^\circ - 52^\circ = 15^\circ$
ACE is a triangle. Which one of the following gives the area of ADE?
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In the diagram below, ABC is an equilateral triangle. BD and EF are straight line. Find $\angle$y
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WXYZ is a square with an area of 64cm$^2$. Give that M is the midpoint of XY and N is the midpoint of ZY, Find the area of the shaded triangle WMN.
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The shaded figure is made up of 6 equilateral triangle. The length of straight line is 21cm. find the perimeter of the shaded figure.
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Mysha cut out three identical right-angled triangles. She joined them to form a figure PQRS as shown below. SR = 20cm and QR = 8cm. The perimeter of the figure PQRS is 44cm.Find the area of the figure PQRS.
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$44 – 30 = 14$
$14 – 8 = 6$
$\frac{1}{2} \times 8 \times 6 = 24$
$24 \times 3 = 72cm^2$
You are Right
$44 – 30 = 14$
$14 – 8 = 6$
$\frac{1}{2} \times 8 \times 6 = 24$
$24 \times 3 = 72cm^2$
The figure below shows 3 overlapping identical triangle. If the area of each big triangle is 80cm$^2$ and the area of the figure is 210cm$^2$, Find the area of the shaded triangle.
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The figure below is made up of two identical triangle. TSR and QSR. $\angle$QST = 148$^\circ$. Find $\angle$QSR.
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106$^\circ$ $\rightarrow$ $(360 -148) ÷ 2 = 106^\circ$
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106$^\circ$ $\rightarrow$ $(360 -148) ÷ 2 = 106^\circ$
The figure is made up of 1 square and 2 isosceles triangle. Find$\angle$ b.
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In the figure below, two comers of a triangle piece of paper are folded inward, then outward to form a symmetrical shape as shown in Figure B. $\angle$PQR = 11$^\circ$, and $\angle$PQS = 23$^\circ$. Find$\angle$TQR.
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22$^\circ$ $\rightarrow$ $11^\circ + 23^\circ = 34^\circ$
$34^\circ \times 2 = 68^\circ$
$\angle$ TQR $\rightarrow$ $90^\circ - 58\circ = 22^\circ$
You are Right
22$^\circ$ $\rightarrow$ $11^\circ + 23^\circ = 34^\circ$
$34^\circ \times 2 = 68^\circ$
$\angle$ TQR $\rightarrow$ $90^\circ - 58\circ = 22^\circ$
In the figure below, XYZ is an isosceles triangle. XY = YZ and AB// XY. AWY and AVB are straight line. Find $\angle$ t.
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$\angle$AYZ = $180^\circ – 20^\circ – 124^\circ = 36^\circ$
$\angle$YXZ + $\angle$XYZ = $180^\circ – 20^\circ – 36^\circ = 124^\circ$
$\angle$YXZ = $180^\circ - 62^\circ - 20^\circ = 98^\circ$
You are Right
$\angle$AYZ = $180^\circ – 20^\circ – 124^\circ = 36^\circ$
$\angle$YXZ + $\angle$XYZ = $180^\circ – 20^\circ – 36^\circ = 124^\circ$
$\angle$YXZ = $180^\circ - 62^\circ - 20^\circ = 98^\circ$
The figure below, not drawn to scale, show 3 triangles X, Y and Z overlapping one another. The area of triangle X is $\frac{1}{3}$ the area of triangle Y and the area of triangle Y is $\frac{2}{3}$ the area of triangle Z.
Express the unshaded are of triangle Y as a fraction of the unshaded area of triangle Z.
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In the diagram below, ABCO and FODE are identical rhombuses and AOF and OCD are identical equilateral triangle. What fraction of the figure ABCDEF is shaded?
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In the figure below, AC = 15cm, CD = 9cm and AD = 12cm. What is the area of triangle ABC?
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In the figure below, rectangle ABCD is made up of 7 identical small rectangles.
(a)Find the perimeter of the rectangle ABCD.
(b) Find the area of the shaded triangle.
(b) 65
(b) 70
(b) 75
(b) 78
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(a) $20 ÷ 5 = 4$
$20 ÷ 2 = 10$
$10 + 4 = 40$
$20 + 20 + 14 + 14 = 68$
(b) $\frac{1}{2} \times 14 \times 10 = 70$
(b) 70
You are Right
(a) $20 ÷ 5 = 4$
$20 ÷ 2 = 10$
$10 + 4 = 40$
$20 + 20 + 14 + 14 = 68$
(b) $\frac{1}{2} \times 14 \times 10 = 70$
In the figure below not scale, XYZ is an isosceles triangle where XZ =ZY. XZW is a straight line. Three angles are labelled as a, b and c. Which of the following statement is true?
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The figure below is made up of an equilateral triangle CDE a square DEFG of length 7cm with a quadrant in it. Find the perimeter of the shaded region. $(Take \pi = \frac{22}{7})$.
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Find the area of the shaded triangle.
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The figure below shows a rectangle ABCD. EFG and DFB are straight line. The area of rectangle ABCD is 960cm$^2$ and the total area of triangles DEF and BEG is 336cm$^2$. The ratio of length DG to the length GC is 7:5. What is the area of the triangle DFG?
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112cm$^2$ $\rightarrow$ $\bigtriangleup$ DEG + $\bigtriangleup$ DBG = $\frac{7}{12} \times 960 = 560$
$\bigtriangleup$ DFG = $\frac{560 - 336}{2} = 112$
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112cm$^2$ $\rightarrow$ $\bigtriangleup$ DEG + $\bigtriangleup$ DBG = $\frac{7}{12} \times 960 = 560$
$\bigtriangleup$ DFG = $\frac{560 - 336}{2} = 112$
The figure below is made up of 3 identical isosceles right-angled triangle. What is the area of the figure?
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In the figure, CDE and CDF are isosceles triangle. $\angle$DCF is three times as large as $\angle$EDF. Find $\angle$DFE.
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108$^\circ$ $\rightarrow$ $4U + 3U + 3U = 10U$
$1U = 180^\circ ÷ 10 = 18^\circ$
$3U = 18^\circ \times 3 = 54^\circ$
$1U = 18^\circ$
$\angle$ $DEF = 180^\circ – 18^\circ = 54^\circ = 108^\circ$
You are Right
108$^\circ$ $\rightarrow$ $4U + 3U + 3U = 10U$
$1U = 180^\circ ÷ 10 = 18^\circ$
$3U = 18^\circ \times 3 = 54^\circ$
$1U = 18^\circ$
$\angle$ $DEF = 180^\circ – 18^\circ = 54^\circ = 108^\circ$
The figure below shows two triangle. Which one of the following statements is true about the figure?
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