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Question 1 of 79
In the figure, the circle touches each of the two equilateral triangle at exactly three points. What fraction of the figure is shaded?
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The figure below is formed by of 4 identical quarter circles, 1 semicircle and 1 rectangle. Find the area of the shaded figure. Leave your answer in terms of $\pi$.
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The figure below is made up of semicircles. Find the perimeter of the figure below. $(Take \pi = \frac{22}{7})$.
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The figure below shown the design of a floor rug. The edges of the rug is made up of 4 semicircles and 4 quarter-circle, each of radius 21cm. $(Take\pi = \frac{22}{7})$ .
(a) Find the perimeter of the rug.
(b) Find the area of the rug.
(b) 6479cm$^2$
(b) 7594cm$^2$
(b) 8568cm$^2$
(b) 9625cm$^2$
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(a) $132 \times 3 = 396$
(b) 8568cm$^2$ $\rightarrow$ Total area $\rightarrow$ $2772= 1386 + 882 + 3528 = 8568$
(b) 8568cm$^2$
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(a) $132 \times 3 = 396$
(b) 8568cm$^2$ $\rightarrow$ Total area $\rightarrow$ $2772= 1386 + 882 + 3528 = 8568$
The figure below is made up of 3 different semi-circle. The radius of the smallest semi-circle is 3cm. Using $\pi$ = 3.14, Find the area of the shaded part.
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Big $\rightarrow$ $\frac{1}{2} \times \pi \times 5 \times 5 = 12\frac{1}{2}\pi$
Medium $\rightarrow$ $\frac{1}{2} \times \pi \times 4 \times 4 = 8\pi$
Small $\rightarrow$ $\frac{1}{2} \times \pi \times 3 \times 3 = 4.5\pi$
$\rightarrow$ $(12.5\pi – 4.5\pi) + 8\pi = 16\pi$
You are Right
Big $\rightarrow$ $\frac{1}{2} \times \pi \times 5 \times 5 = 12\frac{1}{2}\pi$
Medium $\rightarrow$ $\frac{1}{2} \times \pi \times 4 \times 4 = 8\pi$
Small $\rightarrow$ $\frac{1}{2} \times \pi \times 3 \times 3 = 4.5\pi$
$\rightarrow$ $(12.5\pi – 4.5\pi) + 8\pi = 16\pi$
Three circle are placed side- by- side as show. PQ is 7.5cm and it cuts through the centers of all circle. Find the circumference of the 3 circle Take$\pi$ = 3.14
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$3.14 \times 7.5 = 23.55cm$
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$3.14 \times 7.5 = 23.55cm$
The figure is made up of four semi – circle and a rectangle ABCD. AB = 9cm, BC = 12cm and AC 15cm. Find the total area of the shaded parts. $ (Take \pi = 3.14)$.
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$\frac{1}{2} \times 9 \times 12 = 54cm^2$
$\frac{1}{2} \times 3.14 \times 4.5 \times 4.5 = 31.79cm^2$
$\frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52cm^2$
$54 + 31.79 + 56.52 = 142.31cm^2$
$\frac{1}{2} \times 3.14 \times 7.5 \times 7.5 = 88.31cm^2$
142.31 – 88.31 = 54cm^2$
$54 \times 2 = 108cm^2$
You are Right
$\frac{1}{2} \times 9 \times 12 = 54cm^2$
$\frac{1}{2} \times 3.14 \times 4.5 \times 4.5 = 31.79cm^2$
$\frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52cm^2$
$54 + 31.79 + 56.52 = 142.31cm^2$
$\frac{1}{2} \times 3.14 \times 7.5 \times 7.5 = 88.31cm^2$
142.31 – 88.31 = 54cm^2$
$54 \times 2 = 108cm^2$
A rectangle is placed in a circle with O as center. Use the calculator value of $\pi$ to find the shaded part of the circle
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radius $\rightarrow$ 10cm; Shaded part $\rightarrow$ $\frac{3}{4} \times \pi \times 10cm \times 10cm = 235.62cm^2$
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radius $\rightarrow$ 10cm; Shaded part $\rightarrow$ $\frac{3}{4} \times \pi \times 10cm \times 10cm = 235.62cm^2$
Regions A, B and C represent the number of point assigned to the three target as shown below. The total points in Region A and B, B and C, and A and C are 11. 19 and 16 respectively. How many points is assigned to region A?
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4 Points $\rightarrow$ $11 + 19 + 16 = 46$
$46 ÷ 2 = 23$
$23 – 19 = 4$
You are Right
4 Points $\rightarrow$ $11 + 19 + 16 = 46$
$46 ÷ 2 = 23$
$23 – 19 = 4$
The figure below is made up of a circle, a parallelogram and two identical semi-circle. The diameter of the circle, PR is 12cm and the diameter of the semi-circle is 6cm. PR is perpendicular to SP and RQ. Find the area of the shaded parts. $(Take \pi = 3.14)$
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69.3cm$^2$ $\rightarrow$ $6cm \times 12cm = 72cm$
$113.04cm^2 - 72cm^2 = 41.04cm$
$\frac{1}{2} \times 3.14 \times 3cm \times 3cm = 14.13cm$
$1413 \times 2 = 28.26$cm
$41.04cm + 2826 = 69.3cm$
You are Right
69.3cm$^2$ $\rightarrow$ $6cm \times 12cm = 72cm$
$113.04cm^2 - 72cm^2 = 41.04cm$
$\frac{1}{2} \times 3.14 \times 3cm \times 3cm = 14.13cm$
$1413 \times 2 = 28.26$cm
$41.04cm + 2826 = 69.3cm$
The shaded figure below is formed by semicircles, quarter circle and squares. ABEF is a square. What is the area of the shaded region?
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Area of rectangle: $30 \times 60 = 1800cm^2$
Area of semicircle: $\frac{1}{2} \times 30 \times 30 \times 3.14 = 141cm^2$
$1800cm^2 - 1413cm^2 = 387cm^2$
Area of circle: $15 \times 15 \times 3.14 = 706.5cm^2$
$706.5cm^2 + 387cm^2 = 1093.5cm^2$
You are Right
Area of rectangle: $30 \times 60 = 1800cm^2$
Area of semicircle: $\frac{1}{2} \times 30 \times 30 \times 3.14 = 141cm^2$
$1800cm^2 - 1413cm^2 = 387cm^2$
Area of circle: $15 \times 15 \times 3.14 = 706.5cm^2$
$706.5cm^2 + 387cm^2 = 1093.5cm^2$
The figure below shows a quadrant and a semicircle enclosed with in a square. O is the center of the square. The square has a length of 12cm.
(a) Find the area of the shaded part.
(b) Find the perimeter of the shaded part $(Take\pi = 3.14)$
(b) 31.22cm
(b) 31.42cm
(b) 33.22cm
(b) 33.42cm
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(a) 43.74cm$^2$ $\rightarrow$ $12 \times 12 = 144$
$144 ÷ 2 = 72$
$3.14 \times (12 ÷ 2) \times (12 ÷ 2)$ = 113.04
$113.04 ÷ 4 = 28.26$
$72 – 28.26 = 43.74$
(b) 33.42cm $\rightarrow$ $3.14 \times 12 = 37.68$
$37.68 ÷ 4 = 9.42$
$9.24 + 6 + 6 + 12 = 33.42$
(b) 33.42cm
You are Right
(a) 43.74cm$^2$ $\rightarrow$ $12 \times 12 = 144$
$144 ÷ 2 = 72$
$3.14 \times (12 ÷ 2) \times (12 ÷ 2)$ = 113.04
$113.04 ÷ 4 = 28.26$
$72 – 28.26 = 43.74$
(b) 33.42cm $\rightarrow$ $3.14 \times 12 = 37.68$
$37.68 ÷ 4 = 9.42$
$9.24 + 6 + 6 + 12 = 33.42$
The figure below is formed by overlapping a square and a circle. O is the center of the circle. The radicle is 14cm. Find the perimeter of the figure. $(Take \pi= \frac{22}{7})$
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The diagram below is made up of a quarter and three- quarter circle with center O. Find the perimeter of the figure. $(Take \pi = \frac{22}{7})$.
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The figure below is made up 1 circle, 3 identical rectangles and 12 identical quarter circle of radius 10cm
(a) Find the perimeter of the unshaded part, X
(b)Find the total area of the shaded parts. $(Take \pi = 3.14)$.
(b) 1512cm$^2$
(b) 1514cm$^2$
(b) 1515cm$^2$
(b) 1516cm$^2$
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(a)82.8 $\rightarrow$ $3.14 \times 20 = 62.8$
$62.8 ÷ 2 = 31.4$
Perimeter = $31.4 + 31.4 + 10 + 10 = 82.8$
(b)1514cm$^2$ $\rightarrow$ Area = $12 \times 10 \times 10 + 3.14 \times 10 \times 10 = 1514$cm$^2$
(b) 1514cm$^2$
You are Right
(a)82.8 $\rightarrow$ $3.14 \times 20 = 62.8$
$62.8 ÷ 2 = 31.4$
Perimeter = $31.4 + 31.4 + 10 + 10 = 82.8$
(b)1514cm$^2$ $\rightarrow$ Area = $12 \times 10 \times 10 + 3.14 \times 10 \times 10 = 1514$cm$^2$
The figure below is made up of a semicircle and a right-angled isosceles triangle ABC. AB= BC and the diameter of the semicircle is 28cm. Find the area of the shaded part of the figure.
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$\frac{1}{2} \times 28 \times 14 = 196$cm$^2$
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$\frac{1}{2} \times 28 \times 14 = 196$cm$^2$
The figure is made up of a rectangle and two quarter circle. The breadth of the rectangle is 10cm. what is the total area of the shaded parts?
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The figure is made of a quadrant and a semicircle. The quadrant has a radius of 21cm. what is the perimeter of the unshaded part? $(Take \pi = \frac{22}{7})$.
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A $\rightarrow$ $\frac{1}{2} \times 21 \times \frac{22}{7} = 33$
B $\rightarrow$ $\frac{1}{4} \times 21 \times \frac{22}{7} = 33$
P $\rightarrow$ $33 + 33 + 21 = 87cm$
You are Right
A $\rightarrow$ $\frac{1}{2} \times 21 \times \frac{22}{7} = 33$
B $\rightarrow$ $\frac{1}{4} \times 21 \times \frac{22}{7} = 33$
P $\rightarrow$ $33 + 33 + 21 = 87cm$
The figure below is made up of three semi-circle and a circles. X is the center of the large semi-circle and WZ is 36cm. Find the area of the shaded part. Express your answer in terms of $\pi$.
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Area of fig $\rightarrow$ $\frac{1}{2} \times 18 \times 18 \times \pi = 162\pi$
$\frac{1}{2} \times 36 \times 18 = 324$
$(162\pi - 324) ÷ 2 = 81\pi - 162$
$(20.25\pi – 40.5) \times 2 = 40.5\pi -81$
$(81\pi - 162) – (40.5\pi - 81) = (81\pi – 162)cm^2$
You are Right
Area of fig $\rightarrow$ $\frac{1}{2} \times 18 \times 18 \times \pi = 162\pi$
$\frac{1}{2} \times 36 \times 18 = 324$
$(162\pi - 324) ÷ 2 = 81\pi - 162$
$(20.25\pi – 40.5) \times 2 = 40.5\pi -81$
$(81\pi - 162) – (40.5\pi - 81) = (81\pi – 162)cm^2$
The figure below is made up of a right- angled triangle and a circle overlapping two semicircles. AC is the diameter of the circle. $ (Take\pi = 3.14)$
(a) Find the perimeter of the shaded parts.
(b) Find the total area of the shaded parts.
(b) 20
(b) 22
(b) 24
(b) 26
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(a) $\frac{1}{2} \times 3.14 \times 10 = 15.7$
$\frac{1}{2} \times 3.14 \times 6 = 9.42$
$\frac{1}{2} \times 3.14 \times 8 – 12.56$
$15.7 + 9.24 + 12.56 = 37.68$
(b) $\frac{1}{2} \times 6 \times 8 = 24$
$\frac{1}{2} \times 3.14 \times 3 \times 3 + \frac{1}{2} \times 3.14 \times 4 \times 4 = 39.25$
$\frac{1}{2} \times 3.14 \times 5 \times 5 = 39.25$
$39.25 + 24 – 39.29 = 24$
(b) 24
You are Right
(a) $\frac{1}{2} \times 3.14 \times 10 = 15.7$
$\frac{1}{2} \times 3.14 \times 6 = 9.42$
$\frac{1}{2} \times 3.14 \times 8 – 12.56$
$15.7 + 9.24 + 12.56 = 37.68$
(b) $\frac{1}{2} \times 6 \times 8 = 24$
$\frac{1}{2} \times 3.14 \times 3 \times 3 + \frac{1}{2} \times 3.14 \times 4 \times 4 = 39.25$
$\frac{1}{2} \times 3.14 \times 5 \times 5 = 39.25$
$39.25 + 24 – 39.29 = 24$
The figure below shows 3 identical overlapping circles. The middle circle touches the center of the other 2 circles. Line AB is the diameter of the middle circle. What fraction is unshaded?
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$\frac{4}{16$} = $\frac{1}{4}$
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$\frac{4}{16$} = $\frac{1}{4}$
Janet cut out circles of the same size and pasted them on a rectangular strip of paper as shown below. AB and CD are diameters of the circles. What is the total area of the shaded parts? $ (Take \pi= 3.14)$
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2322cm$^2$ $\rightarrow 120 ÷ 4 = 30$
$ 3.14 \times 30 \times 30 = 2826$
$30 \times 2 = 60$, $ 60 \times 60 = 3600$
$3600 – 2826 = 774$
$774 ÷ 4 = 1935$
No of Shaded $\rightarrow$ 12
$12 \times 193.5 = 2322$
You are Right
2322cm$^2$ $\rightarrow 120 ÷ 4 = 30$
$ 3.14 \times 30 \times 30 = 2826$
$30 \times 2 = 60$, $ 60 \times 60 = 3600$
$3600 – 2826 = 774$
$774 ÷ 4 = 1935$
No of Shaded $\rightarrow$ 12
$12 \times 193.5 = 2322$
The figure shown three semicircles and a circle. AB = BC = CD = DE = 5cm, find the perimeter of the shaded part. Give your answer in 2 decimal places.
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Circumference of Semi: $10 \times 3.142 \times \frac{1}{2} = 15.71$
Circumference of quarter: $15.71 ÷ 2 = 7.855$
Circumference of big quarter: $20 \times 3.142 \times \frac{1}{4} = 15.71$
$15.71 + 7.855 + 7.855 + 15.71 = 47.13cm$
You are Right
Circumference of Semi: $10 \times 3.142 \times \frac{1}{2} = 15.71$
Circumference of quarter: $15.71 ÷ 2 = 7.855$
Circumference of big quarter: $20 \times 3.142 \times \frac{1}{4} = 15.71$
$15.71 + 7.855 + 7.855 + 15.71 = 47.13cm$
The figure below, not drawn to scale, shows a three- quarter circle. Find its perimeter $(Take \pi = \frac{22}{7})$.
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$\frac{3}{4} \times \frac{22}{7} \times 14 = 33$
$33 + 7 + 7 = 47cm$
You are Right
$\frac{3}{4} \times \frac{22}{7} \times 14 = 33$
$33 + 7 + 7 = 47cm$
The figure below is made up of 2 identical quadrants and 2 identical semicircles. Find the area of the shaded part of the figure. $(Take \pi = \frac{22}{7})$.
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The figure below is made up of a quadrant and 2 identical semicircles. AC is twice of AD. What is the perimeter of the shaded figure below? $(Take \pi = \frac{22}{7})$.
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The figure shows two quadrants of circles, centered at C and D respectively. Find the difference between the area of the two shaded regions. $(Take \pi = \frac{22}{7})$.
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Area of rectangle $\rightarrow$ $(28\times14) cm^2 = 392cm^2$
Area of quadrent $\rightarrow$ $(28\times 28 \times \frac{22}{7})cm^2 ÷ 4 = 616cm^2$
Area of A,B and small quadrent $\rightarrow$ $(616 - 392)cm^2 ÷ 4 = 224cm^2$
Area of A and B $\rightarrow$ $(224 -154)cm^2 = 70cm^2$
Area of H and B $\rightarrow$ $(616 - 154)cm^2 = 462cm^2$
Difference in two shaded parts $\rightarrow$ $(462-392)cm^2 = 70cm^2$
You are Right
Area of rectangle $\rightarrow$ $(28\times14) cm^2 = 392cm^2$
Area of quadrent $\rightarrow$ $(28\times 28 \times \frac{22}{7})cm^2 ÷ 4 = 616cm^2$
Area of A,B and small quadrent $\rightarrow$ $(616 - 392)cm^2 ÷ 4 = 224cm^2$
Area of A and B $\rightarrow$ $(224 -154)cm^2 = 70cm^2$
Area of H and B $\rightarrow$ $(616 - 154)cm^2 = 462cm^2$
Difference in two shaded parts $\rightarrow$ $(462-392)cm^2 = 70cm^2$
Elvin had a circular piece of paper. She folded it into quarters and than cut away a semicircle, a quarter circle and a quarter circle and a square from the corners as shown below. Find the perimeter of the paper left after Elvina unfolded the paper. $(Take\pi = 3.14)$.
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$6 \times 6 = 36$
$3.14 \times 22 = 69.08$
$2 \times 3.14 \times 6 = 37.68$
$2 \times (\frac{1}{2} \times 3.14 \times 8 \times 8) = 41.12$
$6 \times 4 = 24$
$41.12 + 24 + 37.68 + 69.08 = 171.88cm$
You are Right
$6 \times 6 = 36$
$3.14 \times 22 = 69.08$
$2 \times 3.14 \times 6 = 37.68$
$2 \times (\frac{1}{2} \times 3.14 \times 8 \times 8) = 41.12$
$6 \times 4 = 24$
$41.12 + 24 + 37.68 + 69.08 = 171.88cm$
The figure below shows a semicircle. Fine the perimeter of the semicircle.
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The figure below shows a quarter circle of radius 10cm two identical quarter circle of radius 7cm. Find the perimeter of the figure in terms of $\pi$.
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Diameter of B = $7 \times 2 = 14$
Diamterer of A = $10 \times 2 = 20$
Perimeter = $(\frac{1}{4} \times \pi \times 14) + (\frac{1}{4} \times \pi \times 14) + (\frac{1}{4} \times 20 \times \pi) + 7 + 7 + (10 - 7) + (10 - 7) = 3.5\pi + 3.5\pi + 5\pi + 20 = 12\pi = 20$
You are Right
Diameter of B = $7 \times 2 = 14$
Diamterer of A = $10 \times 2 = 20$
Perimeter = $(\frac{1}{4} \times \pi \times 14) + (\frac{1}{4} \times \pi \times 14) + (\frac{1}{4} \times 20 \times \pi) + 7 + 7 + (10 - 7) + (10 - 7) = 3.5\pi + 3.5\pi + 5\pi + 20 = 12\pi = 20$
The area of rectangle WXYZ is 98cm$^2$. Find the radius of the seim-circle. $(Take \pi = \frac{22}{7})$.
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The figure below is made up of one big semi-circle is 4 identical small semi-circles. The diameter of each small semi-circle is 14cm. Find the perimeter of the shaded area of the figure. $(Take \pi = \frac{22}{7})$.
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$2 \times \frac{22}{7} \times 14 = 88$
$\frac{1}{2} \times \frac{22}{7} \times56 = 88$
$88 \times 2 = 176cm$
You are Right
$2 \times \frac{22}{7} \times 14 = 88$
$\frac{1}{2} \times \frac{22}{7} \times56 = 88$
$88 \times 2 = 176cm$
The figure below is made up of two semicircles. O is the center of the larger semicircle of radius 8cm. Find the perimeter of the figure.
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In the figure below, not drawn to scale, O is the center of the semi-c ircle. PR = RS, QT //RS and $\angle$PRS = 86$^\circ$. Find $\angle$OPT.
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$180^\circ - 86^\circ = 94^\circ $
$180^\circ - 94^\circ = 86^\circ $
$(180^\circ - 86^\circ) ÷ 2 = 46^\circ$
$47^\circ + 86^\circ = 133^\circ$
$\angle$OPT = $(180^\circ - 133^\circ) ÷ 2 = 23.5^\circ$
You are Right
$180^\circ - 86^\circ = 94^\circ $
$180^\circ - 94^\circ = 86^\circ $
$(180^\circ - 86^\circ) ÷ 2 = 46^\circ$
$47^\circ + 86^\circ = 133^\circ$
$\angle$OPT = $(180^\circ - 133^\circ) ÷ 2 = 23.5^\circ$
The figure is made up of 3 identical quarter circle and a right- angled isosceles triangle. $\angle$ABC = 90$^\circ$ and AB = BC The length of AC is 6cm. Find the area of the figure. Take $\pi$ = 3.14
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Area of square $\rightarrow$ $6 \times 6 = 36$
Atra of 1 triangle $\rightarrow$ $36 \div 4 = 9$
Area of 2 triangle $\rightarrow$ $9 \times 2 = 18$
Area of $\frac{3}{4}$ circle $\rightarrow$ $\frac{3}{4} \times \pi \times r \times r$
$\rightarrow$ $\frac{3}{4} \times \pi \times \sqrt(18) \times \sqrt(18)$
$\rightarrow$ $\frac{3}{4} \times 3.14 \times \sqrt(18) \times \sqrt(18)$
$\rightarrow$ 42.39
Area of figure $\rightarrow$ $42.39 + 9 = 51.39cm^2$
You are Right
Area of square $\rightarrow$ $6 \times 6 = 36$
Atra of 1 triangle $\rightarrow$ $36 \div 4 = 9$
Area of 2 triangle $\rightarrow$ $9 \times 2 = 18$
Area of $\frac{3}{4}$ circle $\rightarrow$ $\frac{3}{4} \times \pi \times r \times r$
$\rightarrow$ $\frac{3}{4} \times \pi \times \sqrt(18) \times \sqrt(18)$
$\rightarrow$ $\frac{3}{4} \times 3.14 \times \sqrt(18) \times \sqrt(18)$
$\rightarrow$ 42.39
Area of figure $\rightarrow$ $42.39 + 9 = 51.39cm^2$
The figure is made of 2 similar segments of a circle. Find the perimeter of the figure.
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Perimeter of figure $\rightarrow$ $(\frac{120}{360}\times 2^\pi \times 9) + 9 + 9 + 10 = (6^\pi + 28)$
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Perimeter of figure $\rightarrow$ $(\frac{120}{360}\times 2^\pi \times 9) + 9 + 9 + 10 = (6^\pi + 28)$
The figure below is made up a quadrant and a semi-circle. Find the perimeter of the shaded figure. $(Take \pi = \frac{22}{7})$.
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$\frac{1}{4} \times 28 \times \frac{22}{7} = 22$
$\frac{1}{2} \times 14 \times \frac{22}{7} = 22$
$14 + 22 + 22 = 58cm$
You are Right
$\frac{1}{4} \times 28 \times \frac{22}{7} = 22$
$\frac{1}{2} \times 14 \times \frac{22}{7} = 22$
$14 + 22 + 22 = 58cm$
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 shows a triangle enclosed within a semicircle O is the center of the semicircle. Find the area of the shaded part. $ (Take\pi = 3.14)$.
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Triangle = $\frac{1}{2} \times 16 \times 9.6 = 76.8$
$16 \div 2 = 8$
$3.14 \times 8 \times 8 \frac{1}{2} = 100.48$
$100.48 – 76.8 = 23.68cm^2$
You are Right
Triangle = $\frac{1}{2} \times 16 \times 9.6 = 76.8$
$16 \div 2 = 8$
$3.14 \times 8 \times 8 \frac{1}{2} = 100.48$
$100.48 – 76.8 = 23.68cm^2$
How many more circles must be shaded so that the fraction of the number of unshaded circles is$\frac{1}{4}$ of the total number of circles?
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The figure below is made up of 2 identical smaller semicircles and a bigger semicircle. O is the center of the bigger semicircle of radius 7cm. Find the perimeter of the whole figure. $(Take \pi = \frac{22}{7})$.
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The figure below is not scale. Line AB is 36cm and it cuts through the center of all four circles. Find the perimeter of the figure. $(Take \pi = \frac{22}{7})$.
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The figure below shows two identical semicircles with radius 8cm each. Find the perimeter of the shaded part.
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The figure below is made up of two identical quarter circles of radius 3cm. What is the perimeter of the figure?
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The figure below, not drawn to scale. Is a circle with O as the center. TS and RW are straight line. The ratio of $\angle$RSO to $\angle$OSW is 2 : 1. $\angle$SWO = 73$^\circ$. Find $\angle$TRS.
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$(180^\circ - 90^\circ - 73^\circ) \times 2 = 34^\circ$
$45^\circ + (180^\circ - 34^\circ - 90^\circ) = 101^\circ$
You are Right
$(180^\circ - 90^\circ - 73^\circ) \times 2 = 34^\circ$
$45^\circ + (180^\circ - 34^\circ - 90^\circ) = 101^\circ$
Find the area of the quarter circle below. Take $\pi$ = 3.14.
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In the figure shown below, O is the center of the circle. OS = ST and ROT is a straight line. Find (a) $\angle$ TOS
(b) $\angle$ TRS
30$^\circ$
40$^\circ$
50$^\circ$
60$^\circ$
Sorry. Please check the correct answer below.
(a) 60$^\circ$ $\rightarrow$ $\angle$ $TOS = 180^\circ ÷ 3 = 60^\circ$
(b) 30$^\circ$ $\rightarrow$ $\angle$ $TRS = 180^\circ – 120^\circ ÷ 2 = 30^\circ$
30$^\circ$
You are Right
(a) 60$^\circ$ $\rightarrow$ $\angle$ $TOS = 180^\circ ÷ 3 = 60^\circ$
(b) 30$^\circ$ $\rightarrow$ $\angle$ $TRS = 180^\circ – 120^\circ ÷ 2 = 30^\circ$
The figure below is formed by large semicircle, 2 small identical semicircles and a straight line. The semicircles are formed along the edges of a right- angled triangle. The dimension of the triangle are 3cm, 4cm and 5cm.
(a) Find the perimeter of the figure.
(b) Find the area of the figure, correct to 2 decimal places. Take $\pi$ = 3.14
(b) 13.81cm$^2$
(b) 14.81cm$^2$
(b) 15.81cm$^2$
(b) 16.81cm$^2$
Sorry. Please check the correct answer below.
(a)Diameter of small semi $\rightarrow$ 2
Perimeter of 2 small semi $\rightarrow$ $\pi \times d$
= $3.14 \times 2$
6.28
Perimeter of 1 big semi $\rightarrow$ $\pi \times d \times \frac{1}{2}$
$3.14 \times 5 \times \frac{1}{2} = 7.85$
Perimeter of figure $\rightarrow$ $7.85 + 6.28 + 3 = 17.13cm$ (b) $\bigtriangleup$ $\rightarrow$ $\frac{1}{2} \times 3 \times 4 = 6$
$5 \div 2 = 2.5$
Semi $\rightarrow$ $\pi \times r \times r \times \frac{1}{2}$
= $3.14 \times 2.5 \times 2.5 \times \frac{1}{2} = 9.81$
Total area $\rightarrow$ $6 + 9.81 = 15.81cm^2$
(b) 15.81cm$^2$
You are Right
(a)Diameter of small semi $\rightarrow$ 2
Perimeter of 2 small semi $\rightarrow$ $\pi \times d$
= $3.14 \times 2$
6.28
Perimeter of 1 big semi $\rightarrow$ $\pi \times d \times \frac{1}{2}$
$3.14 \times 5 \times \frac{1}{2} = 7.85$
Perimeter of figure $\rightarrow$ $7.85 + 6.28 + 3 = 17.13cm$ (b) $\bigtriangleup$ $\rightarrow$ $\frac{1}{2} \times 3 \times 4 = 6$
$5 \div 2 = 2.5$
Semi $\rightarrow$ $\pi \times r \times r \times \frac{1}{2}$
= $3.14 \times 2.5 \times 2.5 \times \frac{1}{2} = 9.81$
Total area $\rightarrow$ $6 + 9.81 = 15.81cm^2$
In the figure, OR, OP, PQ and QR are straight line. The points O, P, Q and R are the centers of the four circle. The radius of each is 15cm. Find the total perimeter of the shaded parts.
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Perimeter = 2$\pi$r + $8 \times 15 = 30\pi + 120cm$
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Perimeter = 2$\pi$r + $8 \times 15 = 30\pi + 120cm$
The figure below is made up of 2 semicircles with diameters 7cm and 14cm respectively. What is the perimeter of the figure? $(Take \pi = \frac{22}{7})$
Sorry. Please check the correct answer below.
$\frac{1}{2} \times \frac{22}{7} \times 14 = 22$
$\frac{1}{2} \times \frac{22}{7} \times 7 = 11$
$ 22 + 11 + 3.5 + 3.5 = 40$
You are Right
$\frac{1}{2} \times \frac{22}{7} \times 14 = 22$
$\frac{1}{2} \times \frac{22}{7} \times 7 = 11$
$ 22 + 11 + 3.5 + 3.5 = 40$
The figure below is made up of 3 semi-circles. Find the area of the figure. $(Take \pi= \frac{22}{7})$
Sorry. Please check the correct answer below.
464cm$^2$ $\rightarrow$ $14 ÷ 2 =7$
$\frac{22}{7} \times 7 \times 7 = 154$
$\frac{22}{7} \times 14 \times 14 ÷ 2 = 44 \times 14 ÷ 2 = 616 ÷ 2 = 308$
$308 + 154 = 462$
You are Right
464cm$^2$ $\rightarrow$ $14 ÷ 2 =7$
$\frac{22}{7} \times 7 \times 7 = 154$
$\frac{22}{7} \times 14 \times 14 ÷ 2 = 44 \times 14 ÷ 2 = 616 ÷ 2 = 308$
$308 + 154 = 462$
The figure below is made up of a quadrant and 2 identical semicircles of radius 10.5cm. Find the perimeter of the figure. $(Take \pi = \frac{22}{7})$.
Sorry. Please check the correct answer below.
$\frac{1}{4} (2 \times \frac{22}{7} \times 10.5) \times 1\frac{1}{4} = 82.5$
$82.5 + 10.5 + 10.5 = 103.5cm$
You are Right
$\frac{1}{4} (2 \times \frac{22}{7} \times 10.5) \times 1\frac{1}{4} = 82.5$
$82.5 + 10.5 + 10.5 = 103.5cm$
The figure below shows a small circle of diameter 12cm inside a big circle of radius 12cm. Find the area of the unshaded part of the big circle.
Sorry. Please check the correct answer below.
$\pi \times 12 \times 12 = 144\pi$
$\pi \times 6 \times 6 = 36\pi$
$144\pi – 36\pi = 108\pi$
You are Right
$\pi \times 12 \times 12 = 144\pi$
$\pi \times 6 \times 6 = 36\pi$
$144\pi – 36\pi = 108\pi$
The figure below is made up of 2 identical small circle and a big circle. The dotted line represents the diameter of the big circle. What fraction of the figure is shaded?
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In the figure below which is not dram to scale, EFGH is a parallelogram. O is the center of the circle. Find
(a) $\angle$x
(b) $\angle$y
(b) 100$^\circ$
(b) 120$^\circ$
(b) 140$^\circ$
(b) 180$^\circ$
Sorry. Please check the correct answer below.
(a) 90$^\circ$
(b) $\angle$HFO = $\angle$OEF = $\angle$ OFE = 60$^\circ$
$\angle$HFO = Isosceles
$\angle$ FHO = $\angle$HFO = $(180^\circ - 120^\circ) ÷ 2 = 30^\circ$
$\angle$x = $30^\circ + 60^\circ = 90^\circ$
$\angle$ GFE = $\angle$GHO = $\angle$y = 120$^\circ$
(b) 120$^\circ$
You are Right
(a) 90$^\circ$
(b) $\angle$HFO = $\angle$OEF = $\angle$ OFE = 60$^\circ$
$\angle$HFO = Isosceles
$\angle$ FHO = $\angle$HFO = $(180^\circ - 120^\circ) ÷ 2 = 30^\circ$
$\angle$x = $30^\circ + 60^\circ = 90^\circ$
$\angle$ GFE = $\angle$GHO = $\angle$y = 120$^\circ$
The figure consists of a big semicircle of radius 10cm and 4 small quadrants each of radius 5cm. Find the perimeter of the shaded part. Leave your answer in terms of $\pi$.
Sorry. Please check the correct answer below.
$\pi \times 10 = 10\pi$
$\frac{1}{2} \times \pi \times 20 = 10\pi$
$10\pi + 10\pi = 20\pi$
You are Right
$\pi \times 10 = 10\pi$
$\frac{1}{2} \times \pi \times 20 = 10\pi$
$10\pi + 10\pi = 20\pi$
A path is constructed around a circular garden. The path has a width of 3m. Find the area of the path. $(Take \pi = \frac{22}{7})$.
Sorry. Please check the correct answer below.
$\frac{22}{7} \times 7 \times 7 \times 7 = 154$
$\frac{22}{7} \times 10 \times 10 = 314_\frac{2}{7}$
$314_\frac{2}{7} – 154 = 160_\frac{2}{7}m_2$
You are Right
$\frac{22}{7} \times 7 \times 7 \times 7 = 154$
$\frac{22}{7} \times 10 \times 10 = 314_\frac{2}{7}$
$314_\frac{2}{7} – 154 = 160_\frac{2}{7}m_2$
The figure below is made up of an isosceles triangle and a circle with center O. The area of the isosceles triangle is 81cm$^2$. $(Take\pi = 3.14)$, Find the circumference of the circle.
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$\frac{1}{2} \times 9 \times 18 = 81$
You are Right
$\frac{1}{2} \times 9 \times 18 = 81$
The figure below is made up of big circle, a small circle and a square. The comers of the square touch the circumference of the circle. The radius of the small circle is 8cm. The area of the square is 256cm $^2$. Find the area of the shaded part. $ (Take\pi = 3.14)$.
Sorry. Please check the correct answer below.
$256 \times \frac{1}{4} = 64$
R \times r = 128
BQ$\rightarrow$ $\frac{1}{4} \times 3.14 \times 128 = 100.48$
SQ $\rightarrow$ $\frac{1}{4} \times 3.14 \times 8 \times 8 = 50.24$
$100.48 – 50.24 = 50.24cm^2$
You are Right
$256 \times \frac{1}{4} = 64$
R \times r = 128
BQ$\rightarrow$ $\frac{1}{4} \times 3.14 \times 128 = 100.48$
SQ $\rightarrow$ $\frac{1}{4} \times 3.14 \times 8 \times 8 = 50.24$
$100.48 – 50.24 = 50.24cm^2$
The figure shows a quarter circle in a semicircle. The diameter of the semicircle is 14cm. Find the area of the unshaded parts. $(Take \pi = \frac{22}{7})$.
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The figure below, not drawn to scale, shows a circle overlapping a larger semi circle. If the ratio of the shaded area to unshaded area is 5: 8, Find the ratio of the radius of the circle to the radius of the larger semi- circle.
Sorry. Please check the correct answer below.
Shahed:Unshaed
5:8
$5 + 4 = 9$
$\sqrt(9)=3
$Big Semicircle = 9Units
Small Semicirlce = 4Units
$\sqrt(4)=2$
Ratio = 2:3
You are Right
Shahed:Unshaed
5:8
$5 + 4 = 9$
$\sqrt(9)=3
$Big Semicircle = 9Units
Small Semicirlce = 4Units
$\sqrt(4)=2$
Ratio = 2:3
The figure below is formed by 3 circles with the same center. Their radii is in the ratio 2 : 3 : 4. The diameter of the smallest circle is 4cm. Find the diameter of the smallest circle is 4cm. Find the area of the shaded parts.
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The figure below is made up of a semi- circle, 2 small quadrants and a rectangle. O is the center of the semi- circle. The diameter of the semi – circle is 32cm and the radius of each quadrant is 8cm. Find the area of the shaded parts. $(Take\pi = 3.14)$
Sorry. Please check the correct answer below.
Area of Semi-Circle = $\frac{1}{2} \times \pi \times 16 \times 16 = 128\pi$cm$^2$
Area of 2 quadrant = $\frac{1}{2} \times \pi \times 8 \times 8 = 32\pi$cm$^2$
Area of crescents = $8 \times 8 \times 2 – 32\pi = 128 – 32\pi$cm$^2$
Area of rectangle = $16 \times 8 = 128cm^2$
Area of shaded parts = $128\pi – 128 – (128 – 32\pi) = 160\pi – 256 = 246.4cm^2$
You are Right
Area of Semi-Circle = $\frac{1}{2} \times \pi \times 16 \times 16 = 128\pi$cm$^2$
Area of 2 quadrant = $\frac{1}{2} \times \pi \times 8 \times 8 = 32\pi$cm$^2$
Area of crescents = $8 \times 8 \times 2 – 32\pi = 128 – 32\pi$cm$^2$
Area of rectangle = $16 \times 8 = 128cm^2$
Area of shaded parts = $128\pi – 128 – (128 – 32\pi) = 160\pi – 256 = 246.4cm^2$
The figure below is made up of 4 identical circle, each with a radius of 7cm. The circle overlap at the shaded parts A, B and C. The area of the shaded part is 30cm$^2$. . Find the total area of the unshaded parts. $(Take \pi = \frac{22}{7})$.
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The figure below is made up of a semicircle, a square and 3 quadrants. The side of the square is 20cm. For each of the following, use the calculator value of $\pi$ to find
(a) The area of the shaded part, correct to 2 decimal places.
(b) The perimeter of the shaded part, correct to 2 decimal places.
(b) 166.50cm
(b) 177.50cm
(b) 188.50cm
(b) 199.50cm
Sorry. Please check the correct answer below.
(a) $\frac{3}{4} \times \pi \times 20\times 20 = 300\pi$
$\frac{1}{4} \times \pi \times 40\times 40 = 400\pi$
$20 \times 20 = 400$
$\frac{1}{4} \times \pi \times 20\times 20 = 100\pi$
$400\pi – 100\pi – 400 = 300\pi – 400$
$300\pi – 400 + 300\pi = 1481.96cm^2$
(b) $\pi \times 40 = 40\pi$
$\pi \times \frac{1}{4} \times 80 = 20\pi$
$20\pi + 40\pi = 188.50cm$
(b) 188.50cm
You are Right
(a) $\frac{3}{4} \times \pi \times 20\times 20 = 300\pi$
$\frac{1}{4} \times \pi \times 40\times 40 = 400\pi$
$20 \times 20 = 400$
$\frac{1}{4} \times \pi \times 20\times 20 = 100\pi$
$400\pi – 100\pi – 400 = 300\pi – 400$
$300\pi – 400 + 300\pi = 1481.96cm^2$
(b) $\pi \times 40 = 40\pi$
$\pi \times \frac{1}{4} \times 80 = 20\pi$
$20\pi + 40\pi = 188.50cm$
The figure below is made up of squares EFGH and PQRS, a circle and two identical semicircles. J and K are the midpoints of PQ and RQ respectively. P,Q, R and S are the midpoints of EF, FG, GH and EH respectively. Use the calculator value of $\pi$ to find the total area of the shaded parts, correct to 2 decimal places.
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Area A = $(Area of quadrant – area of triangle)$ ÷ 4 = $(\pi \times 14 \times 14 \times \frac{1}{4} - \frac{1}{2} \times 14 \times 14) = 55.94cm^2$
Area of triangle = $\frac{1}{2} \times 14 \times 7 = 49cm^2$
Total area of shaded parts = $55.94 + 49 = 104.94cm^2$
You are Right
Area A = $(Area of quadrant – area of triangle)$ ÷ 4 = $(\pi \times 14 \times 14 \times \frac{1}{4} - \frac{1}{2} \times 14 \times 14) = 55.94cm^2$
Area of triangle = $\frac{1}{2} \times 14 \times 7 = 49cm^2$
Total area of shaded parts = $55.94 + 49 = 104.94cm^2$
The figure shows 4 identical in a square, ABCD. The area of the square is 64cm$^2$. Find the area of the shaded part. $ (Take\pi = 3.14)$.
Sorry. Please check the correct answer below.
$180 – 105 – 30 = 45$
$180 – 85 – 45 = 50$
$180 – 50 = 130$
$130 \div 2 = 65^\circ$
You are Right
$180 – 105 – 30 = 45$
$180 – 85 – 45 = 50$
$180 – 50 = 130$
$130 \div 2 = 65^\circ$
A circular hoopla hoop of radius 10cm is cut into 4 equal pieces and re-arranged to make the shape as shown. What is the perimeter of the shape? $ (Take\pi = 3.14)$
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The figure shows two identical semicircles. O and P are the centers of the semicircles. $\frac{1}{5}$ of each semicircle is shaded. Find the total area of the shaded parts. $ (Take\pi = 3.14)$.
Sorry. Please check the correct answer below.
1. D $\rightarrow$ $6 \div 2 = 30$
r $\rightarrow$ $30 \div 2 = 15$
$3.14 \times 15 \times 15 = 706.5$
Total $\rightarrow$ 10u
Shaded $\rightarrow$ 2u
$706.5 \div 10 = 70.65$
$70.65 \times 2 = 141.3$
You are Right
1. D $\rightarrow$ $6 \div 2 = 30$
r $\rightarrow$ $30 \div 2 = 15$
$3.14 \times 15 \times 15 = 706.5$
Total $\rightarrow$ 10u
Shaded $\rightarrow$ 2u
$706.5 \div 10 = 70.65$
$70.65 \times 2 = 141.3$
O is the center of the large circle and AO is the diameter of the small circle. The diameter of the large circle is 2 times the diameter of the small circle. The circumferences of the big and small circle meet each other at point A. The perimeter of the shaded figure is 30$\pi$ cm, what is the diameter of the small circle?
Sorry. Please check the correct answer below.
Perimeter of small circle = $\pi$d
Perimeter of big circle = $\pi$ + 2d = 2$\pi$d
Total perimeter of figure = $\pi$ d + 2 $\pi$ d
= 3d – 30$\pi$
d = 10cm
You are Right
Perimeter of small circle = $\pi$d
Perimeter of big circle = $\pi$ + 2d = 2$\pi$d
Total perimeter of figure = $\pi$ d + 2 $\pi$ d
= 3d – 30$\pi$
d = 10cm
The figure shown is made up of three identical small circles and a Larger circle with radius 10cm.Find the ratio of the unshaded area to the shaded area. $ (Take\pi = 3.14)$.
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The figure below comprises 5 circle. Circle A is identical to circle B. Circle C is identical to circle D. The ratio of the area of circle A to the area of circle C to the area of circle E is 1 : 4 : 16. If the diameter of circle C is 8cm, find the total perimeter of shaded regions.
Sorry. Please check the correct answer below.
$8 ÷ 2 = 4$
$pi \times 4 \times 4 = 16\pi$
$16\pi \div 4 = 4\pi$
$4\pi \div \pi = 4$
$2 \times 2 = 4$
$2 + 2 = 4$
$4 \times \pi = 4\pi$
$4\pi \times 2 = 8\pi$
$8\pi \times 2 = 16\pi$
$8\pi + 16\pi = 24\pi$cm
You are Right
$8 ÷ 2 = 4$
$pi \times 4 \times 4 = 16\pi$
$16\pi \div 4 = 4\pi$
$4\pi \div \pi = 4$
$2 \times 2 = 4$
$2 + 2 = 4$
$4 \times \pi = 4\pi$
$4\pi \times 2 = 8\pi$
$8\pi \times 2 = 16\pi$
$8\pi + 16\pi = 24\pi$cm
The figure below shows a circle with center O. Given that AC = AB and $\angle$OBC = 40$^\circ$, Find $\angle$CAB.
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The figure below is created with 4 quarter circles and a square.
(a) The unshaded part marked A is enclosed by the quarter and quarter circle. Find the area of A.
(b) Find the area of the shaded figure. $ (Take\pi = 3.14)$.
(b) 9.6 cm$^2$
(b) 9.12 cm$^2$
(b) 11.6 cm$^2$
(b) 11.12 cm$^2$
Sorry. Please check the correct answer below.
$6cm \times 6cm \times \frac{1}{4} \times 3.14 = 28.26cm^2$
$6cm \times 6cm = 36 cm^2$
(a) $36 cm^2 – 28.26 cm^2 = 7.74 cm^2$
$4cm \times 4cm \times 3.14 \times \frac{1}{4} = 12.56 cm^2$
$4cm \times 4cm = 16 cm^2$
$16 cm^2 – 12.56 cm^2 = 3.44cm^2$
$2cm \times 4cm = 8cm^2$
$36cm^2 – 3.44 cm^2 - 8 cm^2 = 24.56cm^2$
$24.56 cm^2 – 7.74 cm^2 = 16.82 cm^2$
$2cm \times 2cm = 4 cm^2$
$12.56 cm^2- 4 cm^2 = 8.56 cm^2$
$2cm \times 2cm \times 3.14 \times \frac{1}{4} = 3.14 cm^2$
$12.82 cm^2 – 3.14 cm^2 – 8.56 cm^2 = 5.12 cm^2$
(b) $5.12 cm^2 + 4 cm^2 = 9.12 cm^2$
(b) 9.12 cm$^2$
You are Right
$6cm \times 6cm \times \frac{1}{4} \times 3.14 = 28.26cm^2$
$6cm \times 6cm = 36 cm^2$
(a) $36 cm^2 – 28.26 cm^2 = 7.74 cm^2$
$4cm \times 4cm \times 3.14 \times \frac{1}{4} = 12.56 cm^2$
$4cm \times 4cm = 16 cm^2$
$16 cm^2 – 12.56 cm^2 = 3.44cm^2$
$2cm \times 4cm = 8cm^2$
$36cm^2 – 3.44 cm^2 - 8 cm^2 = 24.56cm^2$
$24.56 cm^2 – 7.74 cm^2 = 16.82 cm^2$
$2cm \times 2cm = 4 cm^2$
$12.56 cm^2- 4 cm^2 = 8.56 cm^2$
$2cm \times 2cm \times 3.14 \times \frac{1}{4} = 3.14 cm^2$
$12.82 cm^2 – 3.14 cm^2 – 8.56 cm^2 = 5.12 cm^2$
(b) $5.12 cm^2 + 4 cm^2 = 9.12 cm^2$
Find the perimeter of a $\frac{3}{4}$ circle of radius 28cm. $(Take \pi = \frac{22}{7})$.
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The figure below is formed by a square and 2 semicircles of diameter diameter 6cm. Find the perimeter of the figure. Leave your answer in terms of $\pi$
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A boomerang as shown in the figure below is made up of 3 identical small quadrants and 1 big quadrant. The radius of each small quadrant is 2cm and the radius of the big quadrant is 6cm. For each of the following, use the calculator value of $\pi$ to find.
(a) The perimeter of the shaded figure.
(b) The area of the shaded figure.
(b) 3.8cm$^2$
(b) 4.8cm$^2$
(b) 5.8cm$^2$
(b) 6.8cm$^2$
Sorry. Please check the correct answer below.
$\frac{1}{4} \times \pi \times 12 = 3\pi$
$\frac{1}{4} \times pi \times 4 = 1\pi$
$3\pi \pi + \pi + \pi = 6\pi$
Perimeter = $6\pi$ = 18.8cm
(b) $\frac{1}{4} \times \pi \times 6 \times 6 = 9\pi$
$\frac{1}{4} \times \pi \times 2 \times 2 = 1\pi$
$2 \times 2 = 4$
$4 \times 3 = 12$
Area = $9\pi – 3\pi -12 = 6.8cm^2$
(b) 6.8cm$^2$
You are Right
$\frac{1}{4} \times \pi \times 12 = 3\pi$
$\frac{1}{4} \times pi \times 4 = 1\pi$
$3\pi \pi + \pi + \pi = 6\pi$
Perimeter = $6\pi$ = 18.8cm
(b) $\frac{1}{4} \times \pi \times 6 \times 6 = 9\pi$
$\frac{1}{4} \times \pi \times 2 \times 2 = 1\pi$
$2 \times 2 = 4$
$4 \times 3 = 12$
Area = $9\pi – 3\pi -12 = 6.8cm^2$
The figure below is made up of a big semicircle and two identical small semicircles. What is the perimeter of the figure if the radius of the big semicircle is 14cm? $(Take \pi = \frac{22}{7})$.
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In the figure below, a square lies within a circle. Given that the square has an area of 98cm$^2$, Find the total area of the shaded parts. $(Take \pi = \frac{22}{7})$.
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