Geometry and trigonometry

Review exercises

1. A right-angled triangle has sides $b=12\mathrm{cm},c=37\mathrm{cm}$. Angle $B\hat{C}A$ is the right angle.
   
   (a)Find the side $a$.
   (b)Calculate angles $\alpha =C\hat{A}B$ and $\beta =A\hat{B}C$.
   Solutions:    (a)  $a=35\mathrm{cm}$;     (b)  $\alpha \approx {71.1}^{\circ },\beta \approx {18.9}^{\circ }$
2. An isosceles triangle has sides $a=b=33\mathrm{cm},c=18\mathrm{cm}$.
   
   (a)Find the height $h_c$ giving your result in exact form.
   (b)Calculate area. Round your result to three significant figures.
   (c)Calculate all three angles.
   Solutions:    (a)  $h_c=12\sqrt{7}\mathrm{cm}$;     (b)  $A\approx 286{\mathrm{cm}}^2$;     (c)  $\alpha =\beta \approx {74.2}^{\circ },\gamma \approx {31.7}^{\circ }$
3. Triangle $KLM$ is given:
   
    
   (a)Find the side $k=LM$.
   (b)Calculate the area of this triangle.
   Solutions:    (a)  $k\approx 12.0\mathrm{cm}$;     (b)  $A\approx 97.7{\mathrm{cm}}^2$
4. Triangle $ABC$ has side $b=8\mathrm{cm}$ and angles $\hat{A}={45.6}^{\circ },\hat{C}={76.5}^{\circ }$.
   
   (a)Calculate side $a$.
   (b)Calculate the area.
   (c)Hence, calculate the height $h_a$.
   Solutions:    (a)  $a\approx 6.75\mathrm{cm}$;     (b)  $A\approx 26.2{\mathrm{cm}}^2$;     (c)  $h_a=\frac{2A}{a}\approx 7.78\mathrm{cm}$
5. Triangle $ABC$ has sides $b=17\mathrm{cm},c=22\mathrm{cm}$ and angle $\hat{A}={52}^{\circ }{40}^{\prime }$.
   
   (a)Calculate side $a$.
   (b)Calculate angle $\hat{B}$. Give its value rounded to the nearest minute.
   (c)Calculate the area of this triangle.
   Solutions:    (a)  $a\approx 17.9\mathrm{cm}$;     (b)  $\hat{B}\approx {49}^{\circ }{9}^{\prime }$;     (c)  $A\approx 149{\mathrm{cm}}^2$
6. Triangle $ABC$ has sides $a=11\mathrm{cm},b=20\mathrm{cm},c=13\mathrm{cm}$.
   
   (a)Calculate the largest angle in this triangle.
   (b)Hence or otherwise calculate the area.
   The midpoint of the side $AB$ is labelled $P$.
   (c)Calculate $PC$.
   Solutions:    (a)  $A\hat{B}C\approx {113}^{\circ }$;     (b)  $A=66{\mathrm{cm}}^2$;     (c)  $PC\approx 14.8\mathrm{cm}$
7. Triangle $ABC$ is given (see picture):
   
    
   (a)Find the side $x$.
   (b)Calculate the perimeter of this triangle.
   Solutions:    (a)  $x=14\mathrm{cm}$;     (b)  $P=26+14+30=70\mathrm{cm}$
8. Triangle $RST$ is given (see picture):
   
    
   (a)Find $x$.
   (b)Calculate the sides of this triangle.
   (c)Calculate the area of this triangle.
   Solutions:    (a)  $x=7\mathrm{cm}$;     (b)  $6\mathrm{cm},14\mathrm{cm},10\mathrm{cm}$;     (c)  $A\approx 26.0{\mathrm{cm}}^2$
9. Annie thought that the floor in her room has the shape of a rectangle. She measured all four sides and one of the diagonals. Her results are written in the picture (the picture is not to scale):
    
   (a)(i)Calculate all four angles of her room.
    (ii)Name the type of this quadrilateral. Is it a rectangle?
   (b)Find the area of this room. Write it down in square meters.
   (c)(i)Find the length of the diagonal $BD$.
    (ii)By how many percent the diagonal $BD$ is shorter than $AC$?
   Solutions:    (a)  $\hat{A}=\hat{C}\approx {88.7}^{\circ },\hat{B}=\hat{D}\approx {91.3}^{\circ }$,   it isn't a rectangle, it's a parallelogram.     (b)  area$\approx 9.62{\mathrm{m}}^2$;     (c)  $BD\approx 440\mathrm{cm}$,   it's shorter by $2.18\mathrm{\%}$
10. In year 1980 Miss Jane Marple bought a TV set. The screen had the ratio $w:h=4:3$ and the diagonal was 40 cm long.
    
    (a)Calculate the width $w$ and the height $h$ of the screen.
    (b)Calculate the area of the screen.
    In year 2020, Miss Jane Marple bought a new TV set. The screen has the ratio $w:h=16:9$ and the diagonal is 104 cm long.
    (c)Calculate the width $w$ and the height $h$ of the screen of the new TV set.
    (d)Calculate the area of the screen.
    (e)How many times larger is the area of the new screen?
    Solutions:    (a)  $w=32\mathrm{cm},h=24\mathrm{cm}$;     (b)  $A=768{\mathrm{cm}}^2$;     (c)  $w=90.6\mathrm{cm},h=51.0\mathrm{cm}$;     (d)  $A\approx 4622{\mathrm{cm}}^2$;     (e)  it's six times larger
11. A farmer has a field of triangular shape. Two of the sides have the lengths of 25 m and 70 m. The angle between these two sides is 75°.
    
    (a)Farmer is going to put a fence around his field. The price for 1 m of fence is 15 €.

    (i)Calculate the third side.

    (ii)Calculate the perimeter.

    (iii)Calculate the price he's going to pay for the fence around his triangular field.
    (b)Calculate the area of this field.
    (c)This farmer is thinking about selling the triangular field and buying a square one with the same area.

    (i)Calculate the side of such square field.

    (ii)Calculate the price for the fence around the square field.
    Solutions:    (a)  Third side $\approx 68.0\mathrm{m}$, $P\approx 163\mathrm{m}$, he'll pay 2444.46 € (approx. 2440 €);     (b)  area $\approx 845{\mathrm{m}}^2$;     (c)  side $\approx 29.1\mathrm{m}$, the price would be 1744.32 € (approx. 1740 €)
12. A circular sector is shaded in a circle with the radius $r=15\mathrm{cm}$:
    
     
    Area of the shaded sector is $90{\mathrm{cm}}^2$.
    (a)Find the central angle.
    (b)Find the length of the minor arc $AB$.
    Solutions:    (a)  $\theta \approx {45.8}^{\circ }$;     (b)  $\ell =12\mathrm{cm}$
13. A circle has the centre $C$ and the radius $r=6\mathrm{cm}$. Points $P,Q$ and $R$ lie on the circle (see the diagram):
    
     
    The minor arc $PQ$ has the length $7\mathrm{cm}$.
    (a)Find the central angle $P\hat{C}Q$.
    (b)Hence, find the area of the shaded region.
    Area of the circular sector $CQR$ is $24{\mathrm{cm}}^2$.
    (c)Find the central angle $Q\hat{C}R$.
    (d)Find the length of the minor arc $PR$.
    (e)Find the length of the straight line segment $PR$.
    Solutions:    (a)  $P\hat{C}Q\approx {66.8}^{\circ }$;     (b)  area $\approx 4.45{\mathrm{cm}}^2$;     (c)  $Q\hat{C}R\approx {76.4}^{\circ }$;     (d)  arc $\ell =15\mathrm{cm}$;     (e)  straight line segment $d\approx 11.4\mathrm{cm}$
14. Jenifer is preparing a decorative figure (see the picture below). She started by drawing a square with side $a=12\mathrm{cm}$. She divided each side in three equal parts (so that the length of each part is 4 cm). Then she added several circular arcs in red colour. Centres of these arcs are the vertices of the square and the midpoints of its sides. She coloured the region enclosed by these arcs.
     
    (a)Calculate the total length of the red line.
    (b)Calculate the total area of the figure enclosed by this red line.
    Solutions:    (a)  Length $\approx 101\mathrm{cm}$;     (b)  area $\approx 270{\mathrm{cm}}^2$
15. A triangle has vertices in points $P(-2,3),Q(12,3),R(3,15)$.
    
    (a)Calculate the sides of this triangle.
    (b)Calculate the area.
    (c)Calculate the largest angle in this triangle.
    Solutions:    (a)  $PQ=14,QR=15,RP=13$;     (b)  $A=84$;     (c)  $Q\hat{P}R\approx {67.4}^{\circ }$
16. A quadrilateral $ABCD$ has vertices in points $A(-1,2),B(23,9),C(30,33),D(6,26)$.
    
    (a)(i)Calculate the sides of this quadrilateral.
     (ii)Hence, determine the type of this quadrilateral.
    (b)Calculate the area.
    (c)Find the inradius (radius of the inscribed circle).
    Solutions:    (a)  $AB=BC=CD=DA=25$, it's a rhombus;     (b)  area $=527$;     (c)  $r\approx 10.5$

17. Three towns have the coordinates: $A(5,7),B(95,72),C(182,15)$. A travel sales agent starts his working day in town $A$. Then he travels to town $B$ and later he continues his way to $C$. When he finishes his work in town $C$, he wants to return to the town $A$. His question is, which road should he chose:
    • He can use the highway from $C$ to $B$ and then to $A$. In this case he can drive at $130\mathrm{km}/\mathrm{h}$.
    • He can use a direct road from $C$ to $A$, but in this case he must observe the speed limit of $90\mathrm{km}/\mathrm{h}$.
    (a)Calculate the distances:

    (i)$AB$,

    (ii)$BC$

    (iii)$AC$
    (b)Calculate the time needed for the travel on the highway $C-B-A$.
    (c)Calculate the time needed for the travel on the direct road $C-A$.
    Solutions:    (a)  $AB\approx 111\mathrm{km},BC\approx 104\mathrm{km},AC\approx 177\mathrm{km}$;    (b)  1 hour 39 minutes;    (c)  1 hour 58 minutes
18. An airplane flies in a straight line from town $A(10,10)$ to town $B(107,134)$.
    
    (a)Calculate the distance of this flight.
    (b)Find the bearing.
    Solutions:    (a)  $AB\approx 157\mathrm{km}$;    (b)  $\beta \approx {38.0}^{\circ }$
19. A ship leaves the port $A(100,100)$ and travels in a straight line on a bearing of 225°. The speed of this ship is $35\mathrm{km}/\mathrm{h}$. After 2 hours and 40 minutes the ship reaches the port $B$.
    
    (a)Calculate the distance $AB$.
    (b)Find the coordinates of $B$.
    On the next day the ship returns to the home port $A$.
    (c)Find the bearing on the way from $B$ to $A$.
    Solutions:    (a)  $AB\approx 93.3\mathrm{km}$;    (b)  $B(34,34)$;    (c)  $\beta ={45}^{\circ }$
20. A town has the form of a rectangle $12\times 8\mathrm{km}$. There are three shopping malls in this town. They are located at $A(1,5),B(9,1)$ and $C(11,5)$.
    
    Inhabitants of this town always go shopping to the nearest shopping mall, so the town is divided in three regions, depending on which is the closest shopping mall.
    
    (a)Find the equations of the Voronoi edges representing the borders of these regions.
    (b)Draw these three regions and their borders on the diagram.
    (c)Write down the coordinates of the point where all three edges meet.
    (d)Calculate the area of the region containing $C$.
    Solutions:    (a)  $y=2x-7(\mathrm{for}x⩽6)$, $y=-\frac{1}{2}x+8(\mathrm{for}x⩾6)$, $x=6(\mathrm{for}y⩾5)$;     (c)  $P(6,5)$;     (c)  area $=27$
21. A town has the form of a rectangle $13\times 9\mathrm{km}$. There are four schools in this town. They are located at $A(6,5),B(5,8),C(11,5)$ and $D(4,1)$. The school districts are organized so that every student attends the school nearest to his home. Here is a picture of the school districts:
    
    
    (a)Find the equations of all three borders of the school district of $A$.
    (b)Calculate the area of the school district of $A$.
    Solutions:    (a)  $y=\frac{1}{3}x+\frac{14}{3}(\mathrm{for}1⩽x⩽8.5)$, $y=-\frac{1}{2}x+\frac{11}{2}(\mathrm{for}1⩽x⩽8.5)$, $x=8.5(\mathrm{for}1.25⩽y⩽7.5)$;     (b)  area $\approx 23.4$
22. A cardboard box has the form of a cuboid. The ratio of its sides is $\ell :w:h=6:2:3$ and its volume is $V=4\frac{1}{2}\mathrm{litres}$.
    
    (a)Calculate the length, width and height.
    (b)Calculate the space diagonal.
    (c)Calculate the total surface area of this box.
    Solutions:    (a)  $\ell =30\mathrm{cm},w=10\mathrm{cm},h=15\mathrm{cm}$;     (b)  $D=35\mathrm{cm}$;     (c)  $A=1800{\mathrm{cm}}^2$
23. Egiptologist Dr Henry Walton Jones is investigating the Great Pyramid. For his calculations he introduced a three-dimensional coordinate system and he used meters as units. In this coordinate system vertices of the base are $A(-115,-115,0),B(115,-115,0),$ $C(115,115,0),D(-115,115,0)$ and the apex is $E(0,0,147)$.
    
     
    (a)Calculate the volume of the pyramid.
    (b)Calculate the area of one of the lateral faces.
    Solutions:    (a)  Volume $\approx 2.59\cdot {10}^6{\mathrm{m}}^3$;     (b)  area $\approx 2.15\cdot {10}^4{\mathrm{m}}^2$
24. Authorities of a small town are making plans for Christmas decorations based on silver balls hanging on strings. They bought several cans of silver colour and 480 plastic balls with the radius of 7 cm.
     
    (a)Calculate the surface area of one ball.
    (b)The label on the can says that 1 gram of colour covers $30{\mathrm{cm}}^2$ of surface. How many grams of colour are needed for one ball?
    (c)This colour is sold in 1 kg cans only. Find out how many cans will they have to buy in order to colour all the balls.
    Balls will be hung on strings of different lengths: 100 balls will have strings of the length 40 cm, 160 balls will have strings of the length 50 cm, 140 balls will have strings of the length 60 cm, all other balls will have strings of the length 70 cm.
    (d)(i)Find the total length of string needed.
     (ii)Find the average length of a string.
    Solutions:    (a)  Surface area $\approx 616{\mathrm{cm}}^2$;     (b)  20.5 grams are needed for 1 ball;     (c)  they will need 10 cans;     (d)  total length: 26000 cm = 260 m, average length of one string: 54.2 cm
25. Death Star is a giant space ship in the form of a sphere with diameter of 160 km. Borg Cube is a giant space ship in the form of a cube.
    
    (a)Find the volume of the Death Star.
    (b)Find the surface area of the Death Star.
    Borg Cube has the same volume as the Death Star.
    (c)Find the surface area of the Borg Cube.
    (d)By what percentage is the surface area of the Borg Cube greater than the surface area of the Death Star?
    Solutions:    (a)  $V\approx 2.14\cdot {10}^6{\mathrm{km}}^3$;     (b)  $A_{DS}\approx 8.04\cdot {10}^4{\mathrm{km}}^2$;     (c)  $A_{BC}\approx 9.98\cdot {10}^4{\mathrm{km}}^2$;     (d)  by 24.1% greater

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