Math Guide

Learning high school geometry

Introduction:


            A project for mathematics class x have several topics. Measurements areused in our usual life in several ways. For example, we measure the length of a cloth for stitching, the area of a Room for white washing, the perimeter of a ground for fencing and the volume of a bottle for filling water. Based on the measurements, we do further calculations according to our requirements. The branch of mathematics which handles the measurement of, volumes, angles, areas, circumference, and lengths and of plane and solid figures is called menstruation. In this article we shall discuss mathematics project for menstruation for class x.


Learnhigh school geometry gives a brief introduction on high school geometry. High school geometry is helpful for students practically through out their life.

  • Lines and planes
  • Prism,pyramid,cylinder, cone , sphere
  • Polygon- interior and exterior angle
  • Construction of triangles , circles , tangents to circles
  • Formal and informal Proofs
  • Symmetry, reflection, translation,dialation
  • Postulates based on similarity,congruence of triangles
  • Co-ordinate geometry
  • Circles, tangent , secant
  • Learn different types of quadrilateral - parallelogram , square , rhombus,trapezoid,kite .Properties and uses
  • Parallel lines , transversal and angles
  • Area of geometric figures

Learn examples on high school geometry:


Pro1: A running track of 7m wide is as shown in Figure 2.46. The inside perimeter is 720m and the length of each straight portion is 140m. The curved portions are in the form of semi-circles. Find the area of the track. (use π ≈22/7)


project


Solution:

Let r be the radius of the inner semicircles. Then the inside perimeter is

2 × 140 + 2 × (π × r) or 280 + 2πr. But this is given as 720m.

280 + 2πr = 720 or 2πr = 440 or r =(440/2π)  440x7 / 2x22 =70m

So the radius of the inner semicircle r = 70m.

The radius of the outer semicircle R = 70 + 7 = 77m.

Nowthe area of the running track is equal to the sum of the areas of the semicircular tracks and the areas of the rectangular tracks. But, the area of one semi-circular track

=1/2 π(R2-r2)=`1/2` x`22/7` (772-702)=`11/7` x147x7="1617

The area of one rectangular track = 140 × 7 = 980 sq. m.

Area of the track = 2 × 1617 + 2 × 980 = 3234 + 1960 = 5194 sq.m.


Pro 2: Find the area of Figure


project

Sol:

The figure ABCDE is the combination of ABDE and BCD with the side BD in juxta position. ABDE is a trapezium where the parallel sides `barA` `barE` and  `barB``barD` have lengths 10 cm and 16 cm respectively. The distance between the parallel sides is 9 cm. So, the area of the trapezium ABDE is

`1 / 2` (a + b)x h="1/2(10+16)x9=117cm2

BCD is a triangle whose base  `barB` `barD` is of length 16 cm and height 8 cm. So its area is

`1 / 2 ` b x h =`1/2` x16 x 8 = 64cm2

The area of the combined figure ABCDE is Area of ABDE + Area of BCD= 117+64 = 181 cm2.


Pro 3: Find the area of the shaded region in Figure 2, where the boundaries of the region are quadrants of a circle.


Areaofshape


Sol:The given region is that which remains after cutting away four equal quadrants each of radius 14 cm from a square of side 28 cm.


 Step 1:                 The area of the square = 28 × 28 = 784 cm2.

 Step 2:                  The area of one quadrant circle = `1/4` π × 14 × 14 = `1/4` × `22/7` × 14 × 14 = 154 cm2.

                          Required area = 784 − 4(154) = 784 − 616 = 168 sq. cm.


 

Pro4: The perimeter of a rhombus is 20 cm. One of the diagonals is of length 8 cm. Find the length of the another diagonal and the area of therhombus.


Sol:            Let d1 and d2 be the diagonals' length.

  •                    Then perimeter = 2 `sqrt((d_1)^2 + (d_2)^2)`

  •                    But the perimeter is 20 cm. 

                   2`sqrt((d_1)^2 + (d_2)^2)` = 20 cm or  d12 + d22 = 100.

  •                     Here one of the diagonals is of length 8 cm.
  •                    Take d1 = 8. Then 64 +  d22 = 100 or  d22 = 36.

                                                                          d2 = 6 cm.

  •                    The area of the rhombus is `1 / 2` (d1 × d2) = `1/ 2` × 8 × 6 = 24 cm2.

 

Pro5: A capsule is in the form of a cylinder with hemispherical ends. Thetotal height of the capsule is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the capsule.


Sol:

Diameter of the cylinder = 7cm

Radius of the cylinder = 3.5 cm = `7/2` cm

Radius of the cylinder = Radius of the hemisphere

Total height of the capsule = 19 cm

Height of the cylinder = 19 – (2 × 3.5) = 12 cm

Total surface area of the capsule = CSA of the cylinder + surface area of two hemispheres.

                                                     = 2`pi` rh+2*2`pi` r2 sq. units = 2* `pi` * (3.5 ) * (2+12+3.5)cm2

                                                     = 2*`pi` *`7/2` *19 =133sq .cm2



Practice problems based on high school geometry:


Pro 1: Aplay ground is to be constructed with two straight segments and two semicircular segments as shown in Figure.The radius of each semicircularsegment is 21m. The length of each of the straight segment is 85m. Findthe area of the playground. (Take π ≈ 22/7)


project

Ans: 4,956 sq.m 

Pro 2: Find the area of the design as in Figure (Take π≈722)


 project


Ans: 129 `5/7` cm2


Pro 3: Theperimeter of a rhombus is 20 cm. One of the diagonals is of length 6 cm. Find the length of the another diagonal and the area of the rhombus.

Ans: length = 8 cm, area = 24 cm2