**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**)

**Solution:**

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

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

280 + 2π*r *= 720 or 2π*r *= 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(R^{2}-r^{2})=`1/2` x`22/7` (77^{2}-70^{2})=`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**

**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 = 64cm^{2}

The area of the combined figure *ABCDE *is Area of *ABDE *+ Area of *BCD*= 117+64 = 181 cm^{2}.

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

**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 cm^{2}.

** Step 2:** The area of one quadrant circle = `1/4` π × 14 × 14 = `1/4` × `22/7` × 14 × 14 = 154 cm^{2}.

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 d_{1} and d_{2} 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 d_{1}^{2} + d_{2}^{2 }= 100.

- Here one of the diagonals is of length 8 cm.

- Take d
_{1}= 8. Then 64 + d_{2}^{2}= 100 or d_{2}^{2}= 36.

d_{2} = 6 cm.

- The area of the rhombus is `1 / 2` (d
_{1}× d_{2}) = `1/ 2` × 8 × 6 = 24 cm^{2}.

**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` r^{2} sq. units = 2* `pi` * (3.5 ) * (2+12+3.5)cm^{2}

= 2*`pi` *`7/2` *19 =133sq .cm^{2}

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## 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)

**Ans: **4,956 sq.m^{ }

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

**Ans: **129 `5/7` cm^{2 }

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**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 cm^{2}