Area Of Oblique Triangle

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Any triangle which is not a right angled triangle is generally called an oblique triangle.

An oblique triangle is:

A triangle with all its three angles less than 90° (acute triangle) .

or

A triangle with any of its angle greater than 90° (obtuse triangle).

Area of oblique triangle can be calculated by Heron's (or Hero's) formula

As per Heron's formula the area of a triangle with sides a, b & c is sqrt(s(s-a)(s-b)(s-c))

where s = (a+b+ c)/(2)
Example for Area of Oblique Triangle

Find the area of the a oblique triangle given in figure.

oblique triangle

Answer:

As per Heron's formula the area of a oblique triangle is sqrt(s(s-a)(s-b)(s-c))

where s = (a+b+ c)/(2)

Here a = 10 inches

b = 8 inches

and

c= 7 inches

so s = (10+8+ 7)/(2)

or s = (25)/(2)

? the area of the given oblique triangle =sqrt(25/2(25/2-10)(25/2-8)(25/2-7))

= sqrt(25/2((25-20)/2)((25-16)/2)((25-14)/2))

= sqrt(25/2(5/2)(9/2)(11/2))

= sqrt((25xx 5 xx 9 xx 11)/( 2 xx 2 xx 2 xx 2))

...
... = sqrt(12375/16)

= sqrt(773.44)

= 21.81

? area of the given oblique triangle = 21.81 square inches
Area of Oblique Triangle with Base B and Height H

Area of oblique triangle with base b and height h = half base x height

oblique triangle with base b and height h

Area of the above oblique triangle = 1/2 bh

Example:

Question: Find the area of an oblique triangle in square feet whose base length is 10 metre and height 20 feet.

Answer:

Since we want the area the oblique triangle in square feet, we have to first convert the base length in metres to feet

We know that 1 metre = 3.28 feet

so 10 metre = 3.28 x 10 = 32.8 feet.

Or we have to find the area of an oblique triangle whose base length is 32.8 feet and height 20 feet.

Area of the above oblique triangle = 1/2 x base x height

= 1/2 x 32.8 x 20

= 328 square feet

?area of the above oblique triangle is 32 ft²
Law of Sines and the Law of Cosines in Area of Oblique Triangles

oblique triangle

For solving oblique triangles law of sines and the law of cosines can be used.

As per law of sines: a/(sin A) = b/(sin B) = c/(sin C)

The law of cosines equation which can be given as follows,

c2 = a2 + b2 – 2ab cos C.

The above equation which generally looks like the Pythagorean Theorem and if C is a right angle the cosine value will be 0.
Examples

Problem 1 :

If the three sides of a triangle are given, then you can easily use it to find the angle .The three sides which are a = 6, b = 7, and c = 8

Solution:

Now we are going to solve the area of oblique using “The law of cosines” equation which can be given as follows,

c2 = a2 + b2 – 2ab cos C.

The law of cosines says 64 = 36 + 49 – 84 cos C

So, cos C = 21/84 = 0.25

The value of c = 0.99.

Problem 2 :

Determine the area of an oblique triangle where the base length is 15 cm and height 25 feet.

Solution:

We know that 1 cm = 0.0328 feet

so 15 cm = 0.0328 x 15 = 0.492 feet.

we have to determine the area of an oblique triangle whose base length is 0.492 feet and height 25 feet.

Area of the oblique triangle = (1/2) x base x height

= (1/2) x 0.492 x 20

= 4.92 square feet

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Problem 3 :

The two angles which are given and the opposite side can be calculated as follows if angle A = 45°, angle B = 60°, and side a = 16.Solve and find the angle.

Solution:

Now we are going to solve the area of oblique using “Law of sine”

The law of sine’s which can be given as,

Sin A/a = Sin B/b = Sin C/c.

The law of sines says that

(Sin 45°)/16 = (sin 60°)/b.

Solving for b gives b = 16(sin 60°)/(sin 45°)

= 16 ×(0.866/0.707)

= 16 × 1.22

= 19.52.

The division of a line with two end points is called a line segment. Line segment MN which we denoted by the symbol bar(MN) .

M --|--------------------------------------|---N

Note: We shell denote a line segment bar(MN) by MN only.

Oblique line segments:

Oblique line segments are lining segments, which is not parallel and perpendicular to x and y axis vice versa. It is look like a slope of lines. Here, the slope is not equal to zero.

A oblique line segments:

(a) A line segment has a definite length.

(b) A line segment has two end-points
Draw a Oblique Line Segments by Using Ruler:

Example:

Draw a oblique line segments of length 4.6cm?

Solution:

Given:

length = 4.6 cm.

Steps to draw a oblique line segments:

Oblique line segments

1. Place the ruler on a drawing paper.

2. Mark a point at the zero of the ruler. Here, we mark point N at the zero of the ruler.

3. Count the divisions in the ruler till we reach the required length. Here, we count the divisions in the ruler till we reach 4.6 cm.

4. Mark the required end point. Here, we mark the end point O.

5. Join points N and O.

Therefore, NO is the required line segment of length 4.6 cm. The required oblique line segments represented by bar(NO) .
Draw a Oblique Line Segments by Using Compass:

Example:

Draw a oblique line segments of length 6.6 cm?

Solution:

Given:

length = 6.6 cm.

Steps to draw a oblique line segments:

The pointer and the pencil of a compass must be tightened and levelled properly.
1. Draw a line.

2. Blot a point on a line. Here, we mark a point on the line and name it Y.

3. Set the compass to the desired length. Here, open the compass to measure the length of the line segment by placing the pointer on the 0 mark the ruler.

4. Place the pointer end of the compass at point Y. Here, the pencil point on the 6.6 cm mark.

5. Draw an arc with the other end cutting the line.

6. Name the point of intersection as Z.

Oblique line segments

Therefore, YZ is the required line segment of length 6.6 cm. The required oblique line segments represented by bar(YZ).

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