Heron’s Formula (Part -3)
1. A field in
the shape of quadrilateral ABCD having angle B is 90 where AB = 3 m
BC = 4 m CD = 8 m DA = 9 m
Sol. In
quadrilateral ABCD join AC
In triangle ABC
, by Pythagoras theorem
AC^2 = BC^2 +
AB^2
AC^2 = 4^2 +
3^2
AC^2 = 16 + 9
AC^2 = 25
AC^2 = 5^2
=> AC = 5 m
Area of
triangle ABC = ½ * AB * BC
Area of
triangle ABC = ½ * 3 * 4
Area of
triangle ABC = 6 m^2
In triangle ACD
a = 8
cm b = 9 cm c = 5
S = (a + b + c)
/ 2
S = ( 8 + 9 + 5
) / 2
S = 22 / 2
=> S = 11 m
By Heron’s
Formula
Area
= √ s ( s – a ) ( s – b ) ( s – c )
Area = √
11 ( 11 – 8 ) ( 11 – 9 ) ( 11 – 5 )
Area = √ 11 * 3
* 2 * 6
Area = √ 11 * 3
* 2 * 2 * 3
Area = 2 * 3 √
11
Area = 6 √11
Area = 6 *
3.317
Area = 19.902 m^2
Total area of
field = Area of ABC + Area of ACD
Total area of
field = 6 + 19.902
Total area of
field = 25.902 m^2
2. A field in
the shape of quadrilateral ABCD having angle B is 90 where AB = 8 m
BC = 6 m CD = 7 m DA = 9 m
Sol. In
quadrilateral ABCD join AC
In triangle ABC
, by Pythagoras theorem
AC^2 = BC^2 +
AB^2
AC^2 = 6^2 +
8^2
AC^2 = 36 + 64
AC^2 = 100
AC^2 = 10^2
=> AC = 10 m
Area of
triangle ABC = ½ * AB * BC
Area of
triangle ABC = ½ * 6 * 8
Area of
triangle ABC = 24 m^2
In triangle ACD
a = 7
cm b = 9 cm c = 10
S = (a + b + c)
/ 2
S = ( 7 + 9 +
10 ) / 2
S = 26 / 2
=> S = 13 m
By Heron’s
Formula
Area
= √ s ( s – a ) ( s – b ) ( s – c )
Area = √
13 ( 13 – 7 ) ( 13 – 9 ) ( 13 – 10 )
Area = √ 13 * 6
* 4 * 3
Area = √ 13 * 2
* 3 * 2 * 2 * 3
Area = 2 * 3 √
26
Area = 6 * 5.1
Area = 30.6 m^2
Total area of
field = Area of ABC + Area of ACD
Total area of
field = 24 + 30.6
Total area of
field = 54.6 m^2
3. A field in
the shape of quadrilateral ABCD where AB
= 3 m BC = 4 m AC = 5 cm
CD = 8 m DA = 9 m
Sol. In
quadrilateral ABCD join AC
Area of
triangle ABC = ½ * AB * BC
Area of
triangle ABC = ½ * 3 * 4
Area of
triangle ABC = 6 m^2
In triangle ACD
a = 8
cm b = 9 cm c = 5
S = (a + b + c)
/ 2
S = ( 8 + 9 + 5
) / 2
S = 22 / 2
=> S = 11 m
By Heron’s
Formula
Area
= √ s ( s – a ) ( s – b ) ( s – c )
Area = √
11 ( 11 – 8 ) ( 11 – 9 ) ( 11 – 5 )
Area = √ 11 * 3
* 2 * 6
Area = √ 11 * 3
* 2 * 2 * 3
Area = 2 * 3 √
11
Area = 6 √11
Area = 6 *
3.317
Area = 19.902 m^2
Total area of
field = Area of ABC + Area of ACD
Total area of
field = 6 + 19.902
Total area of
field = 25.902 m^2
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4. A field in
the shape of quadrilateral ABCD where AB = 8 m
BC = 6 m AC = 10 cm CD = 7 m
DA = 9 m
Sol. In
quadrilateral ABCD join AC
Area of
triangle ABC = ½ * AB * BC
Area of
triangle ABC = ½ * 6 * 8
Area of
triangle ABC = 24 m^2
In triangle ACD
a = 7
cm b = 9 cm c = 10
S = (a + b + c)
/ 2
S = ( 7 + 9 +
10 ) / 2
S = 26 / 2
=> S = 13 m
By Heron’s
Formula
Area
= √ s ( s – a ) ( s – b ) ( s – c )
Area = √
13 ( 13 – 7 ) ( 13 – 9 ) ( 13 – 10 )
Area = √ 13 * 6
* 4 * 3
Area = √ 13 * 2
* 3 * 2 * 2 * 3
Area = 2 * 3 √
26
Area = 6 * 5.1
Area = 30.6 m^2
Total area of
field = Area of ABC + Area of ACD
Total area of
field = 24 + 30.6
Total area of
field = 54.6 m^2
5. A triangle
and a parallelogram have the same base and the same area . If the sides of triangles
are 13 cm , 14 cm and 15 cm and the parallelogram stands on the base 14 cm.
Find the height of parallelogram.
Sol. Here the
sides of triangle are
a = 13
cm b = 14 cm c = 15
S = (a + b + c)
/ 2
S = ( 13 + 14 +
15 ) / 2
S = 42 / 2
=> S = 21 m
By Heron’s
Formula
Area
= √ s ( s – a ) ( s – b ) ( s – c )
Area = √
21 ( 21 – 13 ) ( 21 – 14 ) ( 21 – 15 )
Area = √ 21 * 8
* 7 * 6
Area = √ 3 * 7
* 2 * 2 * 2 * 7 * 2 * 3
Area = 2 * 2 *
3 * 7
Area = 84 m^2
Area of parallelogram = Area of triangle
Base * corresponding height = 84
14 * Corresponding height = 84
Corresponding height = 84 / 14
Corresponding height = 6 cm
6. A triangle
and a parallelogram have the same base and the same area . If the sides of
triangles are 6 cm , 8 cm and 10 cm and the parallelogram stands on the base 6
cm. Find the height of parallelogram.
Sol. Here the
sides of triangle are
a = 6
cm b = 8 cm c = 10
S = ( a + b +
c) / 2
S = ( 6 + 8 +
10 ) / 2
S = 24 / 2
=> S = 12 m
By Heron’s
Formula
Area
= √ s ( s – a ) ( s – b ) ( s – c )
Area = √
12 ( 12 – 6 ) ( 12 – 8 ) ( 12 – 10 )
Area = √ 12 * 6
* 4 * 2
Area = √ 2 * 2
* 3 * 2 * 3 * 2 * 2 * 2
Area = 2 * 2 *
2 * 3
Area = 24 m^2
Area of parallelogram = Area of triangle
Base * corresponding height = 24
6 * Corresponding height = 24
Corresponding height = 24 / 6
Corresponding height = 4 cm
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