1. A traffic signal board, indicating ‘SCHOOL AHEAD,’ is an equilateral triangle with side ‘a.’ Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
1. A traffic signal board, indicating ‘SCHOOL AHEAD,’ is an equilateral triangle with side ‘a.’ Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Explanation:
A signal board is an equilateral triangle,
Since an equilateral triangle is equal side
Let us assume, One side of the equilateral triangle = y
The perimeter of that triangle = 3y,
y = 180/3
∴ y = 60 cm
Semi-perimeter of the that triangle (s) = 3y/2 = 3x60/2 = 90
According to Heron’s formula,
2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m (see Fig. 12.9). The advertisements yield on earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m (see Fig. 12.9). The advertisements yield on earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
Explanation:
Given that, the sides of the wall are 22 m,120 m, and 122 m
Since,
The semi-parameter of the wall (s) = (22+120+122)/2 = 132 m
According to Heron’s formula,
Area of the triangle =
= 132132-22132-120(132-122)
= 1320 m2
Given that the rent of advertising per year = ₹ 5000 per m2
∴ For 3-month rent = (3×5000×1320)/12
= ₹ 1650000
3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN.” If the sides of the wall are 15 m, 11 m, and 6 m, find the area painted in colour
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3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN.” If the sides of the wall are 15 m, 11 m, and 6 m, find the area painted in colour
Explanation:
Given,
Wall sides are11 m, 15 m, and 6 m.
The semi-perimeter of wall (s) = (11+15+6)/2 m
= 16 m
According to Heron’s formula,
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm?
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm?
Explanation:
Let us assume that the third side of the triangle to be ‘y’
Given that triangle having parameter = 42 cm
Hence
y+18+10=42
y=42-28
y=14 cm
∴ The semi-perimeter of triangle(s) is equal to 42/2 = 21 cm
According to Heron’s formula,
5. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.
5. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.
Explanation:
Given that
The ratio of the sides of the triangle are 12: 17: 25
Let us assume that y is common factor of each side
∴ we can say that sides of triangle are 12y, 17y and 25y
The parameter of triangle is
12y+17y+25y = 540 cm
y = 540/54 cm
∴ y = 10 cm
Now,
The sides of triangle are 12y = 120 cm, 17y =170 cm, 25y =250 cm.
So, the semi-perimeter of the triangle (s) = 540/2 = 270 cm
According to Heron’s formula,
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Explanation:
Let us assume that the third side of triangle is y.
Parameter of isosceles triangle is 30
12+12+y = 30
∴ y = 6 cm
The semi-perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm
According to Heron’s formula,
7. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
7. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Explanation:
Construct a quadrilateral ABCD and join BD
AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m
And angle C = 90°
Using Pythagoras theorem in ΔBCD
BD2 = CD2 +BC2
⇒ BD2 = 52+122
⇒ BD2 = 169
⇒ BD = 13 m
The area of ΔBCD = (12×5)/2 = 30 m2
The semi-perimeter of ΔABD(s) = (8+9+13)/2 m
= 30/2 m = 15 m
According to Heron’s formula,
∴ The area of quadrilateral ABCD
= 30 + 35.5 = 65.5 m2
8. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.
8. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.
Explanation:
The semi-perimeter of ΔABC(s) = (3+4+5)/2
= 6 cm
According to Heron’s formula,
The semi-perimeter of ΔACD(s) = (5+4+5)/2
= 7 cm
According to Heron’s formula,
= 9.17 cm2 (approximately)
Area of quadrilateral formed by ABCD = 6 cm2 +9.17 cm2 = 15.17 cm2
9. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.
9. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.
Explanation:
It is an isosceles triangle having sides are 5 cm, 1 cm, and 5 cm
Since, Perimeter (s) = 5+5+1 = 11 cm
The semi-perimeter = 11/2 cm = 5.5 cm
According to Heron’s formula,
For the quadrilateral II section:
The quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.
∴ Area = 1x6.5 cm2=6.5 cm2
For the quadrilateral III section:
This is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.
The Area of the trapezoid = Area of the equilateral triangle
+ Area of the parallelogram
The perpendicular height of the parallelogram will be=(1-(0.5)2)1/2
= 0.86 cm
Also, area of the equilateral triangle is equal to (√3/4×y2) = 0.43
∴ The Area of the trapezoid = 0.86+0.43 = 1.3 cm2 (approximately)
For triangle IV and V:
In the diagram triangles are 2 congruent right-angled triangles having the base as 6 cm and height 1.5 cm
The area of both triangle = (6×1.5)/2 cm2 = 4.5 cm2
∴ The total area of the paper used = (2.488+6.5+1.3+2x4.5) cm2 = 19.3 cm2
10. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
10. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Explanation:
Given that the sides of the triangle are 26 cm, 28 cm, and 30 cm.
The perimeter (s) = 26+28+30 = 84 cm
∴ semi-perimeter = 84/2 cm = 42 cm
According to Heron’s formula,
= 336 cm2
Let us assume the height of parallelogram be r
Hence
area of parallelogram = area of the triangle
r x 28 = 336 cm2
∴ r = 12 cm
11. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
11. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Explanation:
the triangle ΔBCD
The semi-perimeter (s) = (30 + 30 + 348)/2 = 54 m
According to Heron’s formula,
∴ Area of rectangle = 2 × area of the ΔBCD = 2 × 432= 864 m2
each cow will be getting area is equal to 864/18= 48 m2
12. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?
12. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?
Explanation:
Each triangle having same area
since
The semi-perimeter (s) = (50+20+50)/2 = 60cm
According to Heron’s formula,
∴ The area of an umbrella = 5 × 200√6 = 1000√6 cm2
13. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
13. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
Explanation:
Area of kite A = (1/2) × (one of the diagonals)2
Area of the kite = (32×32)/2 = 512 cm2.
Since
The area of shade I = Area of shade II=512/2 = 256 cm2
So, the total area of the paper that is required in each shade = 256 cm2
In triangle (III),
The semi-perimeter of triangle is (6+6+8)/2 = 10 cm
14. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm, and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.
14. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm, and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.
Explanation:
One triangular shape having the semi-parametric = (9+28+35)/2 = 36 cm
According to Heron’s formula,
In figure, Total 16 shape having same area,
Hence total area = 16×88.2 = 1411.2 cm2
∴ The total polishing cost of the tiles = 0.5×1411.2 = Rs. 705.6
15. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
15. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Explanation:
In figure we draw the perpendicular for B to side EC another line form BE such that parallel to AD
After that,
We get ABED as parallelogram
Since
AD = BE = 13 m
AB = ED = 10 m
CE = 25-DE= 25-10 = 15 m
In triangle ΔBCE
The semi-perimeter (s) = (15+14+13)/2 = 21 m
According to Heron’s formula,
Also, we get area of triangle ΔBCE = (base of triangle) x (height of triangle)/2
Hence,
(BF x 15)/2 = 84
BF = 11.2 m
Since
Area of ABED = DE x BF
= 11.2 x 10 = 112 m2
∴ Total area of field (ABCD) = 112 +84 = 196 m2
Also Read: Class 9 Herons Formula Extra Questions
