Sponsor Area
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?
'a' = a
'b' = a
'c' = a
Area of the signal board
Perimeter = 180 cm
'a' + 'b' + 'c' = 180
a + a + a = 180
3a = 180
a = 60 cm
Area of the signal board 
Alternatively,
Area of the signal board
a =122 m
b = 22 m
c = 120 m
1 year =12 months
V Rent for 12 months per m2= र 5000
Rent for 1 month per m2 = र
Rent for 3 months of 1320 m2
Rent for 3 months of 1320 m2
= र (1250 x 1320)
= र 1650000.
a = 15 m
b = 11 m
c = 6 m
Area painted in colour
a = 18 cm
b = 10 cm Perimeter - 42 cm
⇒ a + b + c = 42
⇒ 18 + 10 + c = 42
⇒ 28 + c = 42
⇒ c = 42 – 28
⇒ c = 14 cm
Area of the triangle
Let the sides of the triangle be 12k, 17k and 25 k cm.
Then,
Perimeter = 12k+ 17k + 25k = 54k
According to the question,
54k = 540
k = 10
a = 12k = 12 x 10 = 120 cm
b = 17k = 17 x 10 = 170 cm
c = 25k = 25 x 10 = 250 cm
a = 12 cm
b = 12 cm
Perimeter = 30 cm
⇒ a + b + c = 30
⇒ 12 + 12 + c = 30
⇒ 24 + c = 30
⇒ c = 30 – 24
⇒ c = 6 cm
Area of the triangle
In right triangle PSQ,
PQ2 = PS2 + QS2
| By Pythagoras Theorem
= (12)2 + (16)2
= 144 + 256 = 400
New, for 
a = 20 cm
b = 48 cm
c = 52 cm 
∴ Area of the shaded portion
= Area of ΔPQR – Area of ΔPSQ
= 480 – 96 = 384 cm2.
Let ABC be the right angled triangle right angled at B. Let O be the centre of the circumcircle.
Then, O is the mid-point of the hypotenuse
AC. | by geometry
OA = OB = OC
= radius of the circumcircle = 3 cm
∴ Hypotenuse AC = Diameter of the circle
= 2 x radius of the circumcircle
= 2 x 3 = 6
cm Let BM be the perpendicular from B on AC.
∴ BM = 2 cm
∴ Area of the right angled triangle ABC
Side = 8 cm
∴ Perimeter = 6 x 8 = 48 cm
Area of equilateral triangle OAB
∴ Area of the regular hexagon
= 6 x Area of equilateral triangle OAB
Let the sides of the triangle be 13k, 14k and 15k (in cm). Then,
perimeter = a + b + c
= 13k + 14k + 15k
= 42k cm
According to the questions,
42k = 84
∴ Sides are 26 cm, 28 cm and 30 cm.
∴ Perimeter = a + b + b
= 24 + 13 + 13
= 50 cm
b + b + 6 = 24
b = 9 cm
= 1.73).
For triangular ground
a = 5 m
b = 7 m
c= 8 m
Cost of levelling
= 17.3 x 10 = र 173
Let the third side be x cm. Then,
12 + 12 + x = 30
24 + x = 30
x = 6 cm
So, a = 12 cm
b = 12 cm
c = 6 cm
Let the sides of the triangle be 3k, 5k and 7k (in m). Then,
a = 3k
b = 5k
c = 7k
∴ Perimeter = a + b + c
= 3k + 5k + 7k = 15k
According to the question,
15k = 600
⇒ k = 40
∴ a = 120 m
b = 200 m
c = 280 m
Area of the triangle
a + b + 10 = 24
⇒ a + b = 14 ...(1)
Also, a2 + b2 = (10)2
⇒ a2 + b2 = 100 ...(2)
We know that
(a + b)2 = a2 + b2 + 2ab
⇒ (14)2 = 100 + lab
⇒ 2ab = 96
⇒ ab = 48 ...(3)
Also, (a – b)2 = a2 + b2 – 2ab
= 100 – 2 x 48 = 100 – 9b
= 4 | if a > b
⇒ a – b = 2 ....(4)
Solving (1) and (4), we get a = 8 cm b = 6 cm
Let the length of the perpendicular from the opposite vertex to the side whose length is 130 cm be h cm. Then,
Area of the triangular field
From (1) and (2),
65h = 3000
Sponsor Area
For triangular park
∵ 62 + 82 = 102
∴ Angular between sides of length 6 m and
8 m = 90°
∴ Area of the triangular park
Radius of circular area (r)
∴ Circular area = 
∴ Area used for growing uses
= Area of the triangular part
- Circular area
= 24 – 3.14 = 20.86 m2
Here, a = 24 cm
b = 13 cm
Solution not provided.
30 cm2
Solution not provided.
Ans. 12 cm2
Solution not provided.
Ans. 384 cm2
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided
Ans. 21 cm2
Solution not provided
Ans. 84 cm2
Solution not provided
Ans. 337550 cm2
Solution not provided
Ans. 
Solution not provided
Ans. 336 cm2
Solution not provided
Ans. 
Join BD.
Area of right triangle BCD
In right triangle BCD,
BD2 = BC2 + CD2
| BY Pythagoras Thereom
= (12)2 + (5)2 = 144 + 25 = 169
For ΔABD
a = 13 m
b = 8 m
c = 9 m
∴ Area of the ΔABD
∴ Area of the quadrilateral ABCD
= Area of ΔBCD + Area of ΔABD
= 30 m2 + 35.5 m2
= 65.5 m2 (approx.)
Hence, the park occupies the area 65.5 m2. (approx.)
Tips: -
For ΔABC
a = 4 cm
b = 5 cm
c = 3 cm
∵ a2 + c2 = b2
∴ ΔABC is right angled with ∠B = 90°.
∴ Area of right triangle ABC
For ΔACD
a = 4 cm b = 5 cm
c = 5 cm
∴ Area of the ΔACD
= 2 x 4.6 cm2 (approx.)
= 9.2 cm2 (approx.)
∴ Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 6 cm2 + 9.2 cm2
= 15.2 cm2, (approx.)
For Triangular Area I
a = 5 cm b = 5 cm c = 1 cm
Area II = 6.5 x 1 = 6.5 cm2
For Area III
Total area of the paper used
= Area I + Area II + Area III + Area IV + Area V
= 2.5 cm2 + 6.5 cm2 + 1.3 cm2 + 4.5 cm2 + 4.5 cm2
= 19.3 cm2.
For triangle
a = 26 cm
b = 28 cm
c = 30 cm
Area of the triangle
Let the height of the parallelogram be h cm.
Then, area of the parallelogram
= Base x Height = 28 x h cm2
According to the question,
Hence, the height of the parallelogram is 12 cm.
For ΔABC
a = 30 m
b = 48 m
c = 30 m
Area of 
Area of the rhombus
= 2 Area of ΔABC = 2 x 432 = 864 m2
∴ Area of grass for 18 cows = 864 m2. ∴ Area of grass for 1 cow
Aliter : Draw BE ⊥ AC. Then E is the midpoint of AC.
In right triangle AEB.
AB2 = AE2 + BE2
I By Pythagoras Theorem ⇒ (30)2 = (24)2 + BE2
⇒ 900 = 576 + BE2
⇒ BE2 = 900 – 576
⇒ BE2 = 324
BD = 2 BE = 2 x 18 = 36 m Area of rhombus ABCD
Product of diagonals
Area of right 
∴ Area of rhombus ABCD = 4 Area of ΔAEB = 4 x 216 = 864 m2
∴ Area of grass for 18 cows = 864 m2
Area of grass for 1 cow = 
Sponsor Area
For one triangular piece
a = 20 cm b = 50 cm c = 50 cm
Area of one triangle
∴ Area of five triangles of one colour
Hence, 1000
cm2 of each colour is required for the umbrella.
Area of paper of shade II = 256 cm2 For paper of shade III
a = 8 cm b = 6 cm c = 6 cm
For one tile
a = 9 cm b = 28 cm c = 35 cm
Area of one tile
Area of 16 tiles
∴ Cost of polishing the tiles at the rate of 50 p per cm2.
Let the given field be in the shape of a trapezium ABCD in which AB = 25 m, CD = 10 m, BC = 13 m and AD = 14 m.
From D, draw DE || BC meeting AB at E. Also, draw DF ⊥ AB.
∴ DE = BC = 13 m
AE = AB – EB = AB – DC
= 25 – 10= 15 m
For ΔAED
a = 14 m b = 13 m c = 15 m
∴ Area of the ΔAED
⇒ Height of the trapezium is 11.2 m. ∴ Area of parallelogram EBCD = Base x Height
Area of the field = Area of ∴AED + Area of parallelogram EBCD
= 84 m2 + 112 m2 = 196 m2.
Area of 
+ Area of 
= 84 m2
For ΔABC
a = 36.5 cm b = 55 cm c = 36.5 cm
Area of the ΔABC
∴ Area of the rhombus ABCD
= 2 Area of the ΔABC = 2 x 660 = 1320 cm2
∴ Length of the other diagonal is 48 cm.
For ΔACD
a = 28 m b = 41 m c = 15 m
Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 180 m2 + 126 m2 = 306 m2.
For triangle
a = 15 cm b = 14 cm c = 13 cm
Let the height of the parallelogram be h cm. Then, area of the parallelogram
= Base x Height = 15 x h = 15h cm2 According to the question,
Area of the parallelogram
= Area of the triangle
Hence, the height of the parallelogram is 5.6 cm.
For ΔABC, a = 34 cm b = 42 cm c = 20 cm
∴ Area of ΔABC
∴ Area of parallelogram ABCD
= 2 Area of triangle ABC
= 2 x 336 cm2 = 672 cm2
Let each of the equal sides of the rhombus be a cm. Then,
Perimeter = a + a + a + a = 4a m According to the question,
4a = 200
∴ Area of the rhombus
For ∴ABD
a = 13 cm b = 14 cm c = 15 cm
For ΔBCD
a = 9 cm b = 12 cm c = 15 cm
Now, area of quadrilateral ABCD
= area of ΔABD + area of ΔBCD = 84 cm2 + 54 cm2
= 138 cm2
Solution not provided.
Ans. 2.26 hectares (nearly), 1.92 hectares (nearly), 1.92 hectares (nearly)
Solution not provided.
Ans. First group cleaned more area by 54 m2 ; 306 m2.
Area of a quadrilateral =
x a diagonal x sun of the perpendiculats on the diagonal
a diagonal x sum of the perpendiculars on the diagonal
x a diagonal x sun of the perpendiculats on the diagonal
x a diagonal x sun of the perpendiculats on the diagonal
A.
x a diagonal x sun of the perpendiculats on the diagonal
Base x Height
Base x Height
Base x Height
Base x Height
A.
Base x Height
x product of diagonals
product of diagonals
x product of diagonals
x product of diagonals
A.
x product of diagonals
x sum of parallel sides x distance between the parallel sides.
x sum of the parallel sides x distance between the parallel sides
None of these
A.
x sum of parallel sides x distance between the parallel sides.
Sponsor Area
Solution not provided.
Ans. 
Solution not provided.
Ans. र 10500
Solution not provided.
Ans. 288 
cm2
Solution not provided.
Ans. 360 cm2
Solution not provided.
Ans. 
Sponsor Area
Solution not provided.
Ans. र 6705
Solution not provided.
Ans. र 5376
Solution not provided.
Ans. 864 cm2
Solution not provided.
Ans. 480 cm2
Sponsor Area
Sponsor Area
