Q.1. Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
Given: Base = 24 cm
Height = 14.5 cm
We know that,
Area of a triangle = 1/2 × Base × Height
= 1/2 × 24 cm × 14.5 cm
= 174 cm^{2}
Q.2. Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm. Also, find the height corresponding to the longest side.
Given: Side 1 = a (let) = 42 cm
Side 2 = b (let) = 34 cm
Side 3 = c (let) = 20 cm
We know that,
Area of a scalene triangle = √(s(sa) (sb) (sc))
Where,
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (4842)cm × (4834)cm × (4820)cm)
= √(48cm × 6cm × 14cm × 28cm)
= √112896 cm^{2}
= 336 cm^{2}
Clearly,
Length of longest side = 42 cm
Now,
We know that,
Area of a triangle = 1/2 × Base × Height
⇒ 336 cm^{2} = 1/2 × 42 cm × Height
⇒ 336 cm^{2} = 21 cm × Height
⇒ Height = 16 cm
Q.3. Find the area of the triangle whose sides are 18cm, 24cm and 30 cm. Also, find the height corresponding to the smallest side.
Given: Side 1 = a (let) = 18 cm
Side 2 = b (let) = 24 cm
Side 3 = c (let) = 30 cm
We know that,
Area of a scalene triangle = √(s(sa)(sb)(sc))
Where,
⇒ s = 72/2 cm
⇒ s = 36 cm
Now,
Area of a scalene triangle = √(36cm × (3618)cm × (3624)cm × (3630)cm)
= √(36cm × 18cm × 12cm × 6cm)
= √46656 cm^{2}
= 216 cm^{2}
Clearly,
Length of smallest side = 18 cm
Now,
We know that,
Area of a triangle = 1/2 × Base × Height
⇒ 216 cm^{2}= 1/2 × 18 cm × Height
⇒ 216 cm^{2}= 9 cm × Height
⇒ Height = 24 cm
Q.4. The sides of a triangle are in the ratio 5 : 12 : 13, and its perimeter is 150 cm. Find the area of the triangle.
Given: Ratio of Sides = 5 : 12 : 13
Perimeter = 150 cm
Let the sides be,
a = 5x cm
b = 12x cm
c = 13x cm
We know that,
Perimeter of a triangle = a + b + c
⇒ 150 cm = 5x cm + 12x cm + 13x cm
⇒ 150 cm = 30x cm
⇒ x = 5
Therefore,
a = 5x cm = 5 × 5 cm = 25 cm
b = 12x cm = 12 × 5 cm = 60 cm
c = 13x cm = 13 × 5 cm = 65 cm
Now,
We know that,
Area of a scalene triangle = √(s(sa)(sb)(sc))
⇒ s = 75cm
Now,
Area of a scalene triangle = √(75cm × (7525)cm × (7560)cm × (7565)cm)
= √(75cm × 50cm × 15cm × 10cm)
= √562500 cm^{2}
= 750 cm^{2}
Q.5. The perimeter of a triangle fields is 540 m, and its sides are in the ratio 25 : 17 : 12. Find the area of the fields. Also, find the cost of ploughing the field at Rs 40 per 100 m^{2}.
Given: Ratio of Sides = 25 : 17 : 12
Perimeter = 540 m
Let the sides be,
a = 25x m
b = 17x m
c = 12x m
We know that,
Perimeter of a triangle = a + b + c
⇒ 540 m = 25x m + 17x m + 12x m
⇒ 540 m = 54x m
⇒ x = 10
Therefore,
a = 25x m = 25 × 10 m = 250 m
b = 17x m = 17 × 10 m = 170 m
c = 12x m = 12 × 10 m = 120 m
Now,
We know that,
Area of a scalene triangle = √(s(sa)(sb)(sc))
⇒ s= 540/2 m
⇒ s = 270 m
Now,
Area of a scalene triangle = √(270m × (270250)m × (270170)m × (270120)m) = √(270m × 20m × 10m × 150cm)
= √81000000 m^{2}
= 9000 m^{2}
Now,
The cost of ploughing 100 m^{2} = Rs 40
Therefore, The cost of ploughing 1 m^{2} = Rs 40/100
Therefore, The cost of ploughing 9000 m^{2} = Rs (40/100) x 9000
= Rs 3600
Q.6. The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.
Given: Perimeter = 40 cm
Hypotenuse = 17 cm
The diagram is given as:
Let the sides be a, b and c(hypotenuse).
Therefore, a + b + c = 40 cm
⇒ a + b + 17 = 40 cm
⇒ a + b = 40  17 cm
⇒ a + b = 23 cm
⇒ a = (23b) cm
Now we know that,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ a^{2} + b^{2} = c^{2}
⇒ (23b)^{2} + b^{2} = 172
⇒ 23^{2} + b^{2}46b + b^{2} = 289
⇒ 529 + b^{2}46b + b^{2} = 289
⇒ 2b^{2}46b + 240 = 0
⇒ b^{2}23b + 120 = 0
⇒ b^{2}8b15b + 120 = 0
⇒ b(b8)15(b8) = 0
⇒ (b8)(b15) = 0
This gives us two equations,
i. b8 = 0
⇒ b = 8
ii. b15 = 0
⇒ b = 15
Let b = 8 cm
⇒ a = (23b) cm
⇒ a = (238) cm
⇒ a = 15 cm
Now,
Area of triangle = 1/2 × base × height
= 1/2 × 8 × 15
= 60 cm^{2}
Q.7. The difference between the sides at right angles in a rightangled triangle is 7 cm. The area of the triangle is 60 cm^{2}. Finds its perimeter.
Let the sides at right angles be a and b
And, the third side be c.
Given: ab = 7 cm
Area of triangle = 60 cm^{2}
Now, since ab = 7
⇒ a = b + 7
Now we know that,
Area of triangle = 1/2 × base × height
⇒ 60 = 1/2 × b × (b + 7)
⇒ 60 × 2 = b^{2} + 7b
⇒ b^{2} + 7b = 120
⇒ b^{2 }+ 7b – 120 = 0
⇒ b^{2} + 15b  8b – 120 = 0
⇒ b(b + 15)  8(b + 15) = 0
⇒ (b + 15)(b8) = 0
This gives us two equations,
i. b – 8 = 0
⇒ b = 8
ii. b + 15 = 0
⇒ b = 15
Since, the side of the triangle cannot be negative
Therefore, b = 8 cm
⇒ a = (b + 7) cm
⇒ a = (8 + 7) cm
⇒ a = 15 cm
Now we know that,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ a^{2} + b^{2} = c^{2}
⇒ 15^{2} + 8^{2} = c^{2}
⇒ c^{2} = 225 + 64
⇒ c^{2} = 289
⇒ c = 17
Now,
Perimeter of triangle = a + b + c
⇒ Perimeter of triangle = 15 + 8 + 17
⇒ Perimeter of triangle = 40 cm
Q.8. The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangles 24 cm^{2}, find the perimeter of the triangle.
Let the sides at right angles be a and b
And, the third side be c.
Given: ab = 2 cm
Area of triangle = 24 cm^{2}
Now, since ab = 2
⇒ a = b + 2
Now we know that,
Area of triangle = 1/2 × base × height
⇒ 24 = 1/2 × b × (b + 2)
⇒ 24 × 2 = b^{2} + 2b
⇒ b^{2} + 2b = 48
⇒ b^{2 }+ 2b48 = 0
⇒ b^{2} + 8b6b48 = 0
⇒ b(b + 8)6(b + 8) = 0
⇒ (b + 8)(b6) = 0
This gives us two equations,
i. b + 8 = 0
⇒ b = 8
ii. b6 = 0
⇒ b = 6
Since, the side of the triangle cannot be negative
Therefore, b = 6 cm
⇒ a = (b + 2) cm
⇒ a = (6 + 2) cm
⇒ a = 8 cm
Now we know that,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ a^{2 }+ b^{2} = c^{2}
⇒ 8^{2 }+ 6^{2} = c^{2}
⇒ c^{2} = 64 + 36
⇒ c^{2} = 100
⇒ c = 10
Now,
Perimeter of triangle = a + b + c
⇒ Perimeter of triangle = 8 + 6 + 10
⇒ Perimeter of triangle = 24 cm
Q.9. Each side of an equilateral triangle is 10 cm. Find (i) the area of the triangle and (ii) the height of the triangle.
Given: Side of an equilateral triangle = 10 cm
(i) Area of equilateral triangle =
= 25 x 1.732
= 43.3 cm^{2}
(ii) Height of equilateral triangle =
= 5√3
= 5 x 1.732
= 8.66 cm^{2}
Q.10. The height of an equilateral triangle is 6 cm. Find its area. [Take √3 = 1.73.].
Given: Height of an equilateral triangle = 6 cm
Let sides of equilateral triangle be a cm
We know that,
Height of equilateral triangle =
⇒ 6 × 2 = √3 × a
⇒ 12 = a√3
⇒ a = 4 × 1.73
= 6.92 cm2
Now,
Area of equilateral triangle =
= 11.98√3 cm^{2}
= 20.76 cm^{2}
Q.11. If the area of an equilateral triangle is 36√3 cm^{2}, find its perimeter.
Given: Area of an equilateral triangle = 36√3 cm^{2}
We know that,
Area of equilateral triangle =
⇒ side^{2} = 36 × 4
⇒ side = 12 cm
Now,
Perimeter of equilateral triangle = 3 × side
= 3 × 12 cm
= 36 cm
Q.12. If the area of an equilateral triangle is 81√3 cm^{2}, find its height.
Given: Area of an equilateral triangle = 81√3 cm^{2}
We know that,
Area of equilateral triangle =
⇒ side^{2} = 81 × 4
⇒ side = 18 cm
Now,
Height of equilateral triangle =
= 9√3 cm
Q.13. The base of a rightangled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.
Given: Base = 48 cm
Hypotenuse = 50 cm
We know that,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ 48^{2} + Perpendicular^{2 }= 50^{2}
⇒ Perpendicular^{2} = 50^{2 } 48^{2}
⇒ Perpendicular^{2} = 2500– 2304
⇒ Perpendicular^{2} = 196 cm^{2}
⇒ Perpendicular = 14 cm
Area of a triangle = 1/2 × Base × Height
= 1/2 × 48 cm × 14 cm
= 336 cm^{2}
Q.14. The hypotenuse of a rightangled triangle is 65 cm and its base is 60 cm. Find the length of perpendicular and the area of the triangle.
Given: Base = 60 cm
Hypotenuse = 65 cm
We know that,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ 60^{2} + Perpendicular^{2} = 65^{2}
⇒ Perpendicular^{2} = 65^{2}  60^{2}
⇒ Perpendicular^{2} = 4225– 3600
⇒ Perpendicular^{2} = 625 cm^{2}
⇒ Perpendicular = 25 cm
Area of a triangle = 1/2 × Base × Height
= 1/2 × 60 cm × 25 cm
= 750 cm^{2}
Q.15. Find the area of a rightangled triangle, the radius of whose circumcircle measures 8 cm and the altitude drawn to the hypotenuse measures 6 cm.
Given: Radius of circle = 8 cm
Altitude = 6 cm
Since, in a rightangled triangle the hypotenuse is the diameter of circumcircle.
Therefore,
Hypotenuse = 2 × Radius
= 2 × 8 cm
= 16 cm
Now, we consider the hypotenuse as base and the altitude to the hypotenuse as height
So,
Area of a triangle = 1/2 × Base × Height
= 1/2 × 16 cm × 6 cm
= 1/2 × 96 cm^{2}
= 48 cm^{2}
Q.16. Find the length of the hypotenuses of an isosceles right –angled triangle whose area is 200 cm^{2}. Also, find its perimeter. [Given: √2 = 1.41.]
Given: Area = 200 cm
Let the equal sides be a.
We know that,
Area of a triangle = 1/2 × Base × Height
⇒ 200 = 1/2 × a × a
⇒ 200 = 1/2 × a^{2}
⇒ a^{2 }= 200 × 2
⇒ a^{2} = 400
⇒ a = 20 cm
Now,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ 20^{2} + 20^{2} = Hypotenuse^{2}
⇒ Hypotenuse^{2} = 400 + 400
⇒ Hypotenuse^{2} = 800 cm^{2}
⇒ Hypotenuse= 20√2 cm
⇒ Hypotenuse= 28.2 cm
Now,
Perimeter of triangle = 20 + 20 + 28.2 cm
= 68.2 cm
Q.17. The base of an isosceles triangle measures 80 cm and its area is 360 cm^{2}. Find the perimeter of the triangle.
Given: Area of isosceles triangle = 360 cm2
Base of triangle = 80 cm
Let a be the equal sides of the triangle
We know that,
Area of isosceles triangle = 1/4 × b√(4a^{2} – b^{2})
⇒ 360 = 1/4 × 80√(4a^{2} – 80^{2})
⇒ 360 = 1/4 × 80√(4a^{2} – 6400)
⇒ 360 = 20√[4(a^{2} – 1600)]
⇒ 360 = 20 × 2√(a^{2} – 1600)
⇒ 9 = √(a^{2} – 1600)
On squaring both sides we get,
⇒ 81 = a^{2} – 1600
⇒ a^{2} = 1600 + 81 = 1681
⇒ a = 41 cm
Now,
Perimeter of triangle = 41 cm + 41 cm + 80 cm
= 162 cm
Q.18. Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measure 12 cm. Find the area of the triangle.
Let height of triangle = h cm
Given: Base of the triangle (b) = 12 cm
Equal sides (a) = h + 2 cm
Now,
Area of a triangle = 1/2 × Base × Height
And,
Area of isosceles triangle = 1/4 × b√(4a^{2} – b^{2})
⇒ 1/2 × Base × Height = 1/4 × b√(4a^{2} – b^{2})
⇒ 1/2 × 12 × h = 1/4 × 12√[4(h + 2)^{2} – 12^{2}]
⇒ 6h = 3√(4h^{2} + 16h + 16144)
⇒ 2h = √(4h^{2} + 16h128)
On squaring both sides we get,
⇒ 4h^{2} = 4h^{2} + 16h – 128
⇒ 16h – 128 = 0
⇒ 16h = 128
⇒ h = 128/16
⇒ h = 8 cm
Now,
Area of a triangle = 1/2 × Base × Height
= 1/2 × 12 cm × 8 cm
= 1/2 × 96 cm^{2}
= 48 cm^{2}
Q.19. Find the area and perimeter of an isosceles right triangle, each of whose equal sides measures 10 cm. [Take √2 = 1.41.]
Given: Equal sides (i.e., base and perpendicular) = 10 cm
We know that,
Area of a triangle = 1/2 × Base × Height
Area of a triangle = 1/2 × 10 cm × 10 cm
Area of a triangle = 50 cm^{2}
Now,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ 10^{2} + 10^{2 }= Hypotenuse^{2}
⇒ Hypotenuse^{2} = 100 + 100
⇒ Hypotenuse^{2} = 200 cm^{2}
⇒ Hypotenuse= 10√2 cm
⇒ Hypotenuse= 14.1 cm
Now,
Perimeter of triangle = 10 + 10 + 14.1 cm
= 24.1 cm
Q.20. In the given figure, ΔABC is an equilateral triangle the length of whose side is equal to 10 cm, and ΔDBC is rightangled at D and BD = 8 cm. Find the area of the shaded region. [Take: √3 = 1.732.].
Given: AB = BC = AC = a (let) = 10 cm
BD = 8 cm
Now,
Area of an equilateral triangle (∆ABC) =
=
= 25√3 cm^{2}
= 43.3 cm^{2}
Now, in ∆DBC
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ DC^{2} + DB^{2} = BC^{2}
⇒ DC^{2} = BC^{2}BD^{2}
⇒ DC^{2} = 10^{2}8^{2}
⇒ DC^{2} = 10064
⇒ DC^{2 }= 36 cm^{2}
⇒ DC = 6 cm
Now,
Area of a triangle (∆DBC) = 1/2 × Base × Height
= 1/2 × DC × BC
= 1/2 × 6 cm × 8 cm
= 1/2 × 48 cm^{2}
= 24 cm^{2}
Now,
Area of shaded region = ∆ABC  ∆DBC
= 43.3 cm^{2} – 24 cm^{2}
= 19.3 cm^{2}
Q.1. The perimeter of a rectangular plot of land is 80 m and its breadth is 16 m. Find the length and area of the plot.
Given: Perimeter = 80 m
Breadth = 16 m
We know that,
Perimeter of a rectangle = 2(length + breadth)
⇒ 80 m = 2(length + 16 m)
⇒ 40m = length + 16 m
⇒ Length = 40 m – 16 m
⇒ Length = 24 m
Now,
Area of rectangle = Length × Breadth
= 24 m × 16 m
= 384 m^{2}
Q.2. The length of a rectangular park is twice its breadth and its perimeter is 840 m. Find the area of the Park.
Given: Length of park (l) = 2 × breadth(b) = 2b
Perimeter of park = 840 m
We know that,
Perimeter of a rectangle = 2(length + breadth)
⇒ 840 m = 2(2b + b)
⇒ 420 m = 3b
⇒ b = 420/3 m
⇒ b = 140m
Now,
l = 2b = 2 × 140 m = 280 m
Hence,
Area of rectangle = Length × Breadth
= 140 m × 280 m
= 39200 m^{2}
Q.3. One side of a rectangle is 12 cm long and its diagonal measures 37 cm. Find the other side and the area of the rectangle.
Given: Breadth (b) = 12 cm
Diagonal = 37 cm
Let length be l cm
We know that,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ l^{2} + 12^{2} = 37^{2}
⇒ l^{2} = 37^{2}12^{2}
⇒ l^{2} = 1369 cm^{2} – 144 cm^{2}
⇒ l^{2} = 1225 cm^{2}
⇒ l = 35 cm
Now,
Area of rectangle = Length × Breadth
= 35 cm × 12 cm
= 420 cm^{2}
Q.4. The area of a rectangular plot is 462 m^{2} and its length is 28 m. Find its perimeter.
Given: Area = 462 m^{2}
Length = 28 m
We know that,
Area of rectangle = Length × Breadth
⇒ 462 m^{2 }= 28 m × Breadth
⇒ Breadth = 16.5 m
Now,
Perimeter of a rectangle = 2(length + breadth)
= 2(28 m + 16.5 m)
= 2 × 44.5 m
= 89 m
Q.5. A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The area of the lawn is 3375 m^{2}. Find the cost of fencing the lawn at RS. 65 per metre.
Given: Cost of fencing lawn = Rs 65 per metre.
Area of lawn = 3375 m^{2}
Length: Breadth = 5: 3
Let,
Length = 5x
Breadth = 3x
We know that,
Area of lawn = Length × Breadth
⇒ 3375 m^{2} = 5x × 3x
⇒ 3375 m^{2 }= 15x^{2}
⇒ x^{2} = 225 m^{2}
⇒ x = 15 m
Therefore,
Length = 5x = 5 × 15 = 75 m
Breadth = 3x = 3 × 15 = 45 m
Now,
Perimeter of lawn = 2(length + breadth)
= 2(75 m + 45 m)
= 2 × 120 m
= 240 m
Hence, Cost of Fencing = 240 m × Rs 65 per meter = Rs 15600.
Q.6. A room is 16 m long and 13.5 m board. Find the cost of covering its floor with 75mwide carpet at RS. 60 per metre.
Given: Cost of covering = Rs 60 per metre.
Breadth of carpet = 75 cm = 0.75 m
Length of room = 16 m
Breadth of room = 13.5 m
We know that,
Area of room = Length × Breadth
= 16 m × 13.5 m
= 216 m^{2}
Now,
= 288 m
Now,
Cost of covering the floor = 288 m × Rs 60 per meter = Rs 17280.
Q.7. The floor of a rectangular hall is 24 m long and 18 m wide. How many carpets, each of length 2.5 m and breath 80 cm, will be required to cover the floor of the hall?
Given: Length of carpet = 2.5 m
Breadth of carpet = 80 cm = 0.8 m
Length of hall = 24 m
Breadth of hall = 18 m
We know that,
Area of hall = Length × Breadth
= 24 m × 18 m
= 432 m^{2}
And,
Area of carpet = Length × Breadth
= 2.5 m × 0.8 m
= 2 m^{2}
Now,
= 216 carpets
Q.8. A 36mlong, 15mbroad verandah is to be paved with stones, each measuring 6 dm by 5 dm. How many stones will be required?
Given: Length of verandah = 36 m
Breadth of verandah = 15 m
Length of stones = 6 dm = 0.6 m
Breadth of stones = 5 dm = 0.5 m
We know that,
Area of verandah = Length × Breadth
= 36 m × 15 m
= 540 m^{2}
And,
Area of stones = Length × Breadth
= 0.6 m × 0.5 m
= 0.3 m^{2}
Now,
= 1800 stones
Q.9. The area of a rectangle is 192 cm^{2} and its perimeter is 56 cm. Find the dimensions of the rectangle.
Given: Area of rectangle = 192 cm^{2}
Perimeter of rectangle = 56 cm
Let,
Length be l cm
And, breadth be b cm
Now,
Area of rectangle = Length × Breadth
⇒ 192 cm^{2} = l cm × b cm
Perimeter of rectangle = 2(length + breadth)
⇒ 56 cm = 2(l cm + b cm)
Now, substituting the value of l in this we get,
⇒ 28b = 19^{2} + b^{2}
⇒ b^{2} 28b + 192 = 0
⇒ b^{2} 16 b – 12 b + 192 = 0
⇒ b(b  16 ) – 12(b  16 ) = 0
⇒ (b  12 ) (b  16 ) = 0
This gives us two equations,
i. b – 12 = 0
⇒ b = 12
ii. b – 16 = 0
⇒ b = 16
Let b = 12 cm
Hence,
Length = 16 cm
Breadth = 12 cm
Q.10. A rectangular park 35 m long and 18 m wide is to be covered with grass, leaving 2.5 m uncovered all around it. Find the area to be laid with grass.
Given:
Length of park = 35 m
Breadth of park = 18 m
Now,
Length to be covered = 35 – (2.5 + 2.5)
= 30 m
Breadth to be covered = 18 – (2.5 + 2.5)
= 13 m
Area of park = Length × Breadth
= 30 m × 13 m
= 390 m^{2}
Q.11. A rectangular plot measures 125 m by 78 m. It has a gravel path 3 m wide all around on the outside. Find the area of the path and the cost of gravelling it at RS. 75 per m^{2}.
Length of plot = 125 m and Breadth of plot = 78 m. It has a gravel path 3 m wide all around on the outside.
The area of the path and the cost of gravelling it at RS. 75 per m^{2}.
Since gravel path is 3 m wide all around,
∴ Length of plot with path = 125 + (3 + 3)= 131 m
Breadth of plot with path = 78 + (3 + 3)= 84 m
Now,
Area of the rectangular plot without path= L × B
⇒ Area of the rectangular plot without path = 125 × 78 = 9750 m^{2}
Area of rectangular plot with path = L × B
⇒ Area of the rectangular plot with path = 131 × 84 = 11004 m^{2}
Area of the path = Area of the rectangular plot with path  Area of the rectangular plot without path
= 11004  9750= 1254 m^{2}
Cost of gravelling 1 m^{2} path = Rs 75
Cost of gravelling 1254 m^{2} path = Rs 75 × 1254 = Rs 94050
Q.12. A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m^{2}, find the width of the path.
Given:
Length of field = 54 m
Breadth of field = 35 m
Let width of the path be x m
Area of field = Length × Breadth
= 54 m × 35 m
= 1890 m^{2}
Therefore,
Length of field without path = 54  (x + x)
= 54  2x
Breadth of field without path = 35  (x + x)
= 35  2x
Therefore,
Area of field without path = Length without path × Breadth without path
= (54  2x) × (35  2x)
= 1890 – 70x – 108x + 4x^{2}
= 1890 – 178x + 4x^{2}
Now,
Area of path = Area of field  Area of field without path
⇒ 420 = 1890 – (1890 – 178x + 4x^{2})
⇒ 420 = 1890 – 1890 + 178x  4x^{2}
⇒ 420 = 178x  4x^{2}
⇒ 4x^{2}  178x + 420 = 0
⇒ 2x^{2}  89x + 210 = 0
⇒ 2x^{2}  84x – 5x + 210 = 0
⇒ 2x(x  42) – 5(x – 42) = 0
⇒ (x  42)(2x – 5) = 0
This gives us two equations,
i. x  42 = 0
⇒ x = 42
ii. 2x – 5 = 0
Since, width of park cannot be more than breadth of field
Therefore, width of park = 42 m.
Q.13. The length and the breadth of a rectangular garden are in the ratio 9 :5. A path 3.5 m wide, running all around inside it has an area of 1911 m^{2}. Find the dimensions of the garden.
Given:
Length : Breadth 9 : 5
Width of the path = 3.5 m
Area of path = 1911 m^{2}
Let,
Length of field = 9x
Breadth of field = 5x
Area of field = Length × Breadth
= 9x × 5x
= 45 x^{2}
Therefore,
Length of field without path = 9x  (3.5 + 3.5)
= 9x  7
Breadth of field without path = 5x  (3.5 + 3.5)
= 5x  7
Therefore,
Area of field without path = Length without path × Breadth without path
= (9x  7) × (5x  7)
= 45x^{2} – 35x – 63x + 49
= 45x^{2} – 98x + 49
Now,
Area of path = Area of field  Area of field without path
⇒ 1911 = 45 x^{2} – (45x^{2} – 98x + 49)
⇒ 1911 = 45 x^{2} – 45x^{2} + 98x  49
⇒ 1911 = 98x  49
⇒ 98x = 1911 + 49
⇒ 98x = 1960
⇒ x = 20
Hence,
Length of field = 9x = 9 × 20 = 180 m
Breadth of field = 5x = 5 × 20 = 100 m
Q.14. A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, finds its cost at RS. 80 per metre.
Given:
Length = 4.9 m
Breadth = 3.5 m
Margin = 25 cm = 0.25 m
Breadth of carpet = 80 cm = 0.8 m
Cost = Rs 80 per meter
Now,
Length to be carpeted = 4.9 m  (0.25 + 0.25) m
= 4.4 m
Breadth to be carpeted = 3.5 m  (0.25 + 0.25) m
= 3 m
Therefore,
Area to be carpeted = Length to be carpeted × Breadth to be carpeted
= 4.4 m × 3 m
= 13.2 m^{2}
Area of carpet = Area to be carpeted = 13.2 m^{2}
Now,
= 16.5 m
Now,
Cost of 1 m carpet = Rs 80
Therefore, Cost of 16.5 m carpet = Rs 80 × 16.5 m
= Rs 1,320
Q.15. A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the boarder is 12 m^{2} find its width.
Given:
Length = 8 m
Breadth = 5 m
Border = 12 m^{2}
Let the width be x m
Area of floor = Length × Breadth
= 8 m × 5 m
= 40 m^{2}
Now,
Length without border = 8 m  (x + x) m
= (8 – 2x) m
Breadth without border = 5 m  (x + x) m
= (5 – 2x) m
Therefore,
Area without border = Length without border × Breadth without border
= (8 – 2x) × (5 – 2x)
= 40 – 16x – 10x + 4x^{2}
Area of border = Area of floor  Area without border
⇒ 12 = 40 – (40 – 16x – 10x + 4x^{2})
⇒ 12 = 40 – 40 + 16x + 10x  4x^{2}
⇒ 12 = 26x  4x^{2}
⇒ 4x^{2} – 26x + 12 = 0
⇒ 4x^{2} – 24x – 2x + 12 = 0
⇒ 4x(x– 6) – 2(x 6) = 0
⇒ (x– 6) (4x 2) = 0
This gives us two equations,
i. x  6 = 0
⇒ x = 6
ii. 4x 2 = 0
⇒ x = 1/2
Since,
Border cannot be greater than carpet
Hence, width of border is 1/2m = 50 cm
Q.16. A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its middle, one parallel to its length and the other parallel to its breadth. Find the cost of gravelling the roads at RS. 40 per m^{2}.
A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its middle, one parallel to its length and the other parallel to its breadth.
The cost of gravelling the roads at RS. 40 per m^{2}.
Length = 80 m
Breadth = 64 m
Width of road = 5 m
Area of horizontal road = 5 m × 80 m = 400 m^{2}
Area of vertical road = 5 m × 64 m = 320 m^{2}
Area of common part to both roads = 5 m × 5 m = 25 m^{2}
Now,
Area of roads to be gravelled = Area of horizontal road + Area of vertical road  Area of common part to both roads
= 400 m^{2 }+ 320 m^{2}  25 m^{2}
= 695 m^{2}
Cost of gravelling = 695 m^{2 }× Rs 40 per m^{2}
= Rs 27800
Q.17. The dimensions of a room are 14 m x 10 m x 6.5 m. There are two doors and 4 windows in the room. Each door measures 2.5 m x 1.2 m and each window measures 1.5 m x 1m. Find the cost of painting the four walls of the room at RS. 35 per m^{2}.
Given:
Length of walls = 14 m
Breadth of walls = 10 m
Height of walls = 6.5 m
Length of windows = 1.5 m
Breadth of windows = 1 m
Length of doors = 2.5 m
Breadth of doors = 1.2 m
Cost = Rs 35 per m^{2}
Now,
Area of four walls = 2(Length of walls × Height of walls) + 2(Breadth of walls × Height of walls)
= 2(14 × 6.5) + 2(10 × 6.5)
= 182 m^{2} + 130 m^{2}
= 312 m^{2}
Area of two doors = 2(Length of doors × Breadth of doors)
= 2(2.5 × 1.2)
= 6 m^{2}
Area of four windows = 4(Length of windows × Breadth of windows)
= 4(1.5 × 1)
= 6 m^{2}
Therefore,
Area to be painted = Area of 4 walls–(Area of 2 doors + Area of 4 windows)
= 312 m^{2 }– (6 m^{2} + 6 m^{2})
= 300 m^{2}
Cost of painting = 300 m^{2 }× Rs 35 per m^{2}
= Rs 10500
Q.18. The cost of painting the four walls of a room 12 m long at RS. 30 per m^{2} is RS. 7560 and the cost of covering the floor with mat at RS. 25 per m^{2} is RS. 2700. Find the dimensions of the room.
Given:
Length = 12 m
Cost per meter = Rs 30
Total cost = Rs 7560
Cost per meter for floor = Rs 25
Total cost for floor = Rs 2700
Let height be h
Now,
= 2700/25
= 108 m^{2}
= 108/12
= 9 m
= 7560/30
= 252 m^{2}
Area of 4 walls = 2(Length of walls × Height of walls) + 2(Breadth of walls × Height of walls)
⇒ 252 = 2(12 × h) + 2(9 × h)
⇒ 252 = 24h + 18h
⇒ 252 = 42h
⇒ h = 6 m
Therefore,
Dimensions = 12 m × 9 m × 6 m
Q.19. Find the area and perimeter of a square plot of land whose diagonal is 24 m long. [ Take: √2 = 1.41.]
Given:
Diagonal = 24 m
Let the side of square be s
Area of square = 1/2 × Diagona^{l2}
= 1/2 × 24^{2}
= 288 m^{2}
Area of square = side^{2}
⇒ 288 m^{2 }= s^{2}
⇒ s = 12√2 m
⇒ s = 16.92 m
Therefore,
Perimeter of square = 4 × 16.92
= 67.68 m
Q.20. Find the length of the diagonal of a square whose area is 128 cm^{2}. Also find the perimeter.
Given:
Area = 128 cm^{2}
Let the side of square be s
Area of square = 1/2 × Diagonal^{2}
⇒ 128 = 1/2 × Diagonal2
⇒ Diagonal^{2} = 2 × 128
⇒ Diagonal^{2} = 256
⇒ Diagonal= 16 cm
Area of square = side^{2}
⇒ 128 m^{2} = s^{2}
⇒ s = 8√2 cm
⇒ s = 11.28 cm
Therefore,
Perimeter of square = 4 × 11.28
= 45.12 cm
Q.21. The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 km per hour?
Given:
Area = 8 hectares = 0.08 km^{2}
Speed = 4 km per hr
Let the side of square be s
Area of square = 1/2 × Diagonal^{2}
⇒ 0.08 = 1/2 × Diagonal^{2}
⇒ Diagonal^{2} = 2 × 0.08
⇒ Diagonal^{2} = 0.16
⇒ Diagonal= 0.04 km
= 0.01 hr
= (0.01 × 60) mins
= 6 mins
Therefore,
Time taken = 6 mins
Q.22. The cost of harvesting a square field at RS. 900 per hectare is RS. 8100. Find the cost of putting a fence around it at RS. 18 per metre.
Given:
Rate = Rs 900 per hectare
Total Cost = Rs 8100
Rate of fencing = Rs 18 per metre
Let the side of square field be s
Now,
= 8100/900
= 9 hectares = 90000 m^{2}
Area = side^{2}
⇒ 90000 m^{2} = side^{2}
⇒ side = 300 m^{2}
Now,
Perimeter = 4 × side
= 4 × 300 m^{2}
= 1200m^{2}
Therefore,
Cost of fencing = 1200 m^{2} × Rs 18 per metre
= Rs 21600
Q.23. The cost of fencing a square lawn at RS. 14 per metre is RS. 28000. Find the cost of mowing the lawn at RS. 54 per 100 m^{2}.
Given:
Rate = RS. 14 per metre
Total Cost = RS. 28000
Rate of mowing = RS. 54 per 100 m^{2}
Let the side of square field be s
Now,
= 28000/14
= 2000 m
Perimeter = 4 × side
⇒ 2000 m = 4 × s
⇒ s = 500 m
Now,
Area = side^{2}
= (500 m) ^{2}
= 250000 m^{2}
Therefore,
Cost of mowing 100 m^{2} = Rs 54
= Rs 135000
Q.24. In the given figure, ABCD is a quadrilateral in which diagonal BD = 24 cm, AL ⊥ BD and CM ⊥BD such that AL = 9 cm and CM = 12 cm. Calculate the area of the quadrilateral.
Given:
BD = 24 cm
AL = 9 cm
CM = 12 cm
In ∆ADB,
Area of ∆ADB = 1/2 × BD × AL
= 1/2 × 24 cm × 9 cm
= 108 cm^{2}
In ∆CDB,
Area of ∆CDB = 1/2 × BD × CM
= 1/2 × 24 cm × 12 cm
= 144 cm^{2}
Now,
Area of quadrilateral ABCD = Area of ∆ADB + Area of ∆ADB
= 108 cm^{2} + 144 cm^{2}
= 252 cm^{2}
Q.25. Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and ΔBCD is an equilateral triangle having each side equal to 26 cm. Also find the perimeter of the quadrilateral. [Give: √3 = 1.73.]
Given:
BC = 26 cm
DC = 26 cm
AD = 24 cm
BD = 26 cm
In ∆BCD,
= 292.37 cm^{2}
In ∆ADB,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
⇒ AB^{2} + AD^{2} = DB^{2}
⇒ AB^{2} = DB^{2}  AD^{2}
⇒ AB^{2} = 26^{2}  24^{2}
⇒ AB^{2} = 676  576
⇒ AB^{2} = 100
⇒ AB= 10 cm
Area of ∆ADB = 1/2 × AB × AD
= 1/2 × 10 cm × 24 cm
= 120 cm^{2}
Now,
Area of quadrilateral ABCD = Area of ∆ADB + Area of ∆BCD
= 120cm^{2} + 292.37 cm^{2}
= 412.37 cm^{2}
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + DA
= 10 cm + 26 cm + 26 cm + 24 cm
= 86 cm
Q.26. Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm.
Given:
AC = 15 cm
AB = 17 cm
AD = 9 cm
CD = 12 cm
In ∆ACB (rightangled),
Base^{2} + Perpendicular^{2 }= Hypotenuse^{2}
⇒ BC^{2} + AC^{2} = AB^{2}
⇒ BC^{2 }= AB^{2}  AC^{2}
⇒ BC^{2} = 17^{2}  15^{2}
⇒ BC^{2 }= 289  225
⇒ BC^{2} = 64
⇒ BC= 8 cm
Area of ∆ACB = 1/2 × BC × AC
= 1/2 × 8 cm × 15 cm
= 60 cm^{2}
In ∆ADC,
Area of ∆ADC = 1/2 × AD × CD
= 1/2 × 9 cm × 12 cm
= 54 cm^{2}
Now,
Area of quadrilateral ABCD = Area of ∆ACB + Area of ∆ADC
= 60 cm^{2 }+ 54 cm^{2}
= 114 cm^{2}
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + DA
= 17 cm + 8 cm + 12 cm + 9 cm
= 46 cm
Q.27. Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, DA = 34 cm and diagonal BD = 20 cm.
Given:
DB = 20 cm
AB = 42 cm
AD = 34 cm
CD = 29 cm
CB = 21 cm
In ∆ABD(scalene),
Area of a scalene triangle = √(s(sAB)(sBD)(sAD))
Where,
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (4842)cm × (4820)cm × (4834)cm)
= √(48 cm × 6 cm × 28 cm × 14 cm)
= √112896 cm^{2}
= 336 cm^{2}
Similarly,
In ∆BCD (scalene),
Area of a scalene triangle = √(s(sBC)(sCD)(sBD))
Where,
⇒ s = 35 cm
Now,
Area of a scalene triangle = √(35 cm × (3529)cm × (3520)cm × (3521)cm)
= √(35 cm × 6 cm × 15 cm × 14 cm)
= √44100 cm^{2}
= 210 cm^{2}
Now,
Area of quadrilateral ABCD = Area of ∆ABD + Area of ∆BCD
= 336 cm^{2} + 210 cm^{2}
= 546 cm^{2}
Q.28. Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.
Given:
Base = 25 cm
Height = 16.8 cm
Now,
Area of parallelogram = Base × Height
= 25 cm × 16.8 cm
= 420 cm^{2}
Q.29. The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance between the longer sides is 17.4 cm, find the distance between the shorter sides.
Given:
Longer side = 32 cm
Shorter side = 24 cm
Distance between Longer sides = 17.4 cm
Now,
Area of parallelogram = Longer side × Distance between Longer sides
= 32 cm × 17.4 cm
= 556.8 cm^{2}
Also,
Area of parallelogram = Shorter side × Distance between Shorter sides
⇒ 556.8 cm^{2} = 24 cm x cm
⇒ x = 23.2 cm
Hence, Distance between Shorter sides = 23.2 cm
Q.30. The area of a parallelogram is 392 m^{2}. If its altitude is twice the corresponding base, determine the base and the altitude.
Given:
Area = 392 m^{2}
Base = b (let)
Height = 2b
Now,
Area of parallelogram = Base × Height
⇒ 392 = b × 2b
⇒ 392 = 2b^{2}
⇒ b^{2} = 196
⇒ b = 14 cm
Hence,
Base = 14 cm
Altitude = 2 × 14 = 28 cm
Q.31. The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.
Given:
AB = 34 cm
BC = 20 cm
AC = 42 cm
In ∆ABC (scalene),
Area of ∆ABC = √(s(sAB)(sBC)(sAC))
Where,
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (4842)cm × (4820)cm × (4834)cm)
= √(48 cm × 6 cm × 28 cm × 14 cm)
= √112896 cm^{2}
= 336 cm^{2}
Now,
Area of parallelogram ABCD = 2 × Area of ∆ABC
= 2 × 336 cm^{2}
= 672 cm^{2}
Q.32. Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16 cm. Also, find the perimeter of the rhombus.
Given:
Length of diagonal 1 (d_{1}) = 30 cm
Length of diagonal 2 (d_{2}) = 16 cm
Area of rhombus = 1/2 × d_{1} × d_{2}
= 1/2 × 30 cm × 16 cm
= 240 cm^{2}
Now,
Side of rhombus = 1/2 × √( d_{1}^{2} + d_{2}^{2})
= 1/2 × √( 30^{2 }+ 16^{2})
= 1/2 × √(900 + 256)
= 1/2 × √1156
= 1/2 × 34
= 17 cm
Therefore,
Perimeter of rhombus = 4 × Side of rhombus
= 4 × 17 cm
= 68 cm
Q.33. The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long, find (i) the length of the other diagonal, and (ii) the area of the rhombus.
Given:
Perimeter of rhombus = 60 cm
Length of diagonal 1 (d_{1}) = 18 cm
Let, Length of diagonal 2 be d_{2}
(i) Perimeter of rhombus = 4 × side
⇒ 60 = 4 × side
Now,
Side of rhombus = 1/2 × √(d12 + d22)
⇒ 15 = 1/2 × √(18^{2} + d_{2}^{2})
⇒ 15 = 1/2 × √(324 + d_{2}^{2})
⇒ 15 × 2 = √(324 + d_{2}^{2})
⇒ 30 = √(324 + d_{2}^{2})
Squaring both sides,
⇒ 900 = 324 + d_{2}^{2}
⇒ 900324 = d_{2}^{2}
⇒ d_{2}^{2 }= 576
⇒ d_{2} = 24
Therefore,
Length of other diagonal = 24 cm
(ii) Area of rhombus = 1/2 × d_{1} × d_{2}
= 1/2 × 18 cm × 24 cm
= 216 cm^{2}
Q.34. The area of a rhombus is 480 cm^{2}, and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter.
Given:
Area of rhombus = 480 cm^{2}
Length of diagonal 1 (d_{1}) = 48 cm
Let, Length of diagonal 2 be d_{2}
(i) Area of rhombus = 1/2 × d_{1} × d_{2}
⇒ 480 = 1/2 × 48 × d_{2}
⇒ d_{2} = 20 cm
Therefore,
Length of other diagonal = 20 cm
(ii) Side of rhombus = 1/2 × √(48^{2} + 20^{2})
= 1/2 × √(2304 + 400)
= 1/2 × √2704
= 1/2 × 52
= 26 cm
Therefore,
Side of rhombus = 26 cm
(iii) Perimeter of rhombus = 4 × side
= 4 × 26 cm
= 104 cm
Therefore, Perimeter of rhombus = 104 cm
Q.35. The parallel sides of a trapezium are 12 cm and 9 cm and the distance between them is 8 cm. Find the area of the trapezium.
Given:
Side 1 = 12 cm
Side 2 = 9 cm
Distance between sides = 8 cm
Now,
Area of trapezium = 1/2 × Sum of parallel sides × Distance between them
= 1/2 × (12 + 9) × 8
= 1/2 × 21 × 8
= 84 cm^{2}
Q.36. The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide at the top 6 m wide at the bottom and the area of its cross section is 640 m^{2}, find the depth of the canal.
Given:
Top width = 10 m
Bottom width = 6 m
Area of cross section = 640 m^{2}
Let the depth be h
Now,
Area of trapezium = 1/2 × Sum of parallel sides × Distance between them
⇒ 640 = 1/2 × (10 + 6) × h
⇒ 640 × 2 = 16 h
⇒ h = 80 m
Q.37. Find the area of a trapezium whose parallel sides are 11 m and 25 m long and the nonparallel sides are 15 m and 13 m long.
Given:
AB (say) = 11 cm
DC (say) = 25 cm
AD (say) = 15 cm
BC (say) = 13 cm
Draw AE ∥ BC
Now the trapezium is divided into a triangle ADE and a parallelogram AECB.
Since, AECB is a parallelogram
Therefore, AE = BC = 13 cm
And, AB = EC
DE = DC – EC( = AB) = 25 – 11 = 14 cm
Now,
We know that,
Area of a scalene triangle (∆AED) = √(s(sAE)(sED)(sAD))
Where,
⇒ s = 21 cm
Now,
Area of a scalene triangle = √(21cm × (2113)cm × (2114)cm × (2115)cm)
= √(21cm × 8cm × 7cm × 6cm)
= √7056 cm^{2}
= 84 cm^{2}
Also,
Area of a triangle = 1/2 × base × height
⇒ 84 = 1/2 × 14 × height
⇒ height = 12 cm
Now,
Area of a parallelogram = base × height
= 11 cm × 12 cm
= 132 cm^{2}
Now,
Area of Trapezium ABCD = Area of ∆ADE + Area of a parallelogram ABCE
= 84 cm^{2} + 132 cm^{2}
= 216 cm^{2}
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