Perimeter And Area Of Plane Figures
Class 10th Mathematics RS Aggarwal Solution
- Find the area of the triangle whose base measures 24 cm and the corresponding…
- Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm. Also,…
- Find the area of the triangle whose sides are 18cm, 24cm and 30 cm. Also, find…
- The sides of a triangle are in the ratio 5 : 12 : 13, and its perimeter is 150…
- The perimeter of a triangle fields is 540 m, and its sides are in the ratio 25…
- The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm.…
- The difference between the sides at right angles in a right-angled triangle is…
- The lengths of the two sides of a right triangle containing the right angle…
- Each side of an equilateral triangle is 10 cm. Find (i) the area of the…
- The height of an equilateral triangle is 6 cm. Find its area. [Take √3 =…
- If the area of an equilateral triangle is 36√3 cm^2 , find its perimeter.…
- If the area of an equilateral triangle is 81√3 cm^2 , find its height.…
- The base of a right-angled triangle measures 48 cm and its hypotenuse measures…
- The hypotenuse of a right-angled triangle is 65 cm and its base is 60 cm. Find…
- Find the area of a right-angled triangle, the radius of whose circumcircle…
- Find the length of the hypotenuses of an isosceles right -angled triangle…
- The base of an isosceles triangle measures 80 cm and its area is 360 cm^2 .…
- Each of the equal sides of an isosceles triangle measures 2 cm more than its…
- Find the area and perimeter of an isosceles right triangle, each of whose…
- In the given figure, ΔABC is an equilateral triangle the length of whose side…
- The perimeter of a rectangular plot of land is 80 m and its breadth is 16 m.…
- The length of a rectangular park is twice its breadth and its perimeter is 840…
- One side of a rectangle is 12 cm long and its diagonal measures 37 cm. Find the…
- The area of a rectangular plot is 462 m^2 and its length is 28 m. Find its…
- A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The…
- A room is 16 m long and 13.5 m board. Find the cost of covering its floor with…
- The floor of a rectangular hall is 24 m long and 18 m wide. How many carpets,…
- A 36-m-long, 15-m-broad verandah is to be paved with stones, each measuring 6…
- The area of a rectangle is 192 cm^2 and its perimeter is 56 cm. Find the…
- A rectangular park 35 m long and 18 m wide is to be covered with grass,…
- A rectangular plot measures 125 m by 78 m. It has a gravel path 3 m wide all…
- A footpath of uniform width runs all around the inside of a rectangular field…
- The length and the breadth of a rectangular garden are in the ratio 9 :5. A…
- A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered…
- A carpet is laid on the floor of a room 8 m by 5 m. There is a border of…
- A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through…
- The dimensions of a room are 14 m x 10 m x 6.5 m. There are two doors and 4…
- The cost of painting the four walls of a room 12 m long at RS. 30 per m^2 is…
- Find the area and perimeter of a square plot of land whose diagonal is 24 m…
- Find the length of the diagonal of a square whose area is 128 cm^2 . Also find…
- The area of a square field is 8 hectares. How long would a man take to cross…
- The cost of harvesting a square field at RS. 900 per hectare is RS. 8100. Find…
- The cost of fencing a square lawn at RS. 14 per metre is RS. 28000. Find the…
- In the given figure, ABCD is a quadrilateral in which diagonal BD = 24 cm, AL…
- Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and…
- Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD…
- Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, DA =…
- Find the area of a parallelogram with base equal to 25 cm and the…
- The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance…
- The area of a parallelogram is 392 m^2 . If its altitude is twice the…
- The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the…
- Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16…
- The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long,…
- The area of a rhombus is 480 cm^2 , and one of its diagonals measures 48 cm.…
- The parallel sides of a trapezium are 12 cm and 9 cm and the distance between…
- The shape of the cross section of a canal is a trapezium. If the canal is 10 m…
- Find the area of a trapezium whose parallel sides are 11 m and 25 m long and…
- The length of a rectangular hall is 5 m more than its breadth. If the area of the hall…
- The length of a rectangular field is 23 m more than its breadth. If the perimeter of…
- The length of a rectangular field is 12 m and the length of its diagonal is 15 m. The…
- The cost of carpeting a room 15 m long with a carpet 75 cm wide, at RS. 70 per meter,…
- The length of a rectangle is thrice its breadth and the length of its diagonal is 8√10…
- On increasing the length of a rectangle by 20% and decreasing its breadth by 20%, what…
- A rectangular ground 80 m x 50 m has a path 1 m wide outside around it. The area of the…
- The length of the diagonal of a square is 10√2 cm. Its area isA. 200 cm^2 B. 100 cm^2…
- The area of a square field is 6050 m^2 . The length of its diagonal isA. 135 m B. 120 m…
- The area of a square field is 0.5 hectare. The length of its diagonal isA. 150 m B.…
- The area of an equilateral triangle is 4√3 cm^2 . Its perimeter isA. 9 cm B. 12 cm C.…
- Each side of an equilateral triangle is 8 cm. Its area isA. 24 cm^2 B. 24√3 cm^2 C.…
- Each side of an equilateral triangle is 6√3 cm. The altitude of the triangle isA. 8 cm…
- The height of an equilateral triangle is 3√3 cm . its area isA. 6√3 cm^2 B. 27 cm^2 C.…
- The base and height of a triangle are in the ratio 3 : 4 and its area is 216 cm^2 .…
- The length of the sides of a triangular field are 20 m, 21 m and 29 m. The cost of…
- The side of a square is equal to the side of an equilateral triangle. The ratio of…
- The sides of an equilateral triangle is equal to the radius of a circle whose area is…
- The area of a rhombus is 480 cm^2 and the length of one of its diagonals is 20 cm. The…
- One side of a rhombus is 20 cm long and one of its diagonals measures 24 cm. The area…
- In the given figure ABCD is quadrilateral in which ∠ABC = 90°, ∠BDC = 90°, AC = 17 cm,…
- In the given figure ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD =…
- The sides of a triangle are in the ratio 12 : 14 :25 and its perimeter is 25.5 cm. The…
- The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance between them…
- Find the area of an equilateral triangle having each side of length 10 cm. [Take √3 =…
- Find the area of an isosceles triangle each of whose equal side is 13 cm and whose base…
- The longer side of rectangular hall is 24 m and the length of its diagonal is 26 m.…
- The length of the diagonal of a square is 24 cm. Find its area.
- Find the area of a rhombus whose diagonal are 48 cm and 20 cm long.…
- Find the area of a triangle whose sides are 42 cm, 34 cm and 20 cm.…
- A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3 and its area…
- Find the area of a rhombus each side of which measurers 20 cm and one of whose…
- Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and…
- The adjacent sides of a llgm ABCD measure 34 cm and 20 cm and the diagonal AC is 42 cm…
- The cost of fencing a square lawn at RS. 14 per metre is RS. 2800. Find the cost of…
- Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm…
- A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure…
- The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the…
- The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance…
- In a four sided field, the length of the longer diagonal is 128 m. The lengths of…
Exercise 17a
Question 1.Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
Answer:
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 cm2
Question 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.
Answer:
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(s-a) (s-b) (s-c))
Where, 
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (48-42)cm × (48-34)cm × (48-20)cm)
= √(48cm × 6cm × 14cm × 28cm)
= √112896 cm2
= 336 cm2
Clearly,
Length of longest side = 42 cm
Now,
We know that,
Area of a triangle = 1/2 × Base × Height
⇒ 336 cm2 = 1/2 × 42 cm × Height
⇒ 336 cm2 = 21 cm × Height
⇒ Height = 16 cm
Question 3.
Find the area of the triangle whose sides are 18cm, 24cm and 30 cm. Also, find the height corresponding to the smallest side.
Answer:
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(s-a)(s-b)(s-c))
Where,
⇒ s = 36 cm
Now,
Area of a scalene triangle = √(36cm × (36-18)cm × (36-24)cm × (36-30)cm)
= √(36cm × 18cm × 12cm × 6cm)
= √46656 cm2
= 216 cm2
Clearly,
Length of smallest side = 18 cm
Now,
We know that,
Area of a triangle = 1/2 × Base × Height
⇒ 216 cm2= 1/2 × 18 cm × Height
⇒ 216 cm2= 9 cm × Height
⇒ Height = 24 cm
Question 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.
Answer:
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(s-a)(s-b)(s-c))
Where,
⇒ s = 75cm
Now,
Area of a scalene triangle = √(75cm × (75-25)cm × (75-60)cm × (75-65)cm)
= √(75cm × 50cm × 15cm × 10cm)
= √562500 cm2
= 750 cm2
Question 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 m2.
Answer:
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(s-a)(s-b)(s-c))
Where,
⇒ s = 270 m
Now,
Area of a scalene triangle =
√(270m × (270-250)m × (270-170)m × (270-120)m) = √(270m × 20m × 10m × 150cm)
= √81000000 m2
= 9000 m2
Now,
The cost of ploughing 100 m2 = Rs 40
Therefore, The cost of ploughing 1 m2 = Rs 
Therefore, The cost of ploughing 9000 m2 = Rs 
= Rs 3600
Question 6.
The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.
Answer:
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 = (23-b) cm
Now we know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ a2 + b2 = c2
⇒ (23-b)2 + b2 = 172
⇒ 232 + b2-46b + b2 = 289
⇒ 529 + b2-46b + b2 = 289
⇒ 2b2-46b + 240 = 0
⇒ b2-23b + 120 = 0
⇒ b2-8b-15b + 120 = 0
⇒ b(b-8)-15(b-8) = 0
⇒ (b-8)(b-15) = 0
This gives us two equations,
i. b-8 = 0
⇒ b = 8
ii. b-15 = 0
⇒ b = 15
Let b = 8 cm
⇒ a = (23-b) cm
⇒ a = (23-8) cm
⇒ a = 15 cm
Now,
Area of triangle = 1/2 × base × height
= 1/2 × 8 × 15
= 60 cm2
Question 7.
The difference between the sides at right angles in a right-angled triangle is 7 cm. The area of the triangle is 60 cm2. Finds its perimeter.
Answer:
Let the sides at right angles be a and b
And, the third side be c.
Given: a-b = 7 cm
Area of triangle = 60 cm2
Now, since a-b = 7
⇒ a = b + 7
Now we know that,
Area of triangle = 1/2 × base × height
⇒ 60 = 1/2 × b × (b + 7)
⇒ 60 × 2 = b2 + 7b
⇒ b2 + 7b = 120
⇒ b2 + 7b – 120 = 0
⇒ b2 + 15b - 8b – 120 = 0
⇒ b(b + 15) - 8(b + 15) = 0
⇒ (b + 15)(b-8) = 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,
Base2 + Perpendicular2 = Hypotenuse2
⇒ a2 + b2 = c2
⇒ 152 + 82 = c2
⇒ c2 = 225 + 64
⇒ c2 = 289
⇒ c = 17
Now,
Perimeter of triangle = a + b + c
⇒ Perimeter of triangle = 15 + 8 + 17
⇒ Perimeter of triangle = 40 cm
Question 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 cm2, find the perimeter of the triangle.
Answer:
Let the sides at right angles be a and b
And, the third side be c.
Given: a-b = 2 cm
Area of triangle = 24 cm2
Now, since a-b = 2
⇒ a = b + 2
Now we know that,
Area of triangle = 1/2 × base × height
⇒ 24 = 1/2 × b × (b + 2)
⇒ 24 × 2 = b2 + 2b
⇒ b2 + 2b = 48
⇒ b2 + 2b-48 = 0
⇒ b2 + 8b-6b-48 = 0
⇒ b(b + 8)-6(b + 8) = 0
⇒ (b + 8)(b-6) = 0
This gives us two equations,
i. b + 8 = 0
⇒ b = -8
ii. b-6 = 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,
Base2 + Perpendicular2 = Hypotenuse2
⇒ a2 + b2 = c2
⇒ 82 + 62 = c2
⇒ c2 = 64 + 36
⇒ c2 = 100
⇒ c = 10
Now,
Perimeter of triangle = a + b + c
⇒ Perimeter of triangle = 8 + 6 + 10
⇒ Perimeter of triangle = 24 cm
Question 9.
Each side of an equilateral triangle is 10 cm. Find (i) the area of the triangle and (ii) the height of the triangle.
Answer:
Given: Side of an equilateral triangle = 10 cm
(i) Area of equilateral triangle = 
(ii) Height of equilateral triangle = 
Question 10.
The height of an equilateral triangle is 6 cm. Find its area. [Take √3 = 1.73.].
Answer:
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 cm2
= 20.76 cm2
Question 11.
If the area of an equilateral triangle is 36√3 cm2, find its perimeter.
Answer:
Given: Area of an equilateral triangle = 36√3 cm2
We know that,
Area of equilateral triangle = 
⇒ side2 = 36 × 4
⇒ side = 12 cm
Now,
Perimeter of equilateral triangle = 3 × side
= 3 × 12 cm
= 36 cm
Question 12.
If the area of an equilateral triangle is 81√3 cm2, find its height.
Answer:
Given: Area of an equilateral triangle = 
cm2
We know that,
Area of equilateral triangle =
⇒ side2 = 81 × 4
⇒ side = 18 cm
Now,
Height of equilateral triangle = 
= 9√3 cm
Question 13.
The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.
Answer:
Given: Base = 48 cm
Hypotenuse = 50 cm
We know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ 482 + Perpendicular2 = 502
⇒ Perpendicular2 = 502 - 482
⇒ Perpendicular2 = 2500– 2304
⇒ Perpendicular2 = 196 cm2
⇒ Perpendicular = 14 cm
Area of a triangle = 1/2 × Base × Height
= 1/2 × 48 cm × 14 cm
= 336 cm2
Question 14.
The hypotenuse of a right-angled triangle is 65 cm and its base is 60 cm. Find the length of perpendicular and the area of the triangle.
Answer:
Given: Base = 60 cm
Hypotenuse = 65 cm
We know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ 602 + Perpendicular2 = 652
⇒ Perpendicular2 = 652 - 602
⇒ Perpendicular2 = 4225– 3600
⇒ Perpendicular2 = 625 cm2
⇒ Perpendicular = 25 cm
Area of a triangle = 1/2 × Base × Height
= 1/2 × 60 cm × 25 cm
= 750 cm2
Question 15.
Find the area of a right-angled triangle, the radius of whose circumcircle measures 8 cm and the altitude drawn to the hypotenuse measures 6 cm.
Answer:
Given: Radius of circle = 8 cm
Altitude = 6 cm
Since, in a right-angled 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 cm2
= 48 cm2
Question 16.
Find the length of the hypotenuses of an isosceles right –angled triangle whose area is 200 cm2. Also, find its perimeter. [Given: √2 = 1.41.]
Answer:
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 × a2
⇒ a2 = 200 × 2
⇒ a2 = 400
⇒ a = 20 cm
Now,
Base2 + Perpendicular2 = Hypotenuse2
⇒ 202 + 202 = Hypotenuse2
⇒ Hypotenuse2 = 400 + 400
⇒ Hypotenuse2 = 800 cm2
⇒ Hypotenuse= 20√2 cm
⇒ Hypotenuse= 28.2 cm
Now,
Perimeter of triangle = 20 + 20 + 28.2 cm
= 68.2 cm
Question 17.
The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle.
Answer:
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√(4a2 – b2)
⇒ 360 = 1/4 × 80√(4a2 – 802)
⇒ 360 = 1/4 × 80√(4a2 – 6400)
⇒ 360 = 20√[4(a2 – 1600)]
⇒ 360 = 20 × 2√(a2 – 1600)
⇒ 9 = √(a2 – 1600)
On squaring both sides we get,
⇒ 81 = a2 – 1600
⇒ a2 = 1600 + 81 = 1681
⇒ a = 41 cm
Now,
Perimeter of triangle = 41 cm + 41 cm + 80 cm
= 162 cm
Question 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.
Answer:
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√(4a2 – b2)
⇒ 1/2 × Base × Height = 1/4 × b√(4a2 – b2)
⇒ 1/2 × 12 × h = 1/4 × 12√[4(h + 2)2 – 122]
⇒ 6h = 3√(4h2 + 16h + 16-144)
⇒ 2h = √(4h2 + 16h-128)
On squaring both sides we get,
⇒ 4h2 = 4h2 + 16h – 128
⇒ 16h – 128 = 0
⇒ 16h = 128
⇒ h = 8 cm
Now,
Area of a triangle = 1/2 × Base × Height
= 1/2 × 12 cm × 8 cm
= 1/2 × 96 cm2
= 48 cm2
Question 19.
Find the area and perimeter of an isosceles right triangle, each of whose equal sides measures 10 cm. [Take √2 = 1.41.]
Answer:
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 cm2
Now,
Base2 + Perpendicular2 = Hypotenuse2
⇒ 102 + 102 = Hypotenuse2
⇒ Hypotenuse2 = 100 + 100
⇒ Hypotenuse2 = 200 cm2
⇒ Hypotenuse= 10√2 cm
⇒ Hypotenuse= 14.1 cm
Now,
Perimeter of triangle = 10 + 10 + 14.1 cm
= 24.1 cm
Question 20.
In the given figure, ΔABC is an equilateral triangle the length of whose side is equal to 10 cm, and ΔDBC is right-angled at D and BD = 8 cm. Find the area of the shaded region. [Take: √3 = 1.732.].
Answer:
Given: AB = BC = AC = a (let) = 10 cm
BD = 8 cm
Now,
Area of an equilateral triangle (∆ABC) = 
= 
= 25√3 cm2
= 43.3 cm2
Now, in ∆DBC
Base2 + Perpendicular2 = Hypotenuse2
⇒ DC2 + DB2 = BC2
⇒ DC2 = BC2-BD2
⇒ DC2 = 102-82
⇒ DC2 = 100-64
⇒ DC2 = 36 cm2
⇒ 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 cm2
= 24 cm2
Now,
Area of shaded region = ∆ABC - ∆DBC
= 43.3 cm2 – 24 cm2
= 19.3 cm2
Exercise 17b
Question 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.
Answer:
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 m2
Question 2.
The length of a rectangular park is twice its breadth and its perimeter is 840 m. Find the area of the Park.
Answer:
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 = 140m
Now,
l = 2b = 2 × 140 m = 280 m
Hence,
Area of rectangle = Length × Breadth
= 140 m × 280 m
= 39200 m2
Question 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.
Answer:
Given: Breadth (b) = 12 cm
Diagonal = 37 cm
Let length be l cm
We know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ l2 + 122 = 372
⇒ l2 = 372-122
⇒ l2 = 1369 cm2 – 144 cm2
⇒ l2 = 1225 cm2
⇒ l = 35 cm
Now,
Area of rectangle = Length × Breadth
= 35 cm × 12 cm
= 420 cm2
Question 4.
The area of a rectangular plot is 462 m2 and its length is 28 m. Find its perimeter.
Answer:
Given: Area = 462 m2
Length = 28 m
We know that,
Area of rectangle = Length × Breadth
⇒ 462 m2 = 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
Question 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 m2. Find the cost of fencing the lawn at RS. 65 per metre.
Answer:
Given: Cost of fencing lawn = Rs 65 per metre.
Area of lawn = 3375 m2
Length: Breadth = 5: 3
Let,
Length = 5x
Breadth = 3x
We know that,
Area of lawn = Length × Breadth
⇒ 3375 m2 = 5x × 3x
⇒ 3375 m2 = 15x2
⇒ x2 = 225 m2
⇒ 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
Question 6.
A room is 16 m long and 13.5 m board. Find the cost of covering its floor with 75-m-wide carpet at RS. 60 per metre.
Answer:
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 m2
Now,
= 288 m
Now,
Cost of covering the floor = 288 m × Rs 60 per meter
= Rs 17280
Question 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?
Answer:
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 m2
And,
Area of carpet = Length × Breadth
= 2.5 m × 0.8 m
= 2 m2
Now,
= 216 carpets
Question 8.
A 36-m-long, 15-m-broad verandah is to be paved with stones, each measuring 6 dm by 5 dm. How many stones will be required?
Answer:
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 m2
And,
Area of stones = Length × Breadth
= 0.6 m × 0.5 m
= 0.3 m2
Now,
= 1800 stones
Question 9.
The area of a rectangle is 192 cm2 and its perimeter is 56 cm. Find the dimensions of the rectangle.
Answer:
Given: Area of rectangle = 192 cm2
Perimeter of rectangle = 56 cm
Let,
Length be l cm
And, breadth be b cm
Now,
Area of rectangle = Length × Breadth
⇒ 192 cm2 = 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 = 192 + b2
⇒ b2- 28b + 192 = 0
⇒ b2- 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
Question 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.
Answer:
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 m2
Question 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 m2.
Answer:
Given:Length of plot = 125 m and Breadth of plot = 78 m. It has a gravel path 3 m wide all around on the outside.
To find: The area of the path and the cost of gravelling it at RS. 75 per m2.
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 m2
Area of rectangular plot with path = L × B
⇒ Area of the rectangular plot with path = 131 × 84 = 11004 m2
Area of the path = Area of the rectangular plot with path - Area of the rectangular plot without path
= 11004 - 9750
= 1254 m2
Cost of gravelling 1 m2 path = Rs 75
Cost of gravelling 1254 m2 path = Rs 75 × 1254
= Rs 94050
Question 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 m2, find the width of the path.
Answer:
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 m2
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 + 4x2
= 1890 – 178x + 4x2
Now,
Area of path = Area of field - Area of field without path
⇒ 420 = 1890 – (1890 – 178x + 4x2)
⇒ 420 = 1890 – 1890 + 178x - 4x2
⇒ 420 = 178x - 4x2
⇒ 4x2 - 178x + 420 = 0
⇒ 2x2 - 89x + 210 = 0
⇒ 2x2 - 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
Question 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 m2. Find the dimensions of the garden.
Answer:
Given:
Length : Breadth 9 : 5
Width of the path = 3.5 m
Area of path = 1911 m2
Let,
Length of field = 9x
Breadth of field = 5x
Area of field = Length × Breadth
= 9x × 5x
= 45 x2
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)
= 45x2 – 35x – 63x + 49
= 45x2 – 98x + 49
Now,
Area of path = Area of field - Area of field without path
⇒ 1911 = 45 x2 – (45x2 – 98x + 49)
⇒ 1911 = 45 x2 – 45x2 + 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
Question 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.
Answer:
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 m2
Area of carpet = Area to be carpeted = 13.2 m2
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
Question 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 m2 find its width.
Answer:
Given:
Length = 8 m
Breadth = 5 m
Border = 12 m2
Let the width be x m
Area of floor = Length × Breadth
= 8 m × 5 m
= 40 m2
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 + 4x2
Area of border = Area of floor - Area without border
⇒ 12 = 40 – (40 – 16x – 10x + 4x2)
⇒ 12 = 40 – 40 + 16x + 10x - 4x2
⇒ 12 = 26x - 4x2
⇒ 4x2 – 26x + 12 = 0
⇒ 4x2 – 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 1/2
m = 50 cm
Question 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 m2.
Answer:
Given: 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.
To find: The cost of gravelling the roads at RS. 40 per m2.
Length = 80 m
Breadth = 64 m
Width of road = 5 m
Area of horizontal road = 5 m × 80 m = 400 m2
Area of vertical road = 5 m × 64 m = 320 m2
Area of common part to both roads = 5 m × 5 m = 25 m2
Now,
Area of roads to be gravelled = Area of horizontal road + Area of vertical road - Area of common part to both roads
= 400 m2 + 320 m2 - 25 m2
= 695 m2
Cost of gravelling = 695 m2 × Rs 40 per m2
= Rs 27800
Question 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 m2.
Answer:
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 m2
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 m2 + 130 m2
= 312 m2
Area of two doors = 2(Length of doors × Breadth of doors)
= 2(2.5 × 1.2)
= 6 m2
Area of four windows = 4(Length of windows × Breadth of windows)
= 4(1.5 × 1)
= 6 m2
Therefore,
Area to be painted = Area of 4 walls–(Area of 2 doors + Area of 4 windows)
= 312 m2 – (6 m2 + 6 m2)
= 300 m2
Cost of painting = 300 m2 × Rs 35 per m2
= Rs 10500
Question 18.
The cost of painting the four walls of a room 12 m long at RS. 30 per m2 is RS. 7560 and the cost of covering the floor with mat at RS. 25 per m2 is RS. 2700. Find the dimensions of the room.
Answer:
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,
= 108 m2
= 9 m
= 252 m2
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
Question 19.
Find the area and perimeter of a square plot of land whose diagonal is 24 m long. [ Take: √2 = 1.41.]
Answer:
Given:
Diagonal = 24 m
Let the side of square be s
Area of square = 1/2 × Diagonal2
= 1/2 × 242
= 288 m2
Area of square = side2
⇒ 288 m2 = s2
⇒ s = 12√2 m
⇒ s = 16.92 m
Therefore,
Perimeter of square = 4 × 16.92
= 67.68 m
Question 20.
Find the length of the diagonal of a square whose area is 128 cm2. Also find the perimeter.
Answer:
Given:
Area = 128 cm2
Let the side of square be s
Area of square = 1/2 × Diagonal2
⇒ 128 = 1/2 × Diagonal2
⇒ Diagonal2 = 2 × 128
⇒ Diagonal2 = 256
⇒ Diagonal= 16 cm
Area of square = side2
⇒ 128 m2 = s2
⇒ s = 8√2 cm
⇒ s = 11.28 cm
Therefore,
Perimeter of square = 4 × 11.28
= 45.12 cm
Question 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?
Answer:
Given:
Area = 8 hectares = 0.08 km2
Speed = 4 km per hr
Let the side of square be s
Area of square = 1/2 × Diagonal2
⇒ 0.08 = 1/2 × Diagonal2
⇒ Diagonal2 = 2 × 0.08
⇒ Diagonal2 = 0.16
⇒ Diagonal= 0.04 km
= 0.01 hr
= (0.01 × 60) mins
= 6 mins
Therefore,
Time taken = 6 mins
Question 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.
Answer:
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,
= 9 hectares = 90000 m2
Area = side2
⇒ 90000 m2 = side2
⇒ side = 300 m2
Now,
Perimeter = 4 × side
= 4 × 300 m2
= 1200m2
Therefore,
Cost of fencing = 1200 m2 × Rs 18 per metre
= Rs 21600
Question 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 m2.
Answer:
Given:
Rate = RS. 14 per metre
Total Cost = RS. 28000
Rate of mowing = RS. 54 per 100 m2
Let the side of square field be s
Now,
= 2000 m
Perimeter = 4 × side
⇒ 2000 m = 4 × s
⇒ s = 500 m
Now,
Area = side2
= (500 m) 2
= 250000 m2
Therefore,
Cost of mowing 100 m2 = Rs 54
= Rs 135000
Question 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.
Answer:
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 cm2
In ∆CDB,
Area of ∆CDB = 1/2 × BD × CM
= 1/2 × 24 cm × 12 cm
= 144 cm2
Now,
Area of quadrilateral ABCD = Area of ∆ADB + Area of ∆ADB
= 108 cm2 + 144 cm2
= 252 cm2
Question 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.]
Answer:
Given:
BC = 26 cm
DC = 26 cm
AD = 24 cm
BD = 26 cm
In ∆BCD,
= 292.37 cm2
In ∆ADB,
Base2 + Perpendicular2 = Hypotenuse2
⇒ AB2 + AD2 = DB2
⇒ AB2 = DB2 - AD2
⇒ AB2 = 262 - 242
⇒ AB2 = 676 - 576
⇒ AB2 = 100
⇒ AB= 10 cm
Area of ∆ADB = 1/2 × AB × AD
= 1/2 × 10 cm × 24 cm
= 120 cm2
Now,
Area of quadrilateral ABCD = Area of ∆ADB + Area of ∆BCD
= 120cm2 + 292.37 cm2
= 412.37 cm2
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + DA
= 10 cm + 26 cm + 26 cm + 24 cm
= 86 cm
Question 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.
Answer:
Given:
AC = 15 cm
AB = 17 cm
AD = 9 cm
CD = 12 cm
In ∆ACB (right-angled),
Base2 + Perpendicular2 = Hypotenuse2
⇒ BC2 + AC2 = AB2
⇒ BC2 = AB2 - AC2
⇒ BC2 = 172 - 152
⇒ BC2 = 289 - 225
⇒ BC2 = 64
⇒ BC= 8 cm
Area of ∆ACB = 1/2 × BC × AC
= 1/2 × 8 cm × 15 cm
= 60 cm2
In ∆ADC,
Area of ∆ADC = 1/2 × AD × CD
= 1/2 × 9 cm × 12 cm
= 54 cm2
Now,
Area of quadrilateral ABCD = Area of ∆ACB + Area of ∆ADC
= 60 cm2 + 54 cm2
= 114 cm2
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + DA
= 17 cm + 8 cm + 12 cm + 9 cm
= 46 cm
Question 27.
Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, DA = 34 cm and diagonal BD = 20 cm.
Answer:
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(s-AB)(s-BD)(s-AD))
Where, 
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (48-42)cm × (48-20)cm × (48-34)cm)
= √(48 cm × 6 cm × 28 cm × 14 cm)
= √112896 cm2
= 336 cm2
Similarly,
In ∆BCD (scalene),
Area of a scalene triangle = √(s(s-BC)(s-CD)(s-BD))
Where, 
⇒ s = 35 cm
Now,
Area of a scalene triangle = √(35 cm × (35-29)cm × (35-20)cm × (35-21)cm)
= √(35 cm × 6 cm × 15 cm × 14 cm)
= √44100 cm2
= 210 cm2
Now,
Area of quadrilateral ABCD = Area of ∆ABD + Area of ∆BCD
= 336 cm2 + 210 cm2
= 546 cm2
Question 28.
Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.
Answer:
Given:
Base = 25 cm
Height = 16.8 cm
Now,
Area of parallelogram = Base × Height
= 25 cm × 16.8 cm
= 420 cm2
Question 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.
Answer:
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 cm2
Also,
Area of parallelogram = Shorter side × Distance between Shorter sides
⇒ 556.8 cm2 = 24 cm × x cm
⇒ x = 23.2 cm
Hence,
Distance between Shorter sides = 23.2 cm
Question 30.
The area of a parallelogram is 392 m2. If its altitude is twice the corresponding base, determine the base and the altitude.
Answer:
Given:
Area = 392 m2
Base = b (let)
Height = 2b
Now,
Area of parallelogram = Base × Height
⇒ 392 = b × 2b
⇒ 392 = 2b2
⇒ b2 = 196
⇒ b = 14 cm
Hence,
Base = 14 cm
Altitude = 2 × 14 = 28 cm
Question 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.
Answer:
Given:
AB = 34 cm
BC = 20 cm
AC = 42 cm
In ∆ABC (scalene),
Area of ∆ABC = √(s(s-AB)(s-BC)(s-AC))
Where, 
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (48-42)cm × (48-20)cm × (48-34)cm)
= √(48 cm × 6 cm × 28 cm × 14 cm)
= √112896 cm2
= 336 cm2
Now,
Area of parallelogram ABCD = 2 × Area of ∆ABC
= 2 × 336 cm2
= 672 cm2
Question 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.
Answer:
Given:
Length of diagonal 1 (d1) = 30 cm
Length of diagonal 2 (d2) = 16 cm
Area of rhombus = 1/2 × d1 × d2
= 1/2 × 30 cm × 16 cm
= 240 cm2
Now,
Side of rhombus = 1/2 × √( d12 + d22)
= 1/2 × √( 302 + 162)
= 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
Question 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.
Answer:
Given:
Perimeter of rhombus = 60 cm
Length of diagonal 1 (d1) = 18 cm
Let, Length of diagonal 2 be d2
(i) Perimeter of rhombus = 4 × side
⇒ 60 = 4 × side
Now,
Side of rhombus = 1/2 × √(d12 + d22)
⇒ 15 = 1/2 × √(182 + d22)
⇒ 15 = 1/2 × √(324 + d22)
⇒ 15 × 2 = √(324 + d22)
⇒ 30 = √(324 + d22)
Squaring both sides,
⇒ 900 = 324 + d22
⇒ 900-324 = d22
⇒ d22 = 576
⇒ d2 = 24
Therefore,
Length of other diagonal = 24 cm
(ii) Area of rhombus = 1/2 × d1 × d2
= 1/2 × 18 cm × 24 cm
= 216 cm2
Question 34.
The area of a rhombus is 480 cm2, 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.
Answer:
Given:
Area of rhombus = 480 cm2
Length of diagonal 1 (d1) = 48 cm
Let, Length of diagonal 2 be d2
(i) Area of rhombus = 1/2 × d1 × d2
⇒ 480 = 1/2 × 48 × d2
⇒ d2 = 20 cm
Therefore,
Length of other diagonal = 20 cm
(ii) Side of rhombus = 1/2 × √(482 + 202)
= 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
Question 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.
Answer:
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 cm2
Question 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 m2, find the depth of the canal.
Answer:
Given:
Top width = 10 m
Bottom width = 6 m
Area of cross section = 640 m2
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
Question 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.
Answer:
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(s-AE)(s-ED)(s-AD))
Where, 
⇒ s = 21 cm
Now,
Area of a scalene triangle = √(21cm × (21-13)cm × (21-14)cm × (21-15)cm)
= √(21cm × 8cm × 7cm × 6cm)
= √7056 cm2
= 84 cm2
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 cm2
Now,
Area of Trapezium ABCD = Area of ∆ADE + Area of a parallelogram ABCE
= 84 cm2 + 132 cm2
= 216 cm2
Multiple Choice Questions (mcq)
Question 1.The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 750 m2 then its length is
A.15 m
B. 20 m
C. 25 m
D. 30 m
Answer:
Given: Length of hall (l) = 5 + breadth(b) = 5 + b
Area of hall = 750 m2
We know that,
Area of rectangle = Length × Breadth
⇒ 750 = (5 + b) × b
⇒ 750 = b2 + 5b
⇒ b2 + 5b – 750 = 0
⇒ b2 + 30b – 25b – 750 = 0
⇒ b(b + 30) – 25(b + 30) = 0
⇒ (b + 30) (b – 25) = 0
This gives us two equations,
i. b + 30 = 0
⇒ b = -30
ii. b – 25 = 0
⇒ b = 25
Since, the length of the rectangle cannot be negative
Therefore, b = 25 m
⇒ l = (b + 5) m
⇒ l = (25 + 5) m
⇒ l = 30 m
Question 2.
The length of a rectangular field is 23 m more than its breadth. If the perimeter of the field is 206m, then its area is
A. 2420 m2
B. 2520m2
C. 2480m2
D. 2620m2
Answer:
Given: Length of field (l) = 23 + breadth(b) = 23 + b
Perimeter of field = 206 m
We know that,
Perimeter = 2(l + b)
⇒ 206 = 2(23 + b + b)
⇒ 206 = 2(23 + 2b)
⇒ 206 = 46 + 4b
⇒ 4b = 206 – 46
⇒ 4b = 160
⇒ b = 40 m
Therefore,
Length of field = 23 + b
= 23 + 40
= 63 m
Now,
Area of rectangle = Length × Breadth
= 63 × 40
= 2520 m2
Question 3.
The length of a rectangular field is 12 m and the length of its diagonal is 15 m. The area of the field is
A. 108 m2
B. 180 m2
C. 30√3 m2
D. 12√15 m2
Answer:
Given:
Length = 12 m
Length of diagonal = 15 m
We know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ 122 + Perpendicular2 = 152
⇒ Perpendicular2 = 152 – 122
⇒ Perpendicular2 = 225– 144
⇒ Perpendicular2 = 81
⇒ Perpendicular2 = 9
That is,
Breadth = 9 m
Now,
Area = Length × Breadth
= 12 m × 9 m
= 108 m2
Question 4.
The cost of carpeting a room 15 m long with a carpet 75 cm wide, at RS. 70 per meter, is RS. 8400. The width of the room is
A. 9 m
B. 8 m
C. 6 m
D. 12 m
Answer:
Given:
Length of room = 15 m
Width of carpet = 75 cm = 0.75 m
Rate = Rs 70
Total cost = Rs 8400
Now,
= 120 m
Therefore,
Area of carpet = Length of carpet × Width of carpet
= 120 m × 0.75 m = 90 m2
We know that,
Area of room = Area of carpet = 90 m2
Now,
Area of room = Length of room × Width of room
⇒ 90 m2 = 15 m × Width of room
⇒ Width of room = 6 m
Question 5.
The length of a rectangle is thrice its breadth and the length of its diagonal is 8√10 cm. The perimeter of the rectangle is
A. 15√10 cm
B. 16√10 cm
C. 24√10 cm
D. 64 cm
Answer:
Given: Length of rectangle (l) = 3 × breadth(b) = 3b
Diagonal of rectangle = 8√10 m
We know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ b2 + (3b)2 = (8√10) 2
⇒ b2 + 9b2 = 640
⇒ 10b2 = 640
⇒ b2 = 64
⇒ b = 8 cm
Therefore,
l = 3b = 24 cm
Hence,
Perimeter of a rectangle = 2(length + breadth)
= 2(24 + 8)
= 64 cm
Question 6.
On increasing the length of a rectangle by 20% and decreasing its breadth by 20%, what is the change in its area?
A. 20% increase
B. 20% decrease
B. No change
D. 4 % decrease
Answer:
Let the length be l
And, breadth be b
Now,
Area = l × b = lb
Increase in length = 20% of length + length
Decrease in breadth = breadth - 20% of breadth
Since,
Therefore,
The area is decreased
= 4%
Hence change in area = 4% decrease
Question 7.
A rectangular ground 80 m x 50 m has a path 1 m wide outside around it. The area of the path is
A. 264 m2
B. 284 m2
C. 400 m2
D. 464 m2
Answer:
Given:
Length = 80 m
Breadth = 50 m
Width of the path = 1m
Area of path = 1911 m2
Length of field with path = 80 + (1 + 1)
= 82 m
Breadth of field with path = 50 + (1 + 1)
= 52 m
Area of field with path = Length of field with path × Breadth of field with path
= 82 m × 52 m
= 4264 m2
Area of field without path = Length without path × Breadth without path
= 80 m × 50 m
= 4000 m2
Now,
Area of path = Area of field - Area of field without path
= 4264 m2 – 4000 m2
= 264 m2
Question 8.
The length of the diagonal of a square is 10√2 cm. Its area is
A. 200 cm2
B. 100 cm2
C. 150 cm2
D. 100√2 cm2
Answer:
Given:
Length of diagonal = 10√2 cm
Let the side of square = x cm
We know that,
Hypotenuse2 = Base2 + Perpendicular2
⇒ (10√2)2 = x2 + x2
⇒ 200 = 2x2
⇒ x2 = 100
⇒ x = 10 cm
Now,
Area of a square = side2
= (10 cm)2
= 100 cm2
Question 9.
The area of a square field is 6050 m2. The length of its diagonal is
A. 135 m
B. 120 m
C. 112 m
D. 110 m
Answer:
Given:
Area of square field = 6050 m2
Let the side of square = x m
We know that,
Area of a square = side2
⇒ 6050 = x2
⇒ x = 55√2
Now,
Hypotenuse2 = Base2 + Perpendicular2
= (55√2)2 + (55√2)2
= 6050 + 6050
= 12100 m2
Therefore,
Diagonal = √12100
= 110 m
Question 10.
The area of a square field is 0.5 hectare. The length of its diagonal is
A. 150 m
B. 100√2 m
C. 100 m
D. 50√2 cm2
Answer:
Given:
Area of square field = 0.5 hectare = 5000 m2
Let the side of square = x m
We know that,
Area of a square = side2
⇒ 5000 = x2
⇒ x = 50√2
Now,
Hypotenuse2 = Base2 + Perpendicular2
= (50√2)2 + (50√2)2
= 5000 + 5000
= 10000 m2
Therefore,
Diagonal = √10000
= 100 m
Question 11.
The area of an equilateral triangle is 4√3 cm2. Its perimeter is
A. 9 cm
B. 12 cm
C. 12√3 cm
D. 6√3 cm
Answer:
Given:
Area of equilateral triangle = 4√3 cm2
We know that,
Area of equilateral triangle =
⇒ side2 = 4 × 4
⇒ side2 = 16
⇒ side = 4 cm
Now,
Perimeter of triangle = 3 × side
= 3 × 4 cm
= 12 cm
Question 12.
Each side of an equilateral triangle is 8 cm. Its area is
A. 24 cm2
B. 24√3 cm2
C. 16√3 cm2
D. 8√3 cm2
Answer:
Given:
Side of equilateral triangle = 8 cm
We know that,
Area of equilateral triangle =
= 16√3 cm2
Question 13.
Each side of an equilateral triangle is 6√3 cm. The altitude of the triangle is
A. 8 cm
B. 9 cm
C. 3√3 cm
D. 6 cm
Answer:
Given: Side of an equilateral triangle = 6√3 cm
Height of equilateral triangle =
= 9 cm2
Question 14.
The height of an equilateral triangle is 3√3 cm . its area is
A. 6√3 cm2
B. 27 cm2
C. 9√3 cm2
D. 27√3 cm2
Answer:
Given:
Height of equilateral triangle = 3√3 cm
We know that,
Height of equilateral triangle =
⇒ Side = 6 cm
Now,
Area of equilateral triangle =
= 9√3 cm2
Question 15.
The base and height of a triangle are in the ratio 3 : 4 and its area is 216 cm2. The height of the triangle is
A. 18 cm
B. 24 cm
C. 21 cm
D. 28 cm
Answer:
Given:
Base: Height = 3: 4
Area = 216 cm2
Let,
Base = 3x
Height = 4x
We know that,
Area of a triangle = 1/2 × base × height
⇒ 216 = 1/2 × 3x × 4x
⇒ 216 × 2 = 12x2
⇒ 12 x2 = 432
⇒ x2 = 36
⇒ x = 6 cm
Therefore,
Height = 4x
= 24 cm
Question 16.
The length of the sides of a triangular field are 20 m, 21 m and 29 m. The cost of cultivating the field at RS. 9 per m2 is
A. RS. 2610
B. 3780
C. RS. 1890
D. 1800
Answer:
Given:
Rate = Rs 9 per m2
Side a = 20 m
Side b = 21 m
Side c = 29 m
Area of a scalene triangle = √(s(s-a)(s-b)(s-c))
Where,
⇒ s = 35 m
Now,
Area of triangular field = √(35 m × (35-20)m × (35-21)m × (35-29)m)
= √(35 m × 15 m × 14 m × 6 m)
= √44100 m2
= 210 m2
Total Cost = 210 m2 × Rs 9 per m2
= Rs 1890
Question 17.
The side of a square is equal to the side of an equilateral triangle. The ratio of their areas is
A. 4 : 3
B. 2 : √3
C. 4 :√3
D. None of these
Answer:
Let the side be x
Now,
Area of equilateral triangle =
And,
Area of square = side2
= x2
Therefore, the ratio is 4:√3
Question 18.
The sides of an equilateral triangle is equal to the radius of a circle whose area is 154 cm2. The area of the triangle is
A. 49 cm2
B. 
C. 
D. 77 cm2
Answer:
Let the side = radius = x
Now,
Area of circle = πr2
⇒ x2 = 49
⇒ x = 7 cm
Area of equilateral triangle =
Question 19.
The area of a rhombus is 480 cm2 and the length of one of its diagonals is 20 cm. The length of each side of the rhombus is
A. 24 cm
B. 30 cm
C. 26 cm
D. 28 cm
Answer:
Given:
Area of rhombus = 480 cm2
Length of diagonal 1 (d1) = 20 cm
Let, Length of diagonal 2 be d2
Area of rhombus = 1/2 × d1 × d2
⇒ 480 = 1/2 × 20 × d2
⇒ d2 = 48 cm
Now,
Side of rhombus = 1/2 × √(482 + 202)
= 1/2 × √(2304 + 400)
= 1/2 × √2704
= 1/2 × 52
= 26 cm
Question 20.
One side of a rhombus is 20 cm long and one of its diagonals measures 24 cm. The area of the rhombus is
A. 192 cm2
B. 480 cm2
C. 240 cm2
D. 384 cm2
Answer:
Given:
Side = 24 cm
Length of diagonal 1 (d1) = 20 cm
Let, Length of diagonal 2 be d2
We know that,
Side of rhombus = 1/2 × √(d12 + d22)
⇒ 20 = 1/2 × √(242 + d22)
⇒ 20 × 2 = √(576 + d22)
⇒ 40 = √(576 + d22)
Squaring both sides,
⇒ 1600 = 576 + d22
⇒ d22 = 1024
⇒ d2 = 32 cm
Now,
Area of rhombus = 1/2 × d1 × d2
= 1/2 × 24 × 32
= 384 cm2
Formative Assessment (unit Test)
Question 1.In the given figure ABCD is quadrilateral in which ∠ABC = 90°, ∠BDC = 90°, AC = 17 cm, BC = 15 cm, BD = 12 cm and CD = 9 cm. The area of quad ABCD is
A. 102 cm2
B. 114 cm2
C. 95 cm2
D. 57 cm2
Answer:
Given:
AC = 17 cm
BC = 15 cm
BD = 12 cm
CD = 9 cm.
∠ABC = 90°
∠BDC = 90°
In ∆ABC,
Using Pythagoras theorem,
AB2 + BC2 = AC2
⇒ AB2 = AC2- BC2
⇒ AB = √( AC2- BC2)
⇒ AB = √( 172- 152)
⇒ AB = √(289-225)
⇒ AB = √64
⇒ AB = 8 cm
Therefore,
Area of ∆ABC = 1/2 × AB × BC
= 1/2 × 8 × 15
= 60 cm2
And,
In ∆BDC,
Area of ∆BDC = 1/2 × BD × DC
= 1/2 × × 12 × 9
= 54 cm2
Therefore,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆BDC
= 60 cm2 + 54 cm2
= 114 cm2
Question 2.
In the given figure ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD = 9 m and CE ⊥ AB. Area of trap. ABCD is
(A) 306 m2
(B) 316 m2
(C) 296 m2
(D) 284 m2
A.306 m2
B. 316 m2
C. 296 m2
D. 284 m2
Answer:
Given: ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD = 9 m and CE ⊥ AB.
To find: Area of trap. ABCD
Solution:
AB = 40 m, BC = 15 m, AD = 9 m and CD = 28 m.
In trapezium ABCD,
Area of trapezium = 1/2 × sum of parallel sides × distance between them
= 1/2 × (28 + 40) × 9
= 1/2 × 68 × 9
= 306 m2
Question 3.
The sides of a triangle are in the ratio 12 : 14 :25 and its perimeter is 25.5 cm. The largest side of the triangle is
A. 7 cm
B. 14 cm
C. 12.5 cm
D. 18 cm
Answer:
Given: Ratio of Sides = 12: 14: 25
Perimeter = 25.5 cm
Let the sides be,
a = 12x cm
b = 14x cm
c = 25x cm
We know that,
Perimeter of a triangle = a + b + c
⇒ 25.5 cm = 12x cm + 14x cm + 25x cm
⇒ 25.5 cm = 51x cm
⇒ x = 0. 5
Therefore,
a = 12x cm = 12 × 0.5 cm = 6 cm
b = 14x cm = 14 × 0.5 cm = 7 cm
c = 25x cm = 25 × 0.5 cm = 12.5 cm
Clearly largest side is c = 12.5 cm
Question 4.
The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. The area of the trapezium is
A. 104 cm2
B. 78 cm2
C. 52 cm2
D. 65 cm2
Answer:
Given:
Side 1 = 9.7 cm
Side 2 = 6.3 cm
Distance between sides = 6.5 cm
Area of trapezium = 1/2 × sum of parallel sides × distance between them
= 1/2 × (9.7 + 6.3) × 6.5
= 1/2 × 16 × 6.5
= 52 cm2
Question 5.
Find the area of an equilateral triangle having each side of length 10 cm. [Take √3 = 1.732.]
Answer:
Given:
Side of an equilateral triangle = 10 cm
Area of equilateral triangle =
= 25√3
= 25 × 1.732
= 43.3 cm2
Question 6.
Find the area of an isosceles triangle each of whose equal side is 13 cm and whose base is 24 cm.
Answer:
Given:
Side AB = Side AC = 13 cm
Base = 24 cm
In ∆ADC (right-angled),
DC = 12 cm
By Pythagoras theorem,
AD2 + DC2 = AC2
⇒ AD2 = AC2 - DC2
⇒ AD2 = 132 - 122
⇒ AD2 = 169 – 144 = 25
⇒ AD = 5 cm
Now,
Area of triangle = 1/2 × base × height
= 1/2 × BC × AD
= 1/2 × 24 × 5
= 60 cm2
Question 7.
The longer side of rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall.
Answer:
Given:
Length (l) = 24 m
Diagonal = 26 m
Let breadth be b
We know that,
Base2 + Perpendicular2 = Hypotenuse2
⇒ 242 + b2 = 262
⇒ b2 = 262 - 242
⇒ b2 = 676 - 576 = 100
⇒ b= 10 m
Area of rectangle = Length × Breadth
= 24 m × 10 m
= 240 m2
Question 8.
The length of the diagonal of a square is 24 cm. Find its area.
Answer:
Given:
Length of diagonal = 24 cm
Let the side of square = x cm
We know that,
Hypotenuse2 = Base2 + Perpendicular2
⇒ 242 = x2 + x2
⇒ 576 = 2x2
⇒ x2 = 288
⇒ x = 12√2 cm
Now,
Area of a square = side2
= (12√2 cm)2
= 288 cm2
Question 9.
Find the area of a rhombus whose diagonal are 48 cm and 20 cm long.
Answer:
Given:
Length of diagonal 1 (d1) = 48 cm
Length of diagonal 2 (d2) = 20 cm
Area of rhombus = 1/2 × d1 × d2
= 1/2 × 48 cm × 20 cm
= 480 cm2
Question 10.
Find the area of a triangle whose sides are 42 cm, 34 cm and 20 cm.
Answer:
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(s-a)(s-b)(s-c))
Where,
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (48-42)cm × (48-34)cm × (48-20)cm)
= √(48cm × 6cm × 14cm × 28cm)
= √112896 cm2
= 336 cm2
Question 11.
A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3 and its area is 3375 m2. Find the cost of fencing the lawn at RS. 20 per metre.
Answer:
Given: Cost of fencing lawn = Rs 20 per metre.
Area of lawn = 3375 m2
Length : Breadth = 5:3
Let,
Length = 5x
Breadth = 3x
We know that,
Area of lawn = Length × Breadth
⇒ 3375 m2 = 5x × 3x
⇒ 3375 m2 = 15x2
⇒ x2 = 225 m2
⇒ 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 20 per meter
= Rs 4800
Question 12.
Find the area of a rhombus each side of which measurers 20 cm and one of whose diagonals is 24 cm.
Answer:
Given:
Length of diagonal 1 (d1) = 24 cm
Side = 20 cm
Let, Length of diagonal 2 be d2
We know that,
Side of rhombus = 1/2 × √(d12 + d22 )
⇒ 20 = 1/2 × √(242 + d22 )
⇒ 20 × 2 = √(576 + d22 )
⇒ 40 = √(576 + d22)
Squaring both sides,
⇒ 1600 = 576 + d22
⇒ d22 = 1600-576
⇒ d22 = 1024
⇒ d2 = 32 cm
Now,
Area of rhombus= 1/2 × d1 × d2
= 1/2 × 24 × 32
= 384 cm2
Question 13.
Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and non-parallel sides are 15 cm and 13 cm.
Answer:
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(s-AE)(s-ED)(s-AD))
Where,
⇒ s = 21 cm
Now,
Area of a scalene triangle = √(21cm × (21-13)cm × (21-14)cm × (21-15)cm)
= √(21cm × 8cm × 7cm × 6cm)
= √7056 cm2
= 84 cm2
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 cm2
Now,
Area of Trapezium ABCD = Area of ∆ADE + Area of a parallelogram ABCE
= 84 cm2 + 132 cm2
= 216 cm2
Question 14.
The adjacent sides of a llgm ABCD measure 34 cm and 20 cm and the diagonal AC is 42 cm long. Find the area of the ll gm.
Answer:
Given:
AB = 34 cm
BC = 20 cm
AC = 42 cm
The diagonal of a parallelogram divides it into two equal triangles.
Therefore,
Area of ABCD = 2 × Area of ∆ABC
Now,
We know that,
Area of a scalene triangle = √(s(s-AC)(s-AB)(s-BC))
Where, 
⇒ s = 48cm
Now,
Area of a scalene triangle = √(48cm × (48-42)cm × (48-34)cm × (48-20)cm)
= √(48cm × 6cm × 14cm × 28cm)
= √112896 cm2
= 336 cm2
Therefore,
Area of ABCD = 2 × 336 cm2
= 672 cm2
Question 15.
The cost of fencing a square lawn at RS. 14 per metre is RS. 2800. Find the cost of mowing the lawn at RS. 54 per 100m2.
Answer:
Given:
Rate = RS. 14 per metre
Total Cost = RS. 2800
Rate of mowing = RS. 54 per 100 m2
Let the side of square field be s
Now,
= 200 m
Perimeter = 4 × side
⇒ 200 m = 4 × s
⇒ s = 50 m
Now,
Area = side2
= (50 m) 2
= 2500 m2
Therefore,
Cost of mowing 100 m2 = Rs 54
= Rs 1350
Question 16.
Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diag. BD = 20 cm.
Answer:
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(s-AB)(s-BD)(s-AD))
Where,
⇒ s = 48 cm
Now,
Area of a scalene triangle = √(48cm × (48-42)cm × (48-20)cm × (48-34)cm)
= √(48 cm × 6 cm × 28 cm × 14 cm)
= √112896 cm2
= 336 cm2
Similarly,
In ∆BCD (scalene),
Area of a scalene triangle = √(s(s-BC)(s-CD)(s-BD))
Where, 
⇒ s = 35 cm
Now,
Area of a scalene triangle = √(35 cm × (35-29)cm × (35-20)cm × (35-21)cm)
= √(35 cm × 6 cm × 15 cm × 14 cm)
= √44100 cm2
= 210 cm2
Now,
Area of quadrilateral ABCD = Area of ∆ABD + Area of ∆BCD
= 336 cm2 + 210 cm2
= 546 cm2
Question 17.
A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one if the sides of the ll gm is 66 m long, find its corresponding altitude.
Answer:
Given:
Diagonal 1 (d1) of rhombus = 120 m
Diagonal 2 (d2) of rhombus = 44 m
Side of parallelogram = 66 m
Area of rhombus = 1/2 × d1 × d2
= 1/2 × 120 m × 44 m
= 2640 m2
Now,
Area of parallelogram = Base × Height
⇒ 2640 m2 = 66 m × Height
⇒ Height = 40 m
Question 18.
The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the rhombus.
Answer:
Given:
Length of diagonal 1 (d1) = 48cm
Length of diagonal 2 (d2) = 20 cm
Side of rhombus = 1/2 × √(d12 + d22 )
= 1/2 × √(482 + 202 )
= 1/2 × √(2304 + 400)
= 1/2 × √2704
= 1/2 × 52
= 26 cm
Therefore,
Perimeter of rhombus = 4 × Side of rhombus
= 4 × 26 cm
= 104 cm
Question 19.
The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, find the distance between the longer sides.
Answer:
Given:
Longer side = 36 cm
Shorter side = 27 cm
Distance between Shorter sides = 12 cm
Let, Distance between Longer sides = x cm
Now,
Area of parallelogram = Shorter Side × Distance between Longer sides
= 27 cm × 12 cm
= 324 cm2
Also,
Area of parallelogram = Longer side × Distance between Longer sides
⇒ 324 cm2 = 36 cm × x cm
⇒ x = 9 cm
Hence,
Distance between Shorter sides = 9 cm
Question 20.
In a four sided field, the length of the longer diagonal is 128 m. The lengths of perpendiculars from the opposite vertices upon this diagonal are 22.7 m and 17.3 m. Find the area of the field.
Answer:
Given:
BD = 128 m
CF = 22.7 m
AE = 17.3 m
Now,
In ∆ABD,
Area of a triangle = 1/2 × base × height
= 1/2 × BD × AE
= 1/2 × 128 × 17.3
= 1107.2 m2
Similarly,
In ∆CBD,
Area of a triangle = 1/2 × base × height
= 1/2 × BD × FC
= 1/2 × 128 × 22.7
= 1452.8 m2
Now,
Area of field = ∆ABD + ∆CBD
= 1107.2 m2 + 1452.8 m2
= 2560 m2