ICSE

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Objective Type Questions

Mental Maths

Question 1.
Fill in the blanks:
(i) The simplest form of the ratio \(\frac { 1 }{ 6 }\) : \(\frac { 1 }{ 4 }\) is ………
(ii) 75 cm : 1.25 m = ……..
(iii) If two ratios are equivalent, then the four quantities are said to be in ………
(iv) If 8, x, 48 and 18 are in proportion then the value of x is ………
(v) If the cost of 10 pencils is ₹ 15, then the cost of 6 pencils is ……..
(vi) If a cyclist is travelling at a speed of 15 km/h, then the distance covered by him in 20 minutes is ……..
Solution:
(i) The simplest form of the ratio \(\frac { 1 }{ 6 }\) : \(\frac { 1 }{ 4 }\) is 2 : 3.
(ii) 75 cm : 1.25 m = 3 : 5
(iii) If two ratios are equivalent,
then the four quantities are said to be in proportion.
(iv) If 8, x, 48 and 18 are in proportion then the value of x is
8 : x :: 48 : 18
x = 3
(v) If the cost of 10 pencils is ₹ 15, then the cost of 6 pencils is
Cost of 10 pencils = ₹ 15
Let cot of 6 pencils = ₹ x
10 : 6 15 : x
10x = 6 × 5
x = 3
Cost of 6 pencils = ₹ 3
(vi) If a cyclist is travelling at a speed of 15 km/ h,
then the distance covered by him in 20 minutes is ……
Speed of a cyclist = 15 km/h
Distance travelled in 20 minutes = \(\frac { 20 }{ 60 }\) × 15 km = 5 km

Question 2.
State whether the following statements are true (T) or false (F):
(i) A ratio is always greater than 1.
(ii) Ratio of half an hour to 20 seconds is 30 : 20.
(iii) The ratio 5 : 7 is greater than the ratio 5 : 6.
(iv) If the numbers 3, 5, 12 and x are in proportion then the value of x is 20.
(v) The ratio 3 : 2 and 4 : 5 are equivalent.
Solution:
(i) A ratio is always greater than 1. (False)
Correct:
Ratio can be less than 1 or equal to 1.
It is not necessary that it is greater than 1.
(ii) Ratio of half an hour to 20 seconds is 30 : 20. (False)
Correct:
= 30 × 60 : 20 = 1800 : 20 = 90 : 1
(iii) The ratio 5 : 7 is greater than the ratio 5 : 6. (False)
Correct:
\(\frac { 5 }{ 7 }\), \(\frac { 5 }{ 6 }\)
\(\frac { 30:35 }{ 42 }\)
30 < 35
5 : 7 is not greater than the ratio 5 : 6
(iv) If the numbers 3, 5, 12 and x are in proportion
then the value of x is 20. (True)
3, 5, 12 and x are in proportion then
x × 3 = 5 × 12
x = 20
(v) The ratio 3 : 2 and 4 : 5 are equivalent. (False)
Correct:
\(\frac { 3 }{ 2 }\) ≠ \(\frac { 4 }{ 5 }\) as 15 ≠ 8

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 14):
Question 3.
A ratio equivalent to 6 : 10 is
(a) 3 : 4
(b) 18 : 30
(c) 12 : 40
(d) 5 : 3
Solution:
6 : 10 = 3 : 5
18 : 30 = 3 : 5
6 : 40 is equivalent to 18 : 30 (b)

Question 4.
A ratio equivalent to the ratio \(\frac { 2 }{ 3 }\) : \(\frac { 3 }{ 4 }\) is
(a) 4 : 6
(b) 8 : 9
(c) 6 : 8
(d) 9 : 8
Solution:
\(\frac { 2 }{ 3 }\) : \(\frac { 3 }{ 4 }\)
\(\frac { 8:9 }{ 12 }\)
= 8 : 9 (b)

Question 5.
The ratio of 75 mL to 3 litres is
(a) 25 : 1
(b) 40 : 1
(c) 1 : 40
(d) 3 : 200
Solution:
75 mL to 3 litres
= 75 : 3000
= 1 : 40 (c)

Question 6.
The ratio of the number of sides of a rectangle to the number of edges of a cuboid is
(a) 1 : 2
(b) 1 : 3
(c) 2 : 3
(d) none of these
Solution:
Ratio of number of sides of a rectangle
to the number of edges of a cuboid = 4 : 12 = 1 : 3 (b)

Question 7.
In a class of 35 students, the number of girls is 20. The ratio of number of boys to the number of girls in the class is
(a) 3 : 4
(b) 4 : 3
(c) 7 : 4
(d) 7 : 3
Solution:
Total number of students = 35
No. of girls = 20
No. of boys = 35 – 20 = 15
Ratio in boys and girls = 15 : 20 = 3 : 4 (a)

Question 8.
The ratio of number of girls to the number of boys in a class is 6 : 7. If there are 21 boys in the class, then the number of girls in the class is
(a) 39
(b) 24
(c) 18
(d) 13
Solution:
Girls : boys = 6 : 7
No. of boys = 21
Let no. of girls = x
6 : 7 :: x : 21
7 × x = 6 × 21
x = 18
No. of boys = 18 (c)

Question 9.
Two numbers are in the ratio 3 : 5. If the sum of the numbers is 144, then the largest number is
(a) 48
(b) 54
(c) 72
(d) 90
Solution:
Ratio between two numbers = 3 : 5
Sum of numbers 144
Then larger number = \(\frac { 144 }{ 3+5 }\) × 5
= \(\frac { 144 }{ 8 }\) × 5 = 90 (d)

Question 10.
If x, 12, 8 and 32 are in proportion, then x is
(a) 6
(b) 4
(c) 3
(d) 2
Solution:
x, 12, 8 and 32 are in proportion, then
x × 32 = 12 × 8
x = 3 (c)

Question 11.
If 3, 12 and x are in continued proportion, then x is
(a) 4
(b) 6
(c) 16
(d) 48
Solution:
3, 12 and x are in continued proportion, then
3 : 12 :: 12 : x
3x = 12 × 12
x = 48 (d)

Question 12.
If the weight of 5 bags of sugar is 27 kg, then the weight of one bag of sugar is
(a) 5.4 kg
(b) 5.2 kg
(c) 5.4 kg
(d) 5.6 kg
Solution:
Weight of 5 bags of sugar = 27 kg
Weight of one bag of sugar
= 5 : 1 :: 27 : x
5x = 1 × 27
x = \(\frac { 27 }{ 5 }\) = 5.4 kg (c)

Question 13.
Sonali bought one dozen notebooks for ₹ 66. What did she pay for one notebook?
(a) ₹ 6.50
(b) ₹ 6.60
(c) ₹ 5.60
(d) ₹ 5.50
Solution:
Cost of 12 books = ₹ 66
Let cost of one book = x
12 : 1 = 66 : x
12 × x = 1 × 66
x = 5.5
Cost of 1 book = ₹ 5.50 (d)

Question 14.
The speed of 90 km/h is equal to
(a) 10 m/sec
(b) 18 m/sec
(c) 25 m/sec
(d) none of these
Solution:
Speed of 90 km/h = \(\frac { 90\times 5 }{ 18 }\) m/sec
= 25 m/sec (c)

Value Based Questions

Question 1.
Sudhanshu divided his property into two parts in the ratio 8 : 5. If the first part is ₹ 1,60,000 and second part is donated to an orphanage, find the amount donated to the orphanage. What values are being promoted?
Solution:
Ratio in two parts of a property = 8 : 5
First part = ₹ 1,60,000
Second part = \(\frac { 160000\times 5 }{ 8 }\) = ₹ 10,00,000
It is a good to donate the needy people and support them.

Higher Order Thinking Skills (HOTS)

Question 1.
Present ages of Rohit and Mayank are in the ratio 11 : 8. 8 years hence the ratio of their ages will be 5 : 4. Find their present ages.
Solution:
Ratio in the present ages of Rohit and Mayank = 11 : 8
Let age of Rohit = 11x, then that of Mayank = 8x
8 years hence, their ages will be
Age of Rohit = 11x + 8 years
and age of Mayank = 8x + 8 years
and the ratio of their ages after 8 years = 5 : 4
\(\frac { 11x+8 }{ 8x+8 }\) = \(\frac { 5 }{ 4 }\)
⇒ 44x + 32 = 40x + 40
⇒ 44x – 40x = 40 – 32
⇒ 4x = 8
⇒ x = 2
Present age of Rohit = 11x = 11 × 2 = 22 years
and age of Mayank = 8x = 8 × 2 = 16 years

Question 2.
Ratio of length and breadth of a rectangle is 3 : 2. If the length of the rectangle is 5 m more than the breadth, find the perimeter of the rectangle.
Solution
Ratio in length and breadth of a rectangle = 3 : 2
Let length = 3x m and Breadth = 2x
Also, l of rectangle is 5 m more than the breadth
i.e. 3x = 2x + 5
⇒ 3x – 2x = 5
⇒ x = 5
Length = 3x = 3 × 5 = 15m
Breadth = 2x = 2 × 5 = 10m
Perimeter = 2(Length + Breadth) = 2 (15 + 10) m = 2 × 25 = 50 m

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Ex 15.3

Question 1.
Count the number of cubes in the following structures:

Solution:
Number of cubes in the given structure havebeen given below:
(i) 5 × 3 × 2 = 30
(ii) 7 × 2 + 6 = 14 + 6 = 20
(iii) 5 × 5 + 17 + 4 = 25 + 17 + 4 = 46

Question 2.
What cross-section is made in the following
(i) vertical cut (ii) horizontal cut?
(a) A brick
(b) A round apple
(c) A die
(d) A circular pipe
(e) An ice cream cone
(f) A square pyramid
Solution:

Question 3.
For each solid given below, the three views (1), (2) and (3) are given. Identify for each solid the corresponding top, front and side views:



Solution:
(a) (1) Side (2) Top (3) Front
(b) (1) Front (Top) (2) Side (3) Top (Front)
(c) (1) Top (2) Side (3) Front
(d) (1) Front (Side) (2) Side (Front) (3) Top
(e) (1) Top (2) Side (3) Front

Question 4.
For the soldis given below sketch the front, side and top view:

Solution:

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 15 Visualising Solid Shapes Check Your Progress

Question 1.
Identify the nets which can be used to make a tetrahedron (cut out copies of the nets and try it):

Solution:
The net/nets for a tetrahedron is (i) and (iii).

Question 2.
If four cubes each with 2 cm edge are placed side by side, what would the dimensions of the resultant cuboid be?
Solution:
Four cubes with 2 cm edge are placed side by side,
then the dimensions of the resulting cuboid will be
Length = 4 × 2 = 8 cm, breadth = 2 cm and height = 2 cm
An isometric sketch and oblique sketches are given below:

Question 3.
Two dice are placed side by side as shown in the given figure. What would be the total on the face opposite to
(i) 5 + 6
(b) 4 + 3?

Solution:
When two dice are placed side by side
as shown in the given figure,
the total would be on face opposite to
(i) 5 + 6 is 3
(ii) 4 + 7 is 7

Question 4.
For the structures given below sketch the front, side, and top view:

Solution:
The front view side view and top view of the given two sketches are given below:

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Ex 16.1

Question 1.
ABCD is a square of side 24 cm. EF is parallel to BC and AE = 15 cm. By how much does
(i) the perimeter of AEFD exceed the perimeter of EBCF?
(ii) the area of AEFD exceed the area of EBCF?

Solution:
Side of the square ABCD = 24 cm
EF || BC || AB is drawn and AE = 15 cm
EB = 24 – 15 = 9 cm

(i) Now perimeter of AEFD = 2(15 + 24) cm = 2 × 39 = 78 cm
and perimeter of EBCF = 2(9 + 24) = 2 × 33 cm = 66 cm
Difference of perimeter = 78 – 66 = 12 cm
(ii) Now Area of AEFD = l × b = 15 × 24 = 360 sq. cm
and area of EBCF = 9 × 24 = 216 sq. cm
Difference = 360 – 216 = 144 sq. cm

Question 2.
Nagma runs around a rectangular park 180 m long and 120 m wide at the rate of 7.5 km/ hour. In how much time will she complete five rounds?
Solution:
Length of rectangular plot (l) = 180 m
and breadth (b) = 120 m
Perimeter = 2(l + b) = 2(180 + 120) m = 2 × 300 = 600 m
Distance travelled in 5 rounds = 600 × 5 = 3000 m = 3 km
Speed = 7.5 km/hr

Question 3.
The area of a rectangular plot is 540 m². if its length is 27 m, find its breadth and perimeter.
Solution:
Area of a rectangular plot = 540 m²
Length (l) = 27 m

and perimeter = 2(l + b) = 2(27 + 20) m = 2 × 47 = 94 m

Question 4.
The perimeter of a rectangular field is 151 m. If its breadth is 32 m, find its length and area.
Solution:
Perimeter of a rectangular field = 151 m
Breadth = 32 m
Length =\(\frac { Perimeter }{ 2 }\) – Breadth
= \(\frac { 151 }{ 2 }\) – 32
= \(\frac { 87 }{ 2 }\)
= 43.5 m
and area = l × b = 43.5 × 32 m² = 1392 m²

Question 5.
The area of a rectangular plot is 340 m² and its breadth is 17 m. Find the cost of surrounding the plot with a fence at ₹ 5.70 per meter.
Solution:
Area of plot = 340 m²
and breadth (b) = 17 m
Length = \(\frac { A }{ b }\) = \(\frac { 340 }{ 17 }\) = 20 m
Perimeter = 2(l + b) = 2(20 + 17) = 2 × 37 = 74 m
Rate of fencing around it = ₹ 5.70 per m
Total cost = ₹5.70 × 74 = ₹ 421.80

Question 6.
The area of a square park is the same as that of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:
Side of a square park = 60 m
Area = (Side)² = 60 × 60 = 3600 m²
Area of rectangular park = 3600 m²
and length (l) = 90 m

Question 7.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also, find which shape encloses more area and by how much?
Solution:
A wire in shape of a rectangle whose length (l) = 40 m
and breadth (b) = 22 m
Perimeter (Length) of wire = 2(l + b) = 2(40 + 22) cm = 2 × 62 cm = 124 cm
Perimeter of square wire = 124 cm
Then side = \(\frac { Perimeter }{ 4 }\) = \(\frac { 124 }{ 4 }\) = 31 m
Now area of rectangle = l × b = 40 × 22 = 880 cm²
and area of square = (Side)² = (31)² cm² = 961 cm²
Difference in area = 961 – 880 = 81 cm²
Area of 81 cm² is more of square shaped wire.

Question 8.
A door of breadth 1 m and height 2 m is fitted in a wall. The length of the wall is 4.5 m and the height is 3.6 m. Find the cost of whitewashing the wall, if the rate of whitewashing the wall is ₹ 20 per m².
Solution:
Breadth of door = 1 m and height = 2 m
Area of door = l × b = 1 × 2 = 2 m²
Length of wall = 4.5 m and height = 3.6 m
Area = 4.5 × 3.6 m² = 16.2 m²
Area of wall excluding area of door = 16.2 – 2 = 14.2 m²
Rate of white washing = ₹ 20 per m²
Total cost = 14.2 × 20 = ₹ 284

Question 9.
A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
Solution:
Length of a rectangular park (l) = 45 m
and breadth (b) = 30 m
Width of path outside the park = 2.5 m
Outer length (L) = 45 + 2 × 2.5 m = 45 + 5 = 50 m

and width (B) = 30 + 2 × 2.5 = 30 + 5 = 35 m
Area of park = L × B – l × b
= 50 × 35 – 45 × 30 m²
= 1750 – 1350
= 400 m²

Question 10.
A carpet of size 5 m × 2 m has 25 cm wide red border. The inner part of the carpet is blue in colour. Find the area of the blue portion. What is the ratio of the areas of red portion to blue portion?
Solution:
Length of blue carpet (l) = 5 m
Breadth (b) = 2 m
Width of red border = 25 cm

Inner length = 5 – \(\frac { 2\times 25 }{ 100 }\) = 5 – 0.5 = 4.5 m
and breadth = 2 – 0.5 = 1.5 m
Now area of carpet = 4.5 × 1.5 m² = 6.75 m²
and area of border = 5 × 2 – 6.75 m² = 10 – 6.75 = 3.25 m²
Now ratio between border and carpet (blue part) = 3.25 : 6.75 = 13 : 27

Question 11.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2.
Solution:
Width of a verandah = 2.25 m
Length of room (i) = 5.5 m
and breadth (b) = 4.0 m
Outer length (L) = 5.5 + 2 × 2.25 m = 5.5 + 4.5 = 10 m
and outer breadth = 4 + 4.5 = 8.5 m
(i) Area of verandah = Outer area – Inner area
= 10 × 8.5 – 5.5 × 4 m²
= 85 – 22 m²
= 63 m²
(ii) Rate of cementing the floor of verandah = ₹ 200 per m²
Total cost = ₹ 63 × 200 = ₹ 12600

Question 12.
Two crossroads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of ₹ 105 per m2.
Solution:
Length of rectangular park (l) = 70 m
and breadth (ft) = 45 m
Width of each road = 5 m

(i) Area of roads = 70 × 5 + 45 × 5 – (5)² m²
= 350 + 225 – 25 = 550 m²
(ii) Rate of constructing the roads = ₹105 per m²
Total cost = ₹ 105 × 550 = ₹ 57750

Question 13.
A rectangular room is 10 m long and 7.5 m wide. Find the cost of covering the floor with carpet 1.25 m wide at ₹ 250 per metre.
Solution:
Length of rectangular room (l) = 10 m
and breadth (b) = 7.5 m
Area of floor of the room = l × b = 10 × 7.5 = 75 m²
Width of carpet = 1.25 m

Cost of 1 m carpet = ₹ 250
Total cost = ₹ 250 × 60 = ₹ 15000

Question 14.
Find the cost of flooring a room 6.5 m by 5 m with square tiles of side 25 cm at the rate of ₹ 9.40 per tile.
Solution:
Length of floor of a room (l) = 6.5 m
and breadth (b) = 5 m
Area = 6.5 × 5 m² = 32.5 m²

Cost of tile at the rate of ₹ 9.40 per tile = ₹ 9.40 × 520 = ₹ 4888

Question 15.
The floor of a room is in the shape of a square of side 4.8 m. The floor is to be covered with square tiles of perimeter 1.2 m. Find the cost of covering the floor if each tile costs ₹ 27.
Solution:
Side of square room = 4.8 m
Area = (4.8)² m² = 23.04 m²
Perimeter of one tile = 1.2 m
Side = \(\frac { 1.2 }{ 4 }\) = 0.3 m
Area of one tile = (0.3)² = 0.09 m²

Cost of one tile = ₹ 27
Total cost = 256 × ₹ 27 = ₹ 6912

Question 16.
A rectangular plot of land is 50 m wide. The cost of fencing the plot at the rate of ₹ 18 per metre is ₹ 4680. Find:
(i) the length of the plot.
(ii) the cost of leveling the plot at the rate of ₹ 7.6 per m².
Solution:
Breadth of a plot = 50 m
Cost of fencing around it = ₹ 4680
Rate of fencing = ₹ 18 per m
Perimeter = \(\frac { 4680 }{ 18 }\) = 260 m
Length =\(\frac { Perimeter }{ 2 }\) – Breadth
= \(\frac { 260 }{ 2 }\) – 50
= 130 – 50
= 80 m
(ii) Now area of plot = l × b = 80 × 50 m² = 4000 m²
Rate of leveling the plot = ₹ 7.6 per m²
Total cost = ₹ 4000 × ₹ 7.6 = ₹ 30400

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Ex 16.2

Question 1.
Find the area of each of the following parallelogram:

Solution:
(i) Base of the parallelogram (b) = 8 cm and height (h) = 4.5 cm
Area = b × h = 8 × 4.5 = 36 cm²
(ii) Base of the parallelogram (b) = 2 cm and height (h) = 4.4 cm
Area = b × h = 2 × 4.4 = 8.8 cm²
(iii) Base of the parallelogram (b) = 2.5 cm and height (h) = 3.5 cm
Area = b × h = 2.5 × 3.5 = 8.75 cm²

Question 2.
Find the area of each of the following triangles:

Solution:
(i) Base of the triangle (b) = 6.4 cm and height (b) = 6 cm
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 6.4 × 6 cm²
= 19.2 cm²
(ii) Base of triangle (b) = 5 cm and height (h) = 6 cm
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 5 × 6 = 15 cm²
(iii) Base of the triangle (b) = 4.5 cm and altitude (h) = 6 cm
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 4.5 × 6 cm²
= 13.5 cm²

Question 3.
Find the missing values:

Solution:
Area of ||gm = b × h

Question 4.
Find the missing values:

Solution:

Question 5.
In the given figure, ABCD is a parallelogram whose two adjacent sides are 6 cm and 4 cm. If the height corresponding to the base AB is 3 cm, find:
(i) the area of parallelogram ABCD
(ii) the height corresponding to the base AD.

Solution:
In ||gm ABCD,
Base AB (b) = 6 cm
Altitude (h) = 3 cm

(i) Area = b × h = 6 × 3 = 18 cm²
In second case,
(ii) Base = 4 cm
Area= 18 cm²
Altitude (to AD) = \(\frac { Area }{ Base }\)
= \(\frac { 18 }{ 4 }\) cm
= 4.5 cm

Question 6.
In the given figure, ABC is an isosceles triangle with AB = AC = 7.5 cm and BC = 9 cm. If the height AD from A to BC is 6 cm, find:
(z) the area of ∆ABC
(ii) the height CE from C to AB.

Solution:
In an isosceles ∆ABC
AB = AC = 7.5 cm, BC = 9 cm
Height AD to BC = 6 cm
(i) Area of ∆ABC = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × BC × AD
= \(\frac { 1 }{ 2 }\) × 9 × 6 = 27 cm²
(ii) Area of ∆ABC = 27 cm²
Base AB = 7.5 cm

Question 7.
If the base of a right-angled triangle is 8 cm and the hypotenuse is 17 cm, find its area.
Solution:
Base of a right angled triangle = 8 cm
and hypotenuse = 17 cm
Height² = (Hypotenuse)² – (Base)² = 17² – 8² = 289 – 64 = 225 = (15)²
Height = 15 cm
Now are of ∆ = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 8 × 15 cm² = 60 cm²

Question 8.
In the given figure, ∆ABC is right-angled at B. Its legs are 8 cm and 6 cm. Find the length of perpendicular BN on the side AC.

Solution:
In the given figure,
In ∆ABC,
Base BC = 8 cm
and height AB = 6 cm
Area = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 8 x 6
= 24 cm²
Now, BN ⊥ AC
AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100 = (10)²
AC = 10 cm

Question 9.
In the given figure, the area of ∆ABE is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the length of the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ∆AMD, where M is mid-point of side DC?

Solution:
In the given figure, M is mid-point of DC
Area of ∆ABE = Area of ||gm ABCD
Now base of ∆ABC = 10 cm and height = 16 cm
Area = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 10 × 16 = 80 cm²
Now area of ||gm = Area of ∆ = 80 cm²
Base = 10 cm
Length of altitude = \(\frac { Area }{ Base }\) = \(\frac { 80 }{ 10 }\) = 8 cm
and Area of ∆AMD = \(\frac { 1 }{ 2 }\) × Base MD × Altitude
= \(\frac { 1 }{ 2 }\) × \(\frac { 10 }{ 2 }\) × \(\frac { 16 }{ 2 }\)
= 20 cm²

Question 10.
In the given figure, ABCD is a rectangle of size 18 cm by 10 cm. In ABEC, ∠E = 90° and EC = 8 cm. Find the area of the shaded region.

Solution:
In the figure,
ABCD is a rectangle in which
Base (b) = 18 cm
Height (h) = 10 cm
A ∆DEC is cut in which
∠E = 90°, EC = 8 cm
Now area of rectangle = l × b = 18 × 10 = 180 cm²
In ∆EBC,
BC = 10 cm, EC = 8 cm
EB² = BC² – EC² (Pythagoras Theorem)
= 10² – 8² = 100 – 64 = 36 = (6)²
EB = 6 cm
Now area of right ∆EBC = \(\frac { 1 }{ 2 }\) × EB × EC
= \(\frac { 1 }{ 2 }\) × 6 × 8 = 24 cm²
Area of shaded portion = 180 – 24 = 156 cm²

Question 11.
In the following figures, find the area of the shaded regions:


Solution:
(i) ABCD is a rectangle in which
Length (l) = 18 cm
Breadth (b) = 10 cm
Area of rectangle = l × b = 18 × 10 = 180 cm²
DE = 10 cm
EC = 18 – 10 = 8 cm
Now area of ∆BCE = \(\frac { 1 }{ 2 }\) × BC × EC
= \(\frac { 1 }{ 2 }\) × 10 × 8
= 40 cm²
and area of ∆FDE = \(\frac { 1 }{ 2 }\) × DC × DF
= \(\frac { 1 }{ 2 }\) × 10 × 6 = 30 cm²
Area of shaded portion = Area of rectangle – Area of ∆BCE – Area of ∆FDE
= 180 – (40+ 30)
= 180 – 70
= 110 cm²
(ii) In the given figure,
ABCD is a square whose each side = 20 cm
E and F are mid-points of AB and AD respectively
EC and FC are joined
Area of square ABCD = (Side)² = (20)² = 400 cm²
Area of ∆EBC = \(\frac { 1 }{ 2 }\) × EB × BC
= \(\frac { 1 }{ 2 }\) × 10 × 20 = 100 cm²
Area of ∆FDC = \(\frac { 1 }{ 2 }\) × FD × DC
= \(\frac { 1 }{ 2 }\) × 10 × 20 = 100 cm²
Area of ∆AEF = \(\frac { 1 }{ 2 }\) × 10 × 10 = 50 cm²
Area of shaded portion = 400 – (100 + 100 + 50) cm² = 400 – 250 = 150 cm²

ML Aggarwal Class 7 Solutions for ICSE Maths