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Page 1
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 1.
Solution:
We know:
Area of a triangle = 1/2×Base×Height
(i) Base = 42 cm
Height = 25 cm
?
Area of the triangle =
(1/2×42×25)
cm
2
= 525 cm
2
(ii) Base = 16.8 m
Height = 75 cm = 0.75 m
[since 100 cm = 1 m]
?
Area of the triangle =
(1/2×16.8×0.75)
m
2
= 6.3 m
2
(iii) Base = 8 dm = (8
×
10) cm = 80 cm
[since 1 dm = 10 cm]
Height = 35 cm
?
Area of the triangle = (1/2×80×35)
cm
2
= 1400 cm
2
Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
Page 2
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 1.
Solution:
We know:
Area of a triangle = 1/2×Base×Height
(i) Base = 42 cm
Height = 25 cm
?
Area of the triangle =
(1/2×42×25)
cm
2
= 525 cm
2
(ii) Base = 16.8 m
Height = 75 cm = 0.75 m
[since 100 cm = 1 m]
?
Area of the triangle =
(1/2×16.8×0.75)
m
2
= 6.3 m
2
(iii) Base = 8 dm = (8
×
10) cm = 80 cm
[since 1 dm = 10 cm]
Height = 35 cm
?
Area of the triangle = (1/2×80×35)
cm
2
= 1400 cm
2
Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m
Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm
Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
Page 3
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 1.
Solution:
We know:
Area of a triangle = 1/2×Base×Height
(i) Base = 42 cm
Height = 25 cm
?
Area of the triangle =
(1/2×42×25)
cm
2
= 525 cm
2
(ii) Base = 16.8 m
Height = 75 cm = 0.75 m
[since 100 cm = 1 m]
?
Area of the triangle =
(1/2×16.8×0.75)
m
2
= 6.3 m
2
(iii) Base = 8 dm = (8
×
10) cm = 80 cm
[since 1 dm = 10 cm]
Height = 35 cm
?
Area of the triangle = (1/2×80×35)
cm
2
= 1400 cm
2
Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m
Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm
Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Solution:
Area of the right angled triangle = 129.5 cm²
Question 7.
Solution:
In right angled ?ABC,
Base BC = 1.2 m
Question 6.
Page 4
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 1.
Solution:
We know:
Area of a triangle = 1/2×Base×Height
(i) Base = 42 cm
Height = 25 cm
?
Area of the triangle =
(1/2×42×25)
cm
2
= 525 cm
2
(ii) Base = 16.8 m
Height = 75 cm = 0.75 m
[since 100 cm = 1 m]
?
Area of the triangle =
(1/2×16.8×0.75)
m
2
= 6.3 m
2
(iii) Base = 8 dm = (8
×
10) cm = 80 cm
[since 1 dm = 10 cm]
Height = 35 cm
?
Area of the triangle = (1/2×80×35)
cm
2
= 1400 cm
2
Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m
Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm
Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Solution:
Area of the right angled triangle = 129.5 cm²
Question 7.
Solution:
In right angled ?ABC,
Base BC = 1.2 m
Question 6.
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
and hypotenuse AC = 3.7 m
But AC² = AB² + BC² (Pythagoras Theorem)
? (3.7)² = AB² + (1.2)²
? 13.69 = AB² + 1.44
? AB² = 13.69 – 1.44
? AB² = 12.25 = (3.5)²
? AB = 3.5 m
Now, area of ?ABC = ½ x base x altitude
= ½ x 1.2 x 3.5 m² = 2.1 m²
Question 8.
Solution:
Legs of a right angled triangle = 3 : 4
Let one leg (base) = 3x
Then second leg (altitude) = 4x
Area = ½ x base x altitude
= ½ x 3x x 4x = 6x²
6x² = 1014
? x² = 1014/6 = 169 = (13)²
x = 13
one leg'(Base) = 3x = 3 x 13 = 39 cm
and second leg (altitude) = 4x = 4 x 13 = 52 cm
Page 5
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 1.
Solution:
We know:
Area of a triangle = 1/2×Base×Height
(i) Base = 42 cm
Height = 25 cm
?
Area of the triangle =
(1/2×42×25)
cm
2
= 525 cm
2
(ii) Base = 16.8 m
Height = 75 cm = 0.75 m
[since 100 cm = 1 m]
?
Area of the triangle =
(1/2×16.8×0.75)
m
2
= 6.3 m
2
(iii) Base = 8 dm = (8
×
10) cm = 80 cm
[since 1 dm = 10 cm]
Height = 35 cm
?
Area of the triangle = (1/2×80×35)
cm
2
= 1400 cm
2
Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m
Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm
Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Solution:
Area of the right angled triangle = 129.5 cm²
Question 7.
Solution:
In right angled ?ABC,
Base BC = 1.2 m
Question 6.
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
and hypotenuse AC = 3.7 m
But AC² = AB² + BC² (Pythagoras Theorem)
? (3.7)² = AB² + (1.2)²
? 13.69 = AB² + 1.44
? AB² = 13.69 – 1.44
? AB² = 12.25 = (3.5)²
? AB = 3.5 m
Now, area of ?ABC = ½ x base x altitude
= ½ x 1.2 x 3.5 m² = 2.1 m²
Question 8.
Solution:
Legs of a right angled triangle = 3 : 4
Let one leg (base) = 3x
Then second leg (altitude) = 4x
Area = ½ x base x altitude
= ½ x 3x x 4x = 6x²
6x² = 1014
? x² = 1014/6 = 169 = (13)²
x = 13
one leg'(Base) = 3x = 3 x 13 = 39 cm
and second leg (altitude) = 4x = 4 x 13 = 52 cm
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration (Ex 20D) Exercise 20D
Question 9.
Solution:
One side BC of a right triangular scarf = 80 cm
and longest side AC = 1 m = 100 cm
By Pythagoras Theorem,
AC² = AB² + BC²
? (100)² = AB² + (80)²
? 10000 = AB² + 6400
? AB² = 10000 – 6400
? AB² = 3600 = (60)²
? AB = 60
Second side = 60 cm
Area of the scarf = ½ x b x h
= ½ x 80 x 60 cm
2
= 2400 cm²
Rate of cost = Rs. 250 per m²
Total cost =2400/200 x 250 = Rs. 60
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