Mathematics Part ii Solutions Solutions for Class 7 Math Chapter 5 Pythagoras Theorem are provided here with simple step-by-step explanations. These solutions for Pythagoras Theorem are extremely popular among class 7 students for Math Pythagoras Theorem Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 7 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.
Page No 139:
Question 1:
Can you draw a square whose area is double the area of the square below?
Answer:
We know that the area of a square is double the area of another square if the length of the side of the former is equal to the length of the diagonal of the latter.
Let the length of each side of a given square be x units.
Length of diagonal of this square = units = units
So, the length of each side of the required square is units.
The following steps should be followed to draw the required square.
(1) Draw two squares of side x units, adjacent to each other.
(2) Cut the two squares along their diagonals.
(3) Place the triangles in such a way that the obtained quadrilateral is also a square.
The square so obtained has each side equal to the diagonal of the given square.
Thus, the obtained square has an area double the area of the given square.
Page No 139:
Question 2:
How do we draw a square of area 8 square centimeters?
Answer:
Area of the given square = 8 cm2
⇒ (Side)2 = 8 cm2
⇒ Side =
As , we can say that is the diagonal of a square of side 2 cm.
We know that the area of a square is double the area of another square if the length of side of the former is equal to the length of diagonal of the latter.
Here, we will draw a square of side with the help of a square of side 2 cm.
The following steps should be followed to draw the required square.
(1) Draw two squares of each side 2 cm, adjacent to each other.
(2) Cut the two squares along their diagonals.
(3) Place the triangles in such a way that the obtained quadrilateral is also a square.
The square so obtained has each side equal to.
Thus, the obtained square has an area of 8 cm2.
Page No 143:
Question 1:
Now can’t we draw a square of area 20 square centimeters?
Answer:
Area of the given square = 20 cm2
Also, 20 = 22 + 42
We know that the area of a square drawn on the diagonal of a rectangle is equal to the sum of the areas of the squares drawn on two adjacent sides of the rectangle.
The following steps should be followed to draw the required square.
(1) Draw a rectangle of length 4 cm and breadth 2 cm.
(2) Join the diagonal of the rectangle.
(3) Draw a square on the diagonal of the above rectangle.
The obtained square has an area of 20 cm2.
Page No 143:
Question 2:
How about a square of area 41 square centimeters?
Answer:
Area of the given square = 41 cm2
Also, 41 = 52 + 42
We know that the area of a square drawn on the diagonal of a rectangle is equal to the sum of the areas of the squares drawn on two adjacent sides of the rectangle.
The following steps should be followed to draw the required square.
(1) Draw a rectangle of length 5 cm and breadth 4 cm.
(2) Join the diagonal of the rectangle.
(3) Draw a square on the diagonal of the above rectangle.
The obtained square has area 41 cm2.
Page No 145:
Question 1:
Now can you draw a square of area 11 square centimeters?
Answer:
Area of the given square = 11 cm2
Also, 11 = 62− 52
We know, by Pythagoras’ Theorem, that in a right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
Using the Pythagorean Theorem, we can draw the required square.
The following steps should be followed to draw the required square.
(1) Draw a right-angled triangle of base length 5 cm and hypotenuse of length 6 cm.
(2) Draw a square on the third side of the triangle drawn above.
Page No 147:
Question 1:
Some triangles are given below with the lengths of two sides specified. Can you find the length of the third side?
Answer:
(i) The given triangle is
By Pythagoras’ Theorem, we have
(AB)2 = (AC)2 − (BC)2
⇒ (AB)2 = (17 cm)2 − (15 cm)2
⇒ (AB)2 = (289 − 225) cm2
⇒ (AB)2 = 64 cm2
⇒ AB =
⇒ AB = 8 cm
Thus, the length of the third side of the given right-angled triangle is 8 cm.
(ii) The given triangle is
By Pythagoras’ Theorem, we have
(AC)2 = (AB)2 + (BC)2
⇒ (AC)2 = (12 cm)2 + (5 cm)2
⇒ (AC)2 = (144 + 25) cm2
⇒ (AC)2 = 169 cm2
⇒ AC =
⇒ AC = 13 cm
Thus, the length of the third side of the given right-angled triangle is 13 cm.
(iii) The given triangle is
By Pythagoras’ Theorem, we have
(BC)2 = (AC)2 − (AB)2
⇒ (BC)2 = (20.5 cm)2 − (4.5 cm)2
⇒ (BC)2 = (420.25 − 20.25) cm2
⇒ (BC)2 = 400 cm2
⇒ BC =
⇒ BC = 20 cm
Thus, the length of the third side of the given right-angled triangle is 20 cm.
Page No 148:
Question 1:
A rectangular field is of 80 meters long and 60 meters wide. A man walked from one corner of the field to the opposite corner along the boundary of the field. Another man crossed the field diagonally to reach the opposite corner. Who walked more? How much more?
Answer:
The given rectangular field can be shown as follows.
The first man walked from corner A of the field to corner C along the boundary of the field.
∴ Path covered by the first man = (60 + 80) m = 140 m
The second man crossed the field diagonally to reach corner C from corner A.
Now, by using Pythagoras’ Theorem, we have
(AC)2 = (AD)2 + (DC)2
⇒ (AC)2 = (60 m)2 + (80 m)2
⇒ (AC)2 = (3600 + 6400) m2
⇒ (AC)2 = 10000 m2
⇒ AC =
⇒ AC = 100 m
∴ Path covered by the second man = 100 m
Difference between the paths covered by the men = (140 − 100) m = 40 m
Thus, the first man covered 40 m more than the second man.
Page No 148:
Question 2:
A 32 meter tall areca palm snapped at a point 15 meters from the ground and its top now touches the ground. How far from the foot of the tree is its top now, on the ground?
Answer:
The given tree can be shown by a triangle as follows.
Total height of the tree = AB + AC = 32 m
Given AB = 15 m
⇒ 15 m + AC = 32 m
⇒ AC = 32 m − 15 m = 17 m
Now, by using Pythagoras’ Theorem, we have
(BC)2 = (AC)2 − (AB)2
⇒ (BC)2 = (17 m)2 − (15 m)2
⇒ (BC)2 = (289 − 225) m2
⇒ (BC)2 = 64 m2
⇒ BC =
⇒ BC = 8 m
Thus, the top of the tree is 8 m from the foot of the tree.
Page No 148:
Question 3:
A 10 meter long ladder leans on a wall. The foot of the ladder is 6 meters away from the wall. How high is the top of the ladder?
Answer:
Let AC be the ladder and AB be the wall.
Given AC = 10 m and BC = 6 m
Now, by using Pythagoras’ Theorem, we have
(AB)2 = (AC)2 − (BC)2
⇒ (AB)2 = (10 m)2 − (6 m)2
⇒ (AB)2 = (100 − 36) m2
⇒ (AB)2 = 64 m2
⇒ AB =
⇒ AB = 8 m
Thus, the top of the ladder is 8 m high.
Page No 148:
Question 4:
Compute the length of the fourth side of this figure:
Answer:
Construction: Join the diagonal of the given figure.
By using Pythagoras’ Theorem in ΔBCD, we have
(BD)2 = (BC)2 + (DC)2
⇒ (BD)2 = (15 cm)2 + (20 cm)2
⇒ (BD)2 = (225 + 400) cm2
⇒ (BD)2 = 625 cm2
⇒ BD =
⇒ BD = 25 cm
By using Pythagoras’ Theorem in ΔBAD, we have
(AD)2 = (BD)2 − (AB)2
⇒ (AD)2 = (25 cm)2 − (24 cm)2
⇒ (AD)2 = (625 − 576) cm2
⇒ (AD)2 = 49 cm2
⇒ AD =
⇒ AD = 7 cm
Thus, the fourth side of the given figure is 7 cm long.
Page No 148:
Question 5:
Two poles of heights 8 meters and 15 meters stand upright on the ground. Their feets are 24 meters apart. What is the distance between their tops?
Answer:
Let AB be the pole of height 8 m and DC be the pole of height 15 m.
Let BC be the distance between the feet of the two poles.
Construction: Draw a line segment AE parallel to BC.
In the given figure, AECB is a rectangle.
AE = BC = 24 m and EC = AB = 8 m
Also, DE = DC − EC = (15 − 8) m = 7 m
Now, by using Pythagoras’ Theorem in ΔAED, we have
(AD)2 = (DE)2 + (AE)2
⇒ (AD)2 = (7 m)2 + (24 m)2
⇒ (AD)2 = (49 + 576) m2
⇒ (AD)2 = 625 m2
⇒ AD =
⇒ AD = 25 m
Thus, the distance between the tops of the two poles is 25 m.
Page No 149:
Question 1:
The lengths of the sides of some triangles are given below. Find the right-angled triangles among these.
(i) 7 centimeters, 15 centimeters, 17 centimeters
(ii) 8 centimeters, 24 centimeters, 25 centimeters
(iii) 6 centimeters, 8 centimeters, 10 centimeters
(iv) 1.5 centimeters, 1.6 centimeters, 2.5 centimeters
Answer:
(i)The given sides of the triangle are 7 cm, 15 cm and 17 cm.
Square of the longest side = (17 cm)2 = 289 cm2
Sum of the squares of the other two sides = (7 cm)2 + (15 cm)2
= (49 + 225) cm2
= 274 cm2
Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given sides are not the sides of a right-angled triangle.
(ii) The given sides of the triangle are 8 cm, 24 cm and 25 cm.
Square of the longest side = (25 cm)2 = 625 cm2
Sum of the squares of the other two sides = (8 cm)2 + (24 cm)2
= (64 + 576) cm2
= 640 cm2
Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given sides are not the sides of a right-angled triangle.
(iii) The given sides of the triangle are 6 cm, 8 cm and 10 cm.
Square of the longest side = (10 cm)2 = 100 cm2
Sum of the squares of the other two sides = (6 cm)2 + (8 cm)2
= (36 + 64) cm2
= 100 cm2
Since the square of the longest side is equal to the sum of the squares of the other two sides, the given sides are the sides of a right-angled triangle.
(iv) The given sides of the triangle are 1.5 cm, 1.6 cm and 2.5 cm.
Square of the longest side = (2.5 cm)2 = 6.25 cm2
Sum of the squares of the other two sides = (1.5 cm)2 + (1.6 cm)2
= (2.25 + 2.56) cm2
= 4.81 cm2
Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given sides are not the sides of a right-angled triangle.
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