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Construction: Determination Of Mean Proportional

Class 10th Mathematics West Bengal Board Solution
Let Us Work Out 21
  1. 5cm., 2.5cm Let us draw the mean proportional of the following line segment and let us…
  2. 4cm., 3cm. Let us draw the mean proportional of the following line segment and let us…
  3. 7.5cm., 4cm. Let us draw the mean proportional of the following line segment and let…
  4. 10cm., 4cm. Let us draw the mean proportional of the following line segment and let us…
  5. 9cm., 5cm. Let us draw the mean proportional of the following line segment and let us…
  6. 12cm., 3cm. Let us draw the mean proportional of the following line segment and let us…
  7. 7 Let us determine the square roots of the following numbers by geometric method:…
  8. 8 Let us determine the square roots of the following numbers by geometric method:…
  9. 24 Let us determine the square roots of the following numbers by geometric method:…
  10. 28 Let us determine the square roots of the following numbers by geometric method:…
  11. 13 Let us determine the square roots of the following numbers by geometric method:…
  12. 29 Let us determine the square roots of the following numbers by geometric method:…
  13. √14 cm Let us draw the squared figures by taking the following lengths as sides.…
  14. √22 cm Let us draw the squared figures by taking the following lengths as sides.…
  15. √31 cm Let us draw the squared figures by taking the following lengths as sides.…
  16. √33 cm Let us draw the squared figures by taking the following lengths as sides.…
  17. 8cm., 6cm. Let us draw the squares whose areas are equal to the areas of the…
  18. 6cm., 4cm. Let us draw the squares whose areas are equal to the areas of the…
  19. 4.2cm., 3.5cm Let us draw the squares whose areas are equal to the areas of the…
  20. 7.9cm., 4.1cm. Let us draw the squares whose areas are equal to the areas of the…
  21. The lengths of three sides of a triangles are 10cm., 7cm. and 5cm. respectively. Let…
  22. An isosceles triangle whose base is 7cm. length and the length of each of two equal…
  23. An equilateral triangle whose side is 6cm. in length. Let us draw the squares whose…
Let Us Work Out 21
Question 1.

Let us draw the mean proportional of the following line segment and let us measure the values of the mean proportionals in each case with help of a scale:

5cm., 2.5cm
Answer:

Construction steps:


1. Draw a line segment AX with length greater than (5 + 2.5 = 7.5 cm).



2. From A, mark a point B on AX such that AB = 5 cm and from B mark a point C such that BC = 2.5 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 5 cm and BC = 2.5 cm.


With the help of scale, we measure BD = 3.5 cm [Appx]
Question 2.

Let us draw the mean proportional of the following line segment and let us measure the values of the mean proportionals in each case with help of a scale:

4cm., 3cm.
Answer:

Construction steps:


1. Draw a line segment AX with length greater than (4 + 3 = 7 cm).



2. From A, mark a point B on AX such that AB = 4 cm and from B mark a point C such that BC = 3 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 4 cm and BC = 3 cm.


With the help of scale, we measure BD = 3.4 cm [Appx]
Question 3.

Let us draw the mean proportional of the following line segment and let us measure the values of the mean proportionals in each case with help of a scale:

7.5cm., 4cm.
Answer:

Construction steps:


1. Draw a line segment AX with length greater than (7.5 + 4 = 11.5 cm).



2. From A, mark a point B on AX such that AB = 7.5 cm and from B mark a point C such that BC = 4 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 7.5 cm and BC = 4 cm.


With the help of scale, we measure BD = 5.5 cm [Appx]
Question 4.

Let us draw the mean proportional of the following line segment and let us measure the values of the mean proportionals in each case with help of a scale:

10cm., 4cm.
Answer:

Construction steps:


1. Draw a line segment AX with length greater than (10 + 4 = 14 cm).



2. From A, mark a point B on AX such that AB = 10 cm and from B mark a point C such that BC = 4 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 10 cm and BC = 4 cm.


With the help of scale, we measure BD = 6.3 cm [Appx]
Question 5.

Let us draw the mean proportional of the following line segment and let us measure the values of the mean proportionals in each case with help of a scale:

9cm., 5cm.
Answer:

Construction steps:


1. Draw a line segment AX with length greater than (9 + 5 = 11.5 cm).



2. From A, mark a point B on AX such that AB = 9 cm and from B mark a point C such that BC = 5 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 9 cm and BC = 5 cm.


With the help of scale, we measure BD = 6.7 cm [Appx]
Question 6.

Let us draw the mean proportional of the following line segment and let us measure the values of the mean proportionals in each case with help of a scale:

12cm., 3cm.
Answer:

Construction steps:


1. Draw a line segment AX with length greater than (12 + 3 = 15 cm).



2. From A, mark a point B on AX such that AB = 12 cm and from B mark a point C such that BC = 3 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 12 cm and BC = 3 cm.


With the help of scale, we measure BD = 6 cm [Appx]
Question 7.

Let us determine the square roots of the following numbers by geometric method:

7
Answer:

We know, that mean proportion of two number ‘a’ and ‘b’ is . In order to find the square root of each number, we will factorize them into two factors and then we will find the mean proportion of two numbers.


7 = 3.5 × 2


Construction steps:


1. Draw a line segment AX with length greater than (3.5 + 2 = 5.5 cm).



2. From A, mark a point B on AX such that AB = 3.5 cm and from B mark a point C such that BC = 2 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 3.5 cm and BC = 2 cm.


With the help of scale, we measure BD = 2.6 cm [Appx]
Question 8.

Let us determine the square roots of the following numbers by geometric method:

8
Answer:

We know, that mean proportion of two number ‘a’ and ‘b’ is . In order to find the square root of each number, we will factorize them into two factors and then we will find the mean proportion of two numbers.


8 = 4 × 2


Construction steps:


1. Draw a line segment AX with length greater than (4 + 2 = 6 cm).



2. From A, mark a point B on AX such that AB = 4 cm and from B mark a point C such that BC = 2 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 4 cm and BC = 2 cm.


With the help of scale, we measure BD = 2.8 cm [Appx]
Question 9.

Let us determine the square roots of the following numbers by geometric method:

24
Answer:

We know, that mean proportion of two number ‘a’ and ‘b’ is . In order to find the square root of each number, we will factorize them into two factors and then we will find the mean proportion of two numbers.


24 = 6 × 4


Construction steps:


1. Draw a line segment AX with length greater than (6 + 4 = 10 cm).



2. From A, mark a point B on AX such that AB = 6 cm and from B mark a point C such that BC = 4 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 6 cm and BC = 4 cm.


With the help of scale, we measure BD = 4.9 cm [Appx]
Question 10.

Let us determine the square roots of the following numbers by geometric method:

28
Answer:

We know, that mean proportion of two number ‘a’ and ‘b’ is . In order to find the square root of each number, we will factorize them into two factors and then we will find the mean proportion of two numbers.


28 = 7 × 4


Construction steps:


1. Draw a line segment AX with length greater than (7 + 4 = 11 cm).



2. From A, mark a point B on AX such that AB = 7 cm and from B mark a point C such that BC = 4 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 7 cm and BC = 4 cm.


With the help of scale, we measure BD = 5.3 cm [Appx]
Question 11.

Let us determine the square roots of the following numbers by geometric method:

13
Answer:

We know, that mean proportion of two number ‘a’ and ‘b’ is . In order to find the square root of each number, we will factorize them into two factors and then we will find the mean proportion of two numbers.


13 = 5 × 2.6


Construction steps:


1. Draw a line segment AX with length greater than (5 + 2.6 = 7.6 cm).



2. From A, mark a point B on AX such that AB = 5 cm and from B mark a point C such that BC = 2.6 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 5 cm and BC = 2.6 cm.


With the help of scale, we measure BD = 3.6 cm [Appx]
Question 12.

Let us determine the square roots of the following numbers by geometric method:

29
Answer:

We know, that mean proportion of two number ‘a’ and ‘b’ is . In order to find the square root of each number, we will factorize them into two factors and then we will find the mean proportion of two numbers.


29 = 5.8 × 5


Construction steps:


1. Draw a line segment AX with length greater than (5.8 + 5 = 10.8 cm).



2. From A, mark a point B on AX such that AB = 5.8 cm and from B mark a point C such that BC = 5 cm



3. Draw the perpendicular bisector of AC, such that it cuts AC at O.



4. Taking OA = OC as radius and O as center draw a semicircle.



5. Draw BD ⊥ AC such that, point D lies on the semicircle and



Here, BD is the mean proportional of AB and BC, and By construction AB = 5.8 cm and BC = 5 cm.


With the help of scale, we measure BD = 5.4 cm [Appx]
Question 13.

Let us draw the squared figures by taking the following lengths as sides.

√14 cm
Answer:

We know that, area of a square with side √a will be a.


[Area of square = (side)2]


Also, we can draw a square with area equal to a rectangle, we will use the area of rectangle in order to make the area equal to square.


side = √14 cm


Area of square = 14 cm2


Area of rectangle = length × breadth = 14 cm2


Here, let’s choose length = 7 cm and breadth = 2 cm


Construction steps:


1. Draw a rectangle ABCD with length, AB = 7 cm and breadth, BC = 2 cm.



2. Extend CD to CE, such that CE = BC = 2 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD = 14 cm2.
Question 14.

Let us draw the squared figures by taking the following lengths as sides.

√22 cm
Answer:

We know that, area of a square with side √a will be a.


[Area of square = (side)2]


Also, we can draw a square with area equal to a rectangle, we will use the area of rectangle in order to make the area equal to square.


side = √22 cm


Area of square = 22 cm2


Area of rectangle = length × breadth = 22 cm2


Here, let’s choose length = 5.5 cm and breadth = 4 cm


Construction steps:


1. Draw a rectangle ABCD with length, AB = 5.5 cm and breadth, BC = 4 cm.



2. Extend CD to CE, such that CE = BC = 4 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD = 22 cm2.
Question 15.

Let us draw the squared figures by taking the following lengths as sides.

√31 cm
Answer:

We know that, area of a square with side √a will be a.


[Area of square = (side)2]


Also, we can draw a square with area equal to a rectangle, we will use the area of rectangle in order to make the area equal to square.


side = √31 cm


Area of square = 31 cm2


Area of rectangle = length × breadth = 31 cm2


Here, let’s choose length = 6.2 cm and breadth = 5 cm


Construction steps:


1. Draw a rectangle ABCD with length, AB = 6.2 cm and breadth, BC = 5 cm.



2. Extend CD to CE, such that CE = BC = 5 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD = 31 cm2.
Question 16.

Let us draw the squared figures by taking the following lengths as sides.

√33 cm
Answer:

We know that, area of a square with side √a will be a.


[Area of square = (side)2]


Also, we can draw a square with area equal to a rectangle, we will use the area of rectangle in order to make the area equal to square.


side = √33 cm


Area of square = 33 cm2


Area of rectangle = length × breadth = 33 cm2


Here, let’s choose length = 6.6 cm and breadth = 5 cm


Construction steps:


1. Draw a rectangle ABCD with length, AB = 6.6 cm and breadth, BC = 5 cm.



2. Extend CD to CE, such that CE = BC = 5 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD = 33 cm2.
Question 17.

Let us draw the squares whose areas are equal to the areas of the rectangles by taking following lengths as its sides:

8cm., 6cm.
Answer:

Construction steps:


1. Draw a rectangle ABCD with length, AB = 8 cm and breadth, BC = 6 cm.



2. Extend CD to CE, such that CE = BC = 6 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD
Question 18.

Let us draw the squares whose areas are equal to the areas of the rectangles by taking following lengths as its sides:

6cm., 4cm.
Answer:

Construction steps:


1. Draw a rectangle ABCD with length, AB = 6 cm and breadth, BC = 4 cm.



2. Extend CD to CE, such that CE = BC = 4 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD
Question 19.

Let us draw the squares whose areas are equal to the areas of the rectangles by taking following lengths as its sides:

4.2cm., 3.5cm
Answer:

Construction steps:


1. Draw a rectangle ABCD with length, AB = 4.2 cm and breadth, BC = 3.5 cm.



2. Extend CD to CE, such that CE = BC = 3.5 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD
Question 20.

Let us draw the squares whose areas are equal to the areas of the rectangles by taking following lengths as its sides:

7.9cm., 4.1cm.
Answer:

Construction steps:


1. Draw a rectangle ABCD with length, AB = 7.9 cm and breadth, BC = 4.1 cm.



2. Extend CD to CE, such that CE = BC = 4.1 cm.



3. Draw the perpendicular bisector of DE which bisects DE at O.



4. Taking O as center and OD = OE as radius, draw a semicircle.



5. Extend BC which intersects semicircle at F.



6. Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD
Question 21.

Let us draw the squares whose areas equal to the areas of the following triangles:

The lengths of three sides of a triangles are 10cm., 7cm. and 5cm. respectively.
Answer:

Steps of construction:


1) Draw a triangle PQR with sides QR = 10 cm, PQ = 7 cm and PR = 5 cm.



2) Draw a rectangle ARCD with area equal to that of triangle PQR, by choosing length of rectangle as half of base and height same as height of triangle.


[Here, area of rectangle = length × breadth


= 1/2 × base × height


= area of triangle]


Now, we can make a square with area equal to that of rectangle and hence equal to area of triangle.



3) Extend CD to CE, such that CE = BC



4) Draw the perpendicular bisector of DE which bisects DE at O.



5) Taking O as center and OD = OE as radius, draw a semicircle.



6) Extend BC which intersects semicircle at F.



7) Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD


= area of ΔPQR
Question 22.

Let us draw the squares whose areas equal to the areas of the following triangles:

An isosceles triangle whose base is 7cm. length and the length of each of two equal sides is 5cm.
Answer:

Steps of construction:


1) Draw a triangle PQR with sides QR = 7cm cm, PQ = PR = 5 cm.



2) Draw a rectangle ARCP with area equal to that of triangle PQR, by choosing length of rectangle as half of base and height same as height of triangle.


[Here, area of rectangle = length × breadth


= 1/2 × base × height


= area of triangle]


Now, we can make a square with area equal to that of rectangle and hence equal to area of triangle.



3) Extend CP to CE, such that CE = BC



4) Draw the perpendicular bisector of PE which bisects PE at O.



5) Taking O as center and OD = OE as radius, draw a semicircle.



6) Extend BC which intersects semicircle at F.



7) Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD


= area of ΔPQR
Question 23.

Let us draw the squares whose areas equal to the areas of the following triangles:

An equilateral triangle whose side is 6cm. in length.
Answer:

Steps of construction:


1) Draw a triangle PQR with sides PQ = QR = PR = 6 cm.



2) Draw a rectangle ARCP with area equal to that of triangle PQR, by choosing length of rectangle as half of base and height same as height of triangle.


[Here, area of rectangle = length × breadth


= 1/2 × base × height


= area of triangle]


Now, we can make a square with area equal to that of rectangle and hence equal to area of triangle.



3) Extend CP to CE, such that CE = BC



4) Draw the perpendicular bisector of PE which bisects PE at O.



5) Taking O as center and OD = OE as radius, draw a semicircle.



6) Extend BC which intersects semicircle at F.



7) Draw a square CFGH taking CF as side.



Here,


Area of square CFGH = area of rectangle ABCD


= area of ΔPQR