West Bengal Board Class 10 Math Book Solution In English Chapter 21 Determination Of Mean Proportional Exercise 21.1
Application 1. I determined the values of √21 and √15 in the geometric method. Or, in the geometric method, I determine the square roots of 21 and 15.
Hints: 217 x 3. ∴ In this case, I take two line segments a and b whose lengths are 7 units and 3 units respectively, and in the same method, I shall determine to mean proportional of a and b. Again, 15 = 3 x 5, I draw it by taking the lengths of a and b properly.
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Application 2. I determine the value of √23 in the geometric method.
Solution: 23 = 5 x 4.6
1. I take two line segments a and b whose lengths are 5 cm and 4.6 cm respectively.
Measuring with a scale; I am observing that BD = 4.79 cm. (Approx)
∴ √23 = 4.79 (Approx)
I have understood, i.e., if there is a two-digit prime number, for example, 17, 19, 29, 37, etc. then these numbers will be divided by 5,
Class 10 WBBSE Math Solution In English For example: 175 x 3.4, 19=5x 3.8, 29 = 5 x 5.8, 37 = 5 x 7.4, etc.
Class 10 WBBSE Math Solution In English Application 3. Let us draw a square whose area is equal to the area of a rectangle of 7 cm in length and 4cm in breadth.
Solution: Here area of the rectangle ABCD = the area of square CEFG.
Construction: Draw a rectangle ABCD whose AB = 7cm & BC = 4cm From produce DC, cut CM equal to BC.
Now draw a semi-circle with a diameter DM.
Produce BC which cut the semi-circle at G.
Now draw a square with CG as one side.
∴ CEFG is the required square.
Question 1. I draw a square whose area is equal to the area of an equilateral triangle having a side 7 cm in length.
Solution: First draw an equilateral triangle ABC whose each side is 7 cm. Now bisect the base BC at D.
Next, draw a straight line through A parallel to BC and bisect BC at D.
Now draw a rectangle ADCE whose area is equal to the ABC, now from produce AE cut AM equal to CE.
Next, draw a semi-circle with a diameter AM.
Produce CE which cuts the circle at P.
Draw the square EPQR, whose area is equal to the ΔABC.
WBBSE Solutions Guide Class 10 Question 2. Let us draw the mean proportional of the following line segments and let us measure the values of the mean proportionals in each case with the help of a scale
1. 5 cm, 2.5 cm.
Solution: Take a line segment AX > 5 cm.
From AX cut AB = 5 cm & BC = 2.5 cm. Now, with AC as a diameter, draw a semicircle.
Now at B, draw a perpendicular on BC which cuts the semicircle at D.
The length of BD is the required mean proportional BD = 3 cm (by measuring with a scale).
2. 4 cm, 3 cm.
Solution: Take a line segment AX > 4 cm. From AX, cut AB = 4 cm. & BC = 3 cm.
Now, with AC as a diameter, draw a semicircle.
Now at B, draw a perpendicular on BC, which cuts the semicircle at D.
The length of BD (= 3.5 cm, by measuring with a scale) is the required mean proportional.
3. 7.5 cm, 4 cm.
Solution: Take a line segment AX > 7.5 cm. From AX, cut AB = 7.5 cm & BC = 4 cm.
Now with AC as the diameter, draw a semicircle.
Now at B, draw a perpendicular on BC, which cuts the semicircle at D.
The length of BD (= 3.5 cm, by measuring with a scale) is the required mean proportional.
∴ The required mean proportional is BD.
Length of BD = 5.7 cm.
4. 10 cm, 4 cm.
Solution: From line segment AX, cut AB = 10 cm
& BC = 4 cm. Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ The required mean proportional is BD.
Length BD = 6.3 cm. (approx)
WBBSE Solutions Guide Class 10 5. 9 cm, 5 cm.
Solution: From line segment AX, cut AB = 9 cm & BC= 5 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ The required mean proportional is BD.
Length BD 6.7 cm. (approx)
6. 12 cm, 3 cm.
Solution: From line segment AX, cut AB = 12 cm & BC= 3 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ The required mean proportional is BD.
Length BD 6 cm. (approx)
Question 3. Let us determine the square roots of the following numbers by the geometric method :
1. 7
Solution: From line segment AX, cut AB = 7 cm & BC = 1. cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ Required mean proportional is BD.
Length BD 2.6 cm.
∴ √7=2.6 (approx)
2. 8
Solution: From line segment AX, cut AB = 4 cm & BC= 2 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ The required mean proportional is BD.
Length BD 2.8 cm.
∴ √8 = 2.8 (approx)
Ganit Prakash Class 10 Solutions Pdf In English 3. 24
Solution: From line segment AX, cut AB = 4 cm & BC = 6 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ Required mean proportional is BD.
Length BD 4.9 cm.
∴ BD=4.9 (approx)
4. 28
Solution: From line segment AX, cut AB = 4 cm & BC= 7 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ Required mean proportional is BD.
Length BD 5.3 cm.
√28=5.3 (approx)
5. 13
Solution: 5.0 cm x 2.6 cm. From line segment AX, cut AB = 2.6 cm & BC= 5 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
∴ Required mean proportional is BD.
Length BD 3.6 cm.
∴ √13 = 3.6 (approx)
Maths WBBSE Class 10 Solutions 6. 29
Solution: From line segment AX, cut AB = 5 cm & BC= 5.8 cm.
Draw a semicircle with AC as the diameter.
Draw BD perpendicular to B.
The required mean proportional is BD.
Length BD 5.4 cm. (approx)
Question 4. Let us draw the squared figures by taking the following lengths as sides.
1. √14 cm.
Solution: From a line segment AX, cut 2 cm & 7 cm.
Now draw the mean proportional of AB & BC.
BD is the mean proportional of length √14 cm.
Now draw a square whose one side = BD = 3.7 cm.
BDEF is required square.
2. √22 cm.
Solution: From a line segment AX, cut 2 cm & 11 cm. Now draw the mean proportional of AB & BC.
∴ BD is the mean proportional of length √22
Now draw a square whose one side = BD = 4.6 cm.
BDEF is required square.
3. √31 cm.
Solution: From a line segment AX, cut 5 cm & 6.2 cm.
Now draw the mean proportional of AB & BC.
∴ BD is the mean proportional of length √31
Now draw a square whose one side = BD 5.6 cm.
BDEF is required square.
4. √33 cm.
Solution: From a line segment AX, cut 3 cm & 11 cm. Now draw the mean proportional
of AB & BC.
∴ BD is the mean proportional of length √33 cm.
Now draw a square whose one side = BD = 5.7 cm.
BDEF is required square.
Maths WBBSE Class 10 Solutions Question 5. Let us draw the squares whose areas are equal to the areas of the rectangles by taking the following lengths as its sides :
1. 8 cm, 6 cm.
Solution: Draw a rectangle ABCD with 8 cm & 6 cm.
From produced DC, cut CE equals CB.
Now draw a semicircle with DE as the diameter.
CF cuts the semicircle at F.
Now draw a square with CF as one side.
CFGH is the required square.
2. 6 cm, 4 cm.
Solution: Draw a rectangle ABCD with 6 cm & 4 cm.
From produced DC, cut CE equals CB.
Now draw a semicircle with DE as diameter, and CF cuts the semicircle at F.
Now draw a square with CF as one side.
CFGH is the required square.
3. 4.2 cm., 3.5 cm.
Solution: Draw a rectangle with 4.2 cm & 3.5 cm.
From produced DC, cut CE equals CB.
Now draw a semicircle with DE as diameter, and CF cuts the semicircle at F.
Now draw a square with CF as one side.
CFGH is the required square.
4. 7.9 cm., 4.1 cm.
Solution: Draw a rectangle with 7.9 cm & 4.1 cm.
From produced DC, cut CE equals CB.
Now draw a semicircle with DE as diameter, and CF cuts the semicircle at F.
Now draw a square with CF as one side.
CFGH is the required square.
Question 6. Let us draw the squares whose areas are equal to the areas of the following triangles :
1. The lengths of the three sides of a triangle are 10 cm, 7 cm, and 5 cm respectively.
Solution: Draw a triangle ABC, whose AB = 5 cm, BC= 10 cm, & AC = 7 cm.
Bisect BC at D.
Draw a line parallel to BC, through A, which cut the perpendicular bisector of BC at E.
Take EF = DC.
CDEF is the required rectangle whose area is equal to ΔABC.
Now to draw a square from produce EF, cut EM equal to DE.
Draw a semi-circle with diameter EM to produce CF which cut the semi-circle at K.
Now draw the square FGHK, whose area is equal to ΔABC.
Proof : ABC = 1/2 × BC X DE = DC X DE = DC X CF
= EF X FM [DC= EF and FC = FM]
= FK² [ FD, EF & FM proportionate]
= FGHK Sq.
2. An isosceles triangle whose base is 7 cm in length and the length of each of two equal sides is 5 cm.
Solution:
Proof: PERQ sq.
= PE²
= AE.EF [ PE. AE & EF proportionate]
= AE.EC [ EF = EC]
= DC.EC [AE = DC]
= 1/2 × BC x AD = ΔABC
Maths WBBSE Class 10 Solutions 3. An equilateral triangle whose side is 6 cm in length.
Solution:
Proof: AABC =1/2 BCX AD = DC X AD = AE X EC
= AE X EM = PE² [ PE, AE & EM proportionate]
= EPQR Sq.
First, draw an equilateral triangle ABC of side 6 cm. Now bisect the base BC at D.
Next draw a straight line through A, parallel to BC, bisect BC at D draw a rectangle ADCE whose area is equal to the AABC.
From produced AF cuts AM equal to CE.
Next, draw a semi-circle with a diameter AM.
Produce CE which cuts the semi-circle at P.
Draw the square EPQR, whose area is equal to the AABC.
