第 10 类 RD Sharma 解——第 8 章二次方程——练习 8.11

问题1：矩形场的周长为82 m，面积为400 m ² 。求矩形的宽度？

解决方案：

Given:

Perimeter = 82 m and its area = 400 m²

Let the breadth of the rectangle be ‘b’ m.

As we know,

Perimeter of a rectangle = 2×(length + breadth)

82 = 2×(length + b)

41 = (length + b)

Length = (41 – b)m

As we know,

Area of the rectangle = length × breadth

400 = (41 – b)×(b)

400 = 41b – b²

b² – 41b + 400 = 0

b² – 25b – 16b + 400 = 0

b×(b – 25) – 16×(b – 25) = 0

(b – 16)(b – 25) = 0

Now,

either b – 16 = 0

⇒ b = 16

Or, b – 25 = 0

⇒ b = 25

Hence,

The breadth of the rectangle can be either 16 m or 25 m respectively.

问题2：大厅的长度比宽度大5m。如果大厅的地板面积是 84 m ² ，大厅的长宽是多少？

解决方案：

Let the breadth of the rectangle be ‘b’ m

Then,

The length of the hall is 5 m more than its breadth i.e, = (b + 5) m

Given, area of the hall is = 84 m²

As the hall is rectangular,

Area of the rectangular hall = length × breadth

84 = b(b + 5)

b² + 5b – 84 = 0

b² + 12b – 7b – 84 = 0

b(b + 12) – 7(b + 12) = 0

(b + 12)(b – 7) = 0

Hence,

either b + 12 = 0

⇒ b = – 12 m (Side of a rectangle cannot be negative)

Or, b – 7 = 0

⇒ b = 7 m

So, only b = 7 m is considered.

⇒ b + 5 = 12

Therefore,

The length and breadth of the rectangle is 7 m and 12 m respectively.

问题 3：两个正方形的边分别为 x 和 (x + 4) cm。它们的面积之和为 656 cm ² 。找到正方形的边。

解决方案：

Let A and B be the two squares.

And, let ‘s’ cm be the side square A and (s + 4) cm be the side of the square B.

So,

Area of the square A = s² cm²

Area of the square B =(s + 4)² cm²

Given:

Area of the square A + Area of the square B = 656 cm²

⇒ s² cm² + (s + 4)² cm² = 656 cm²

s²+ s² + 16 + 8s – 656 = 0

2s² + 16 + 8s – 656 = 0

2(s² + 4s – 320) = 0

s² + 4s – 320 = 0

s² + 20s – 16s – 320 = 0

s(s + 20) – 16(s + 20) = 0

(s + 20)(s – 16) = 0

Now,

either s + 20 = 0

⇒ s = -20 (Side of a square cannot be negative)

Or, s – 16 = 0 ⇒ s = 16

Hence, the value of s = 16 ⇒ s + 4 = 20

Therefore,

The side of the square A= 16 cm

The side of the square B = 20 cm

问题4：直角三角形的面积是165 cm ² 。如果后者超过前者7m，则确定其基础和高度。

解决方案：

Let the altitude of the right angle triangle be ‘a’ m

Given:

The altitude exceeds the base by 7m ⇒ altitude = (a – 7)m

As we know,

Area of the triangle = 1/2 × base × altitude

⇒ 165 = 1/2 × (a − 7) × a

a(a – 7) = 330

a² – 7a – 330 = 0

a² – 22a + 15a – 330 = 0

a(a – 22) + 15(a – 22) = 0

(a – 22)(a + 15) = 0

Now,

either a – 22 = 0 ⇒ a = 22

Or, a + 15 = 0 ⇒ a = -15 (Altitude of a triangle cannot be negative)

So the value of a = 22 is only considered

⇒ a – 7 = 15

Hence,

The base and altitude of the right angled triangle are 15 cm and 22 cm respectively.

问题5：是否可以设计一个长2倍宽、面积800m ²的长方形芒果园？如果有，求它的长度和宽度。

解决方案：

Let the breadth of the rectangular mango grove be ‘b’ m

Given:

The length of rectangle is twice of its breadth.

So, length = 2b

Area of the grove = 800 m² (given)

As we know,

Area of the rectangle = length × breadth

800 = b × (2b)

2b² – 800 = 0

b² – 400 = 0

⇒ b = √400 = 20 (neglecting the negative sq. root as side can never be negative)

Therefore,

The breadth of the rectangular groove is 20 m.

And, the length of the rectangular groove is 40 m.

Yes, it is possible to design a rectangular groove whose length is twice of its breadth.

问题6：是否可以设计一个周长80m、面积400m²的长方形公园？如果是，求它的长度和宽度：

解决方案：

Perimeter of rectangular park = 80 m

Length + Breadth = 80/2 = 40 m

Let length = x m

Them breadth = 40 – x

According to the condition,

Area = Length x Breadth

x (40 – x) = 400

⇒ 40x – x² = 400

⇒ x² – 40x + 400 = 0

⇒ (x – 20)² = 0

⇒ x – 20 = 0

⇒ x = 20

Yes, it is possible

Length = 20 m

and breadth = 40 – x = 40 – 20 = 20 m

问题 7：两个正方形的面积之和为 640 平方米。如果它们的周长差为 64 m，求两个正方形的边。 [CBSE 2008]

解决方案：

Let the side of the first square = x m and of second squares = y m

Given:

4x – 4y = 64

⇒ x – y = 16 ….(i)

and x² + y² = 640 ….(ii)

From (i), x = 16 + y

In (ii)

(16 + y)² + y² = 640

⇒ 256 + 32y + y² + y² = 640

⇒ 2y² + 32y + 256 – 640 = 0

⇒ y² + 16y – 192 = 0 (Dividing by 2)

⇒ y² + 24y – 8y – 192 = 0

⇒ y (y + 24) – 8 (y + 24) = 0

⇒ (y + 24)(y – 8) = 0

Either y + 24 = 0, then y = -24 (which is not possible as it is negative)

or y – 8 = 0, then y = 8

x = 16 + y = 16 + 8 = 24

Side of first square = 24 m

and side of second square = 8 m