10类RD Sharma解–第8章二次方程式–练习8.10

问题1.直角三角形的斜边为25厘米。三角形的其他两个边的长度之差为5厘米。找到这些边的长度。

解决方案：

Let the length of other two sides of the triangle be x and y.

Therefore, according to the question,

x – y = 5

or,

x = y + 5

Now the two sides are of length y and y + 5.

Applying Pythagoras theorem on the right-angled triangle, we get,

25² = y² + (y + 5)²

625 = y² + y² + 25 + 10y

2y² + 25 + 10y = 625

2y² + 10y = 625 – 25

2y² + 10y – 600 = 0

y² + 5y – 300 = 0

y² + 20y – 15y – 300 = 0 [By middle term splitting]

y(y + 20) – 15(y + 20) = 0

(y – 15)(y + 20) = 0

y -15 = 0 or y + 20 = 0

y = 15 or y = -20

As the length of a side is positive, we will take y = 15.

Therefore,

x = y + 5

x = 15 + 5

x = 20

The two sides of the triangle are 15 cm and 20 cm.

问题2.矩形场的对角线比短边大60米。如果较长的一面比较短的一面多30米，请找到视野的两边。

解决方案：

Let the shorter side of the rectangle be x.

Now, according to the question, length of rectangle is x + 60 and length of the longer side of rectangle is x + 30.

Now, we will apply Pythagoras theorem on the right-angled triangle formed by the diagonal and the two sides,

(x + 60)² = (x + 30)² + x²

x² + 3600 + 120x = x² + 900 + 60x + x²

3600 – 900  = x² + 60x – 120x

2700 = x² – 60x

x² – 60x – 2700 = 0

x² – 90x + 30x – 2700 = 0

x(x – 90) + 30(x – 90) = 0

(x + 30)(x – 90) = 0

x + 30 = 0 or x – 90 = 0

x = -30 or x = 90

We will take positive side, i.e. x = 90

Now, 

length of shorter side is 90 m

length of diagonal is 90 + 60 = 150 m

length of longer side is 90 + 30 = 120 m

问题3.直角三角形的斜边为3√10厘米。如果小腿增加了三倍，长腿增加了两倍，则新的斜边将为9√5厘米。三角形的腿多长?

解决方案：

Let the smaller and longer side of the triangle be x and y.

Now applying Pythagoras theorem, we get,

(3√10)² = x² + y²

x² + y² = 90     ……… (i)

Now the smaller and longer sides are tripled and doubled respectively,

Therefore, the new sides of the triangle will be 3x and 2y.

Applying Pythagoras theorem,

(9√5)² = (3x)² + (2y)²

9x² + 4y² = 405     ……… (ii)

Now multiplying (i) by 4 and then subtracting (i) from (ii), we get

9x² – 5x² + 4y² – 4y²= 405 – 360

5x² = 45

x² = 9

x = 3

Substituting value of x in (i),

9 + y² = 90

y² = 81

y = 9

Length of smaller side is 3 cm.

Length of longer side is 9 cm.

问题4：必须在直径13米的圆形停车场边界上的某个点竖立一根杆，以使其与边界上两个直径相对的固定闸门A和B的距离之差为7米。有可能这样做吗?如果是，应该在两个闸门的距离上竖起电线杆吗?

解决方案：

In the circle, let P be the pole on the circumference of the circle and points A and B be the two diametrically opposite fixed gates.

Let length of PB be x.

Given,

PA – PB = 7

PA = x + PB

PA = x + 7

The triangle PQR is a right-angled triangle at P, as AB is the diameter of the circle.

Now applying Pythagoras theorem,

AB2 = PA2 + PB2

13² = (x + 7)² + x²

169 = x² + 49 + 14x + x²

169 = 2×2 + 14x + 49

2x² + 14x + 49 – 169 = 0

2x² + 14x – 120 = 0

x² + 7x – 60 = 0

x² + 12x – 5x – 60 = 0

x(x + 12) – 5(x + 12) = 0

(x – 5)(x + 12) = 0

x = 5 or x = -12

Only x = 5 is possible.

PB = 5 m

PA = 5 + 7 = 12 m

P should be erected at a distance of 5 m from PB and 12 m from PA.