Selina Concise Class 10 Math Chapter 6 Solving (simple) problems (Based on Quadratic Equations) Exercise 6B Solutions
Solving (Simple) problems
(Based on Quadratic Equations)
Exercise 6B
(Q1) The sides of a right-angled triangle containing the right angle are 4x cm and (2x – 1) cm. If the area of the triangle is 30cm2; calculate the lengths of its sides
Solution:
△ ABC is a right angle triangle.
Right angle
∠B = 90°
The adjacent side of right angle is AB = 4x cm
and the other adjacent side of right is BC = (2x – 1).
Given that the area of triangle is 30 cm2.
By using formula,
Area of triangle = 1/2 × base × height
30 = 1/2 × BC × AB
30 = 1/2 × (2x – 1) × 4x
30 = ½ × (2x – 1) × 4x
30 = 4x2 – 2x
Dividing by ‘2’ on both sides, 30/2 = 4x2/2 – 2x/2
15 = 2x2 – 2x
2x2 – x – 15 = 0
2x2 – 6x + 5x – 15 = 0
2x (x – 3) + 5 (x – 3) = 0
(x – 3) (2x+ 5) = 0
x – 3 = 0 or 2 + 5 = 0
2x = -5
x = 3 or x = -5/2
But the side of right angled triangle cannot be negative.
So, x = 3 is the only solution.
∴ AB = 4x = 4×3 = 12cm
BC = 2x – 1= 2×3 – 1 = 6-1 = 5cm
We have to find hypotenuse
By using Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = (12)2 + (5)2
AC = √(12)2 + (5)2
AC = √144 + 25
AC = √169
AC = 13cm
(Q2) The hypotenuse of a right-angled triangle is 26cm and the sum of other two sides is 34cm. Find the lengths of its sides
Solution:
Given that, the hypotenuse of a right – angled triangle is 26cm
∴ AC = 26cm
and the sum of other two sides is 34cm
AB + BC = 34cm
We don’t know the other two sides AB and BC.
So, let us consider the other two sides are x and (34 – x) cm
by using Pythagoras theorem,
AC2 = AB2 + BC2
(26)2 = x2 + (34 – x)2
676 = x2 + (34)2 – 2×(34) x + x2
676 = x2 + x2 + 1156 – 68x
676 = 2x2 + 1156 – 68x
2x2 – 68x + 1156 – 676 = 0
2x2 – 68x + 480 = 0
Dividing by ‘2’
2x2/2 – 68x/2 + 480/2 = 0
x2 – 34x + 240 = 0
x2 – 10x – 24x + 240 = 0240
x (x – 10) – 24 (x – 10) = 0
(x – 10) (x – 24) = 0
x – 10 = 0 or x – 24 = 0
x = 10 or x = 24
34 – x = 34 – 24 = 0
Therefore, the lengths of the three sides of the right angled triangles are AB = 10cm, BC = 24cm and AC = 26cm
(Q3) The sides of a right – angled triangle are (x – 1) cm, 3x cm and (3x + 1) cm Find:
(i) The value of x
(ii) The lengths of it’s sides
(iii) Its area
Solution:
Given:
The hypotenuse
AC = (3x + 1) cm
and the adjacent two sides are AB = 3x cm and BC = (x – 1) cm
By using Pythagoras theorem
AB2 + BC2 = AC2
AC2 = AB2 + BC2
(3x + 1)2 = (3x)2 + (x – 1)2
(3x)2 + 2×(3x) ×1 + (1)2 = 9x2 + x2 – 2x + 1
9x2 + 6x + 1 = 9x2 + x2 – 2x + 1
(a + b)2 = a2 + 2ab + b2
6x + 1 = x2 – 2x + 1 (a – b)2 = a2 – 2ab + b2
[9x2 are circle because on both side are 9x2 term of same sign]
x2 – 2x – 6x + 1 – 1 = 0
x2 – 8x + 0 = 0
x2 – 8x = 0
x (x – 8) = 0
x = 0 and x – 8 = 0
x = 8
If x = 0 then 3x = 3 (0) = 0
In right angled triangle the side is not zero.
∴ So, x = 8 is the only solution.
∴ If x = 8 then 3x + 1 = 3(8) + 1
= 24 + 1
= 25 cm
If x 8 then x – 1 = 8 – 1
= 7 cm
If x = 8 then 3x = 3×8 = 24cm
∴ The lengths of right angled triangle are –
AB = 3x cm = 24cm
BC = x – 1cm = 7cm
and AC = (3x + 1) cm = 25cm
Area of the triangle =
½ × base × height
= ½ × BC × AB
= ½ × 7 × 24
= 7 × 12
= 84 cm2
∴ The area of right angled triangle is 84 cm2.
(Q4) The hypotenuse of a right –angled triangle exceeds one side by 1 cm and the other side by 18cm; find the lengths of the sides of the triangle.
Solution:
Let us consider the hypotenuse of a right-angled triangle is x cm.
Given:
The hypotenuse of a right-angle triangle exceeds on side by 1 cm.
∴ (x – 1 cm
and the other side by 18cm
(x – 18) cm
By using Pythagoras theorem,
AC2 = AB2 + BC2 (x – 18)
x2 = (x – 18)2 + (x – 1)2
x2 = x2 – 2 × 18 × x + (18)2 + x2 – 2x + 1
= x2 – 36x + 324 + x2 – 2x + 1
x2 = 2x2 – 38x + 325
2x2 – 38x + 325 – x2 = 0
x2 – 38x + 325 = 0
x2 – 13x – 25x + 325 = 0
x (x – 13) – 25 (x – 13) = 0
(x – 13) (x – 25) = 0
(x – 13) = 0 or (x – 25) = 0
x = 13 or x = 25
If x = 1 then x – 18 = 13 – 18
= -5
But the side of triangle is not negative. So x = -5 is not possible
∴ If x = 25 then x – 1 = 25 – 1 = 24cm
If x = 25 then x – 18 = 25 – 18 = 7cm
∴ The lengths of the sides are AB = 7cm, BC = 24cm and AC = 25cm
(Q5) The diagonal of a rectangle is 60m more than it’s shorter side and the larger side is 30 cm more than the shorter side. Find the sides of the rectangle.
Solution:
Given:
The diagonal of a rectangle is 60 cm more than its shorter side.
∴ The larger side is 30cm more than the shorter side.
∴ (x + 30) m
The diagonal of a rectangle i.e. is (x + 60)
by using Pythagoras theorem, In △BDC
(hypotenuse)2 = (Base)2 × (height)2
(BD)2 = (DC)2 × (BC)2
(x + 60)2 = (x + 30)2 × x2
x2 + 2 × (60) × x + (60)2 = x2 + 2 × (30) × x + (30)2 + x2
[∵ (a + b)2 = a2 + 2ab + b2]
x2 + 120x + 3600 = x2 + 60x + 900 + x2
x2 + 120x + 3600 = 2x2 + 60x + 900
2x2 – x2 + 60x – 120x + 900 – 3600 = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x (x – 90) + 30 (x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0 or x + 30 = 0
x = 90 or x = -30
Side of rectangle cannot be negative
∴ x = -30 is not solution.
∴ x = 90 is not only solution
∴ The side of the rectangle are x = 90m and (x + 30)m = 90+30
= 120m
(Q6) The perimeter of a rectangle is 104 m and it’s area is 640 m2, find its length and breadth.
Solution:
Let us consider the length be x m and breadth be y m.
∴ The perimeter of a rectangle is 104m.
∴ by using formula,
The perimeter of a rectangle = 2 (length × breadth)
= 2 (x + y) m
104 = 2 (x + y)
104 = 2x + 2y
dividing 2,
104/2 = 2x + 2y/2
= 2x/2 + 2y/2
52 = x + y
52 – x = y
y = 52 – x
And also given that the area of rectangle is 640 m2
∴ Area of rectangle = length × breadth
640 = x ×
640 = x (52 – x)
640 = 52x – x2
x2 – 52x + 640 = 0
x2 – 32x – 20x + 640 = 0
x (x – 32) – 20 (x – 32) = 0
(x – 32) (x – 20) = 0
x – 32 = 0 or x – 20 = 0
x = 32 or x = 20
If x = 32 then y = 52 – 32
= 20
If x = 20 then y = 52 – 20
= 32
∴ The length of the rectangle is x = 32m and breadth of the rectangle is x = 20m.
Here is your solution of Selina Concise Class 10 Math Chapter 6 Solving (simple) problems (Based on Quadratic Equations) Exercise 6B
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