Question

9. If the perimeter of an isosceles triangle is 83 cm. and its base is 23 cm. Find the measure of its equal sides.
10. The perimeter of a rhombus is 5.6 cm. Find its side.
11. The perimeter of a rectangle is equal to that of a square with side 17 m. If the length of the
rectangle is 26 m. What is its breadth ?
12. What is the cost of fencing a rectangular field of length 16 m and breadth 11 m at the rate of 12 per m. Please solve all questions..step by step.​

Answer 1

Solution :

9 .

Let the equal side of the isosceles triangle be x cm .

Perimeter :

> 2x + 23 = 83

> 2x = 60

> x = 30 .

The measure of its equal sides Is 30 cm.

10. The perimeter of a rhombus is 5.6 cm.

In a rhombus, all the sides are equal .

If a side is a cm .

> 4a = 5.6

> a = 1.4 cm .

The side of the rhombus is 1.4 cm .

11. The perimeter of the rectangle is equal to the perimeter of a square ; the side of the square is 17 m .

Perimeter of the square = 4 × Side = 68 m .

Let the breadth of the rectangle be b m.

> 2[ 26 + b ] = 68

b + 26 = 34

> b = 8 m .

The breadth of the rectangle is 8 m .

12.

Perimeter of the fence :

> 2( L + B )

> 2( 27 )

> 54 m .

Cost -

> Rs. 54 × 12

> Rs. 648 .

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Answer 2

Answer:

Required Answers :-

(1)

As we know that Perimeter is the sum of all sides.

Let the both side be x

83 = 23 + (x + x)

83 - 23 = 2x

60 = 2x

30 = x

Other two sides are 30 cm.

(2)

As we know that

Side of Rhombus = Perimeter/4

Side = 5.6/4

Side = 1.4 cm

Side is 1.4 cm

(3)

Perimeter of square = 4 × s

Perimeter = 4 × 17

Perimeter = 68 cm

Now,

Perimeter of rectangle = 2(l + b)

68 = 2(26 + b)

68/2 = 26 + b

34 = 26 + b

34 - 26 = b

8 = b

(4)

Perimeter = 2(l + b)

Perimeter = 2(16 + 11)

Perimeter = 2(27)

Perimeter = 54 cm

Now,

Total cost = 54 × 12

Total cost = ₹648.