Class 7 Exam  >  Class 7 Notes  >  Mathematics (Maths) Class 7  >  HOTS Question: Perimeter and Area

Class 7 Maths Chapter 9 HOTS Questions - Perimeter and Area

Q1: DL and BM are the heights on sides AB and AD respectively of ABCE parallelogram. Let us suppose the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm. Calculate the length of the BM and the DL.
Ans:

According to the question given above,
The Area of the parallelogram = 1470 cm2
AB = 35 cm
AD = 49 cm
Then
It is clear that
The Area of the parallelogram = base × height
1470 = AB × BM
1470 = 35 × DL
1470/35 = DL
DL = 42 cm
Area of the parallelogram = base × height
1470 = AD × BM
1470 = 49 × BM
BM = 1470/49
BM = 30 cm

Q2: Calculate the area of the shaded portion.
Ans:
To find the area of EFDC, the area of triangle AEF, EBC and Rectangle ABCD has to be calculated.
To calculate the area we will apply the formula, ½ × base × height
Area of AEF = ½ × 6 × 10  = 30 cm2
Area of EBC = ½ × 8 × 10  = 40 cm2
Area of rectangle ABCD = length × breadth = 18 × 10 = 180 cm2
Area of EFDC will be calculated as,
ABCD area – (AEF – EBC)
= 180 – (30 + 40)
= 180 – 70
= 110 cm2

Q3: Find the circumference of the circle with a radius of 14 cm. take the value of =22/7
Ans:

As per the question given, the radius of the circle = 14 cm
Circumference of the circle will be calculated by the formula = 2r
= 2 × 22/7 ×14 = 2 × 22 × 2 = 88 cm 

Q4: The right-angled triangle is ABC given. This is right-angled at A. AD is perpendicular to BC. If AB = 5 cm and BC = 13 cm and AC = 12 cm. Calculate the area of the ABC. Also, calculate the length of the AD.
Ans: 
According to the given question,
AB = 5 cm
BC = 13 cm
AC = 12 cm
We know that
Area of the triangle ABC = ½ × Base × height
= ½ × AB × AC
=½ × 5 × 12 = 1 × 5 × 6 = 30 cm2
Now,
Area of the triangle ABC = ½ × Base × height
30 = ½ × AD × BC
30 = ½ × AD × 13
30 × 2/13 = AD
AD = 60/13 = 4.6 cm

Q5: Calculate the perimetre of the figure. The figure is a semicircle including its diameter.
Ans:

As per the given question,
The Diameter of the semi-circle = 10 cm
Radius = r = d/2 = 10/2 = 5 cm
Circumference of the semi-circle = 22/7 × 5 = 110/7 = 15.71 cm
To calculate the perimeter of the above figure,
Perimeter of the semi-circle = semi-circle circumference + semicircle diameter= 15.71 + 10 = 25.71 cm

Q6: The circumference of a circular sheet is 154 m. calculate the radius. Calculate the area of the sheet as well.
Ans:

As per the question, it is given that
The Circumference of the circle = 154 m
We know the formula
Circumference of the circle will be calculated by the formula = 2 r
154 = 2 × 22/7 × r
154 = 44/7 × r
r = 154 × 7/ 44
r = 14 × 7/ 4
r = 7 × 7/ 2
r = 49/2 = 24.5 m
area of the circle = r2 = 22/7 × (24.5)2  = 22/7 × 600.25
= 22 × 85.75 = 1886.5 m2

Q7: Fill in the blanks:
(i) ________ is the distance around a closed figure.
(ii) _________ is the part of the plane which is occupied by the closed figure.
(iii) The formula to calculate the perimetre of the square will be presented as  _______
(iv) The formula to calculate the perimetre of the rectangle will be presented as _______
(v) The area of the square is calculated as ________
(vi) The area of a rectangle is calculated as ________
(vii) The area of the parallelogram is calculated as _______
(viii) ___________ is the distance around a circular region.

Ans:
(i) Perimeter is the distance around a closed figure.
(ii) Area  is the part of the plane which is occupied by the closed figure.
(iii) The formula to calculate the perimetre of the square will be presented as  4 × side
(iv) The formula to calculate the perimetre of the rectangle will be presented as  2 × (length + breadth)
(v) The area of the square is calculated as side × side
(vi) The area of a rectangle is calculated as length × breadth
(vii) The area of the parallelogram is calculated as base × height
(viii) Circumference is the distance around a circular region.

Q8: The rate of polishing is Rs 15/m2. Find the cost of polishing a circular table – top which has a diameter of 1.6 m.
Ans:
As per the given question,
The diameter of the circular table – top = 1.6 m
As we know the
Radius = r = d/2 = 1.6 / 2 = 0.8 m
Area of the circular table top will be 3.14 × 0.8 × 0.8 = 2.0096 m2
The cost of polishing 1 m2 area = Rs 15
So for calculating the 2.0096 m2 area = Rs 15 × 2.0096 = 30.144
So in order to polish the area of 2.0096 m2, the cost incurred is Rs 30.144

Q9: Find the area of the quadrilateral ABCD. See the figure given below.
Answer. 
As shown in the figure,
AC = 22 cm, BM = 3 cm
DN = 3 cm
BM is perpendicular to AC
DN is perpendicular to AC.
It is clear from the figure that in order to calculate the area of the quadrilateral ABCD, the first step is the calculation of the area of the triangle ABC and ADC.
To calculate the area of the triangle ABC,
= ½ × base × height
= 1 × 11 × 3 = 33 cm2
To calculate the area of the triangle ADC,
= ½ × base × height
= ½ × 22 × 3 = 33 cm2
Area of the quadrilateral = area of ABC + area of ADC
= 33 + 33 = 66 cm2

Q10: Look at the figure given below, the circular card sheet has a radius of 14 cm. Two circles are removed of radius 3.5 cm and a rectangle is also removed, the length and breadth of the rectangle are 3 cm and 1cm respectively. Given the value = 22/7, calculate the area of the remaining sheet by applying the suitable formulas.
Ans:

From the question, it is clear that
The Radius of the circular sheet is 14 cm
The Radius of the two small circles is 3.5 cm
Length of the rectangle is 3 cm
Breadth of the rectangle is 1 cm
To calculate the remaining area,
The Area of the circular card sheet = 22/4 × 14 × 14 = 22 × 2 × 14 = 616 cm2
To calculate the area of the two small circles,
= 2 × (22/7 × 3.5 × 3.5)
=2 × ((22/7) 12.25)
= 2 × 38.5
= 77 cm2
To calculate the area of the rectangle = length × breadth = 3 × 1 = 3 cm2
To calculate the area of the remaining part,
Area of card sheet – ( area of 2 small circles + rectangle area)
= 616 – (77 + 3)
= 616 – 80
= 536 cm2

The document Class 7 Maths Chapter 9 HOTS Questions - Perimeter and Area is a part of the Class 7 Course Mathematics (Maths) Class 7.
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