NCERT Solutions Class 7 Maths Perimeter And Area Exercise 9.1 – Free PDF

NCERT Solutions for Class 7 Maths Chapter 9 – Perimeter and Area Ex 9.1 are given here in easy, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. Perimeter is the length all around a closed shape. Area is the space inside it. You already know how to find the area of a square and a rectangle. Exercise 9.1 takes the next step and teaches you the area of a parallelogram and the area of a triangle.

This exercise has 8 questions. Some ask you to find the area when the base and height are given. Some give you the area and ask you to find a missing base or height. The later questions are word problems where you find a second height of the same shape using its area. Every question below is solved in short steps, with the final answer in bold. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.

NCERT Solutions Class 7 Maths Perimeter And Area Exercise 9.1 – Free PDF

Perimeter and Area

Medium

Q.

Find the missing values :
 

S. NO.	Base	Height	Area of the Parallelogram
a.	20 cm		246 cm2
b.		15 cm	154.5 cm2
c.		8.4 cm	78.72 cm2
d.	15.6 cm		16.38 cm2

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Perimeter and Area

Medium

Q.

Find the missing values :
 

Base	Height	Area of Triangle
15 cm	……………..	84 cm2
……………..	31.4 mm	1256 mm2
22 cm	……………..	170.5 cm2

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Perimeter and Area

Medium

Q.

$\begin{array}{l}\mathrm{DL}\quad \mathrm{and}\quad \mathrm{BM}\quad \mathrm{are}\quad \mathrm{the}\quad \mathrm{heights}\quad \mathrm{on}\quad \mathrm{sides}\quad \mathrm{AB}\quad \mathrm{and}\quad \mathrm{AD}\quad \mathrm{respectively}\\ \mathrm{of}\quad \mathrm{parallelogram}\quad \mathrm{ABCD}.\quad \mathrm{If}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{parallelogram}\quad \mathrm{is}\\ 1470\quad {\mathrm{cm}}^{\mathrm{2}},\mathrm{AB}\quad =\quad 35\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{AD}\quad =\quad 49\quad \mathrm{cm},\quad \mathrm{find}\quad \mathrm{the}\quad \mathrm{length}\quad \mathrm{of}\\ \mathrm{BMand}\quad \mathrm{DL}.\end{array}$

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Perimeter and Area

Medium

Q.

$\begin{array}{l}\triangle \mathrm{ABC}\quad \mathrm{is}\quad \mathrm{right}\quad \mathrm{angled}\quad \mathrm{at}\quad \mathrm{A}\quad .\mathrm{AD}\quad \mathrm{is}\quad \mathrm{perpendicular}\quad \mathrm{to}\quad \mathrm{BC}.\\ \mathrm{If}\quad \mathrm{AB}=5\quad \mathrm{cm},\hspace{0.33em}\mathrm{BC}=13\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{AC}=12\quad \mathrm{cm},\quad \mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\\ \triangle \mathrm{ABC}.\quad \mathrm{Also}\quad \mathrm{find}\quad \mathrm{the}\quad \mathrm{length}\quad \mathrm{of}\quad \mathrm{AD}.\end{array}$

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Perimeter and Area

Medium

Q.

$\begin{array}{l}\triangle \mathrm{ABC}\quad \mathrm{is}\quad \mathrm{isosceles}\quad \mathrm{with}\quad \mathrm{AB}=\mathrm{AC}=7.5\quad \mathrm{cm}\quad \mathrm{and}\quad \mathrm{BC}=9\quad \mathrm{cm}.\quad \\ \mathrm{The}\quad \mathrm{height}\quad \mathrm{AD}\quad \mathrm{from}\quad \mathrm{A}\quad \mathrm{to}\quad \mathrm{BC},\quad \mathrm{is}\quad 6\quad \mathrm{cm}.\quad \mathrm{Find}\quad \mathrm{the}\quad \mathrm{area}\quad \mathrm{of}\quad \triangle \mathrm{ABC}.\quad \\ \mathrm{What}\quad \mathrm{will}\quad \mathrm{be}\quad \mathrm{the}\quad \mathrm{height}\quad \mathrm{from}\quad \mathrm{C}\quad \mathrm{to}\quad \mathrm{AB}\quad \mathrm{i}.\mathrm{e}.,\quad \mathrm{CE}?\end{array}$

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Formulas Used in Exercise 9.1

• Area of a parallelogram = base × height
• Area of a triangle = 1/2 × base × height
• To find the height of a parallelogram: height = area ÷ base
• To find the height of a triangle: height = (2 × area) ÷ base

Remember one thing. The height is always the perpendicular distance from the base to the opposite side or corner. It is not the slanting side.

Exercise 9.1 – Questions and Answers

Exercise 9.1 – Questions and Answers

Q1. Find the area of each of the following parallelograms:

Answer:

(a)
Area of parallelogram = Base × Height
Height = 4 cm
Base = 7 cm
So,
Area of parallelogram = 7 × 4
= 28 cm²

(b)
Area of parallelogram = Base × Height
Height = 3 cm
Base = 5 cm
So,
Area of parallelogram = 5 × 3
= 15 cm²

(c)
Area of parallelogram = Base × Height
Height = 3.5 cm
Base = 2.5 cm
So,
Area of parallelogram = 2.5 × 3.5
= 8.75 cm²

(d)
Area of parallelogram = Base × Height
Height = 4.8 cm
Base = 5 cm
So,
Area of parallelogram = 5 × 4.8
= 24 cm²

(e)
Area of parallelogram = Base × Height
Height = 4.4 cm
Base = 2 cm
So,
Area of parallelogram = 2 × 4.4
= 8.8 cm²

Q2. Find the area of each of the following triangles:

Answer:

(a)
Area of a triangle = ½ × Base × Height
= ½ × 4 × 3
= ½ × 12
= 6 cm²

(b)
Area of a triangle = ½ × Base × Height
= ½ × 5 × 3.2
= ½ × 16
= 8 cm²

(c)
Area of a triangle = ½ × Base × Height
= ½ × 3 × 4
= ½ × 12
= 6 cm²

(d)
Area of a triangle = ½ × Base × Height
= ½ × 3 × 2
= ½ × 6
= 3 cm²

Q3. Find the missing values:

Answer:
Let the height be h and the base be b.

(a) Area of parallelogram = Base × Height
So, we get
246 cm² = 20 cm × h
h = 246 cm² ÷ 20 cm
= 12.3 cm

(b) Area of parallelogram = Base × Height
So, we get
154.5 cm² = b × 15 cm
b = 154.5 cm² ÷ 15 cm
= 10.3 cm

(c) Area of parallelogram = Base × Height
So, we get
48.72 cm² = b × 8.4 cm
b = 48.72 cm² ÷ 8.4 cm
= 5.8 cm

(d) Area of parallelogram = Base × Height
So, we get
16.38 cm² = 15.6 cm × h
h = 16.38 cm² ÷ 15.6 cm
= 1.05 cm

So, we get the completed table:

Q4. Find the missing values:

Answer:
Let the height be h and the base be b.

(a) Area of triangle = ½ × base × height
87 cm² = ½ × 15 cm × h
h = (87 cm² × 2) ÷ 15 cm
= 11.6 cm

(b) Area of triangle = ½ × base × height
1256 mm² = ½ × b × 31.4 mm
b = (1256 mm² × 2) ÷ 31.4 mm
= 80 mm

(c) Area of triangle = ½ × base × height
170.5 cm² = ½ × 22 cm × h
h = (170.5 cm² × 2) ÷ 22 cm
= 15.5 cm

So we get the completed table:

Q5. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm, find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm

Answer:

(a)
Area of a parallelogram = Base × Height
= SR × QM
= 12 × 7.6
= 91.2 cm²

(b)
The area stays the same whichever base and matching height we use.
Area of a parallelogram = Base × Height
= PS × QN
91.2 = 8 × QN
QN = 91.2 ÷ 8
= 11.4 cm

Q6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Answer:
Area of parallelogram = Base × Height
= AB × DL
1470 = 35 × DL
DL = 1470 ÷ 35
= 42 cm

Also, taking AD as the base,
1470 = AD × BM
1470 = 49 × BM
BM = 1470 ÷ 49
= 30 cm

Q7. △ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of △ABC. Also find the length of AD.

Answer:
Since the triangle is right angled at A, the sides AB and AC are perpendicular to each other. So AB can be taken as the base and AC as the height.
Area = ½ × Base × Height
= ½ × 5 × 12
= 30 cm²

Also, taking BC as the base and AD as the height,
Area of triangle = ½ × AD × BC
30 = ½ × AD × 13
AD = (30 × 2) ÷ 13
= 4.6 cm (approximately)

Q8. △ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC is 6 cm. Find the area of △ABC. What will be the height from C to AB, i.e., CE?

Answer:
Area of △ABC = ½ × Base × Height
= ½ × BC × AD
= ½ × 9 × 6
= 27 cm²

Also, taking AB as the base and CE as the height,
Area of △ABC = ½ × Base × Height
= ½ × AB × CE
27 = ½ × 7.5 × CE
CE = (27 × 2) ÷ 7.5
= 7.2 cm

Related Links

• NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.2
• NCERT Solutions Class 7 Maths Chapter 9 – Perimeter and Area