ex 11.2 class 7

Simple solutions for NCERT Solutions for ex 11.2 class 7 are given here in the post. This NCERT Solutions for ex 11.2 class 7 contains topics related to the raw data collection and its organisation. so we definitely want students of Class 7 to solve ex 11.2 class 7 to empower their basics and test the students capability of understanding the concepts. It also helps the students of CBSE Class 7 Maths students.
Question 1.

Find the area of each of the following parallelograms:

(a)

Solution:-

Height of parallelogram = 4 cm

Base of parallelogram = 7 cm

Then,

Area of parallelogram = base × height

= 7 × 4

= 28 cm²

(b)

Solution:-

Height of parallelogram = 3 cm

Base of parallelogram = 5 cm

Area of parallelogram = base × height

= 5 × 3

= 15  cm²

(c)

Solution:-

Height of parallelogram = 3.5 cm

Base of parallelogram = 2.5 cm

Area of parallelogram = base × height

= 2.5 × 3.5

= 8.75  cm²

(d)

Solution:-

Height of parallelogram = 4.8 cm

Base of parallelogram = 5 cm

Area of parallelogram = base × height

= 5 × 4.8

= 24  cm²

(e)

Solution:-

Height of parallelogram = 4.4 cm

Base of parallelogram = 2 cm

Area of parallelogram = base × height

= 2 × 4.4

= 8.8  cm²

Question 2. 

Find the area of each of the following triangles:

(a)

Solution:-

Base of triangle = 4 cm

Height of height = 3 cm

Area of triangle = \(\dfrac{1}{2}\) × base × height

= \(\dfrac{1}{2}\) × 4 × 3

= 1 × 2 × 3

= 6  cm²

(b)

Solution:-

Base of triangle = 3.2 cm

Height of height = 5 cm

Then,

Area of triangle =\(\dfrac{1}{2}\) × base × height

= \(\dfrac{1}{2}\) × 3.2 × 5

= 1 × 1.6 × 5

= 8  cm²

(c)

Solution:-

Base of triangle = 3 cm

Height of height = 4 cm

Then,

Area of triangle = \(\dfrac{1}{2}\) × base × height

= \(\dfrac{1}{2}\) × 3 × 4

= 1 × 3 × 2

= 6  cm²

(d)

Solution:-

From the figure,

Base of triangle = 3 cm

Height of height = 2 cm

Area of triangle = \(\dfrac{1}{2}\) × base × height

= \(\dfrac{1}{2}\) × 3 × 2

= 1 × 3 × 1

= 3  cm²
Question 3. 

Find the missing values:

Solution:-

(a)

Base of parallelogram = 20 cm

Height of parallelogram =?

Area of the parallelogram = 246  cm²

Area of parallelogram = base × height

246 = 20 × height

Height = 246/20

Height = 12.3 cm

∴Height of the parallelogram is 12.3 cm.

(b)

Base of parallelogram =?

Height of parallelogram =15 cm

Area of the parallelogram = 154.5  cm²

Area of parallelogram = base × height

154.5 = base × 15

Base = 154.5/15

Base = 10.3 cm

∴Base of the parallelogram is 10.3 cm.

(c)

Base of parallelogram =?

Height of parallelogram =8.4 cm

Area of the parallelogram = 48.72  cm²

Area of parallelogram = base × height

48.72 = base × 8.4

Base = 48.72/8.4

Base = 5.8 cm

∴Base of the parallelogram is 5.8 cm.

(d)

Base of parallelogram = 15.6 cm

Height of parallelogram =?

Area of the parallelogram = 16.38  cm²

Area of parallelogram = base × height

16.38 = 15.6 × height

Height = 16.38/15.6

Height = 1.05 cm

∴Height of the parallelogram is 1.05 cm.

Question 4. 

Find the missing values:



Solution:-

(a)

Height of triangle =?

Base of triangle = 15 cm

Area of the triangle = 16.38  cm²

Area of triangle = \(\dfrac{1}{2}\) × base × height

87 = \(\dfrac{1}{2}\) × 15 × height

Height =  \(\dfrac{87 × 2}{15}\)

Height = 174/15

Height = 11.6 cm

∴Height of the triangle is 11.6 cm.

(b)

Height of triangle =31.4 mm

Base of triangle =?

Area of the triangle = 1256 mm²

Area of triangle =  \(\dfrac{1}{2}\) × base × height

1256 =  \(\dfrac{1}{2}\)  × base × 31.4

Base =  \(\dfrac{1256 × 2}{31.4}\)

Base =  \(\dfrac{2512}{31.4}\)

Base = 80 mm = 8 cm

∴Base of the triangle is 80 mm or 8 cm.

(c)

Height of triangle =?

Base of triangle = 22 cm

Area of the triangle = 170.5 cm²

Area of triangle = \(\dfrac{1}{2}\) × base × height

170.5 = \(\dfrac{1}{2}\)  × 22 × height

170.5 = 1 × 11 × height

Height = \(\dfrac{170.5}{11}\)

Height = 15.5 cm

∴Height of the triangle is 15.5 cm.

Question 5. 

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



Solution:-

From the question it is given that,

SR = 12 cm, QM = 7.6 cm

(a) Area of the parallelogram = base × height

= SR × QM

= 12 × 7.6

= 91.2 cm²

(b) Area of the parallelogram = base × height

91.2 = PS × QN

91.2 = 8 × QN

QN = 91.2/8

QN = 11.4 cm

Question 6. 

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.




Solution:-

Area of the parallelogram = 1470 cm²

AB = 35 cm

AD = 49 cm

Area of the parallelogram = base × height

1470 = AB × BM

1470 = 35 × DL

DL = \(\dfrac{1470}{35}\)

DL = 42 cm

Area of the parallelogram = base × height

1470 = AD × BM

1470 = 49 × BM

BM = \(\dfrac{1470}{49}\)

BM = 30 cm

Question 7. 

ΔABC 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 ΔABC. Also find the length of AD.



Solution:-

AB = 5 cm, BC = 13 cm, AC = 12 cm

Area of the ΔABC = \(\dfrac{1}{2}\) × base × height

= \(\dfrac{1}{2}\) × AB × AC

= \(\dfrac{1}{2}\) × 5 × 12

= 1 × 5 × 6

= 30 cm²

Area of ΔABC = \(\dfrac{1}{2}\) × base × height

30 = \(\dfrac{1}{2}\) × AD × BC

30 = \(\dfrac{1}{2}\) × AD × 13

\(\dfrac{(30 × 2)}{13}\) = AD

AD = \(\dfrac{60}{13}\)

AD = 4.6 cm

Question 8. 

ΔABC 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 ΔABC. What will be the height from C to AB i.e., CE?




Solution:-

AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm

Area of ΔABC = \(\dfrac{1}{2}\) × base × height

= \(\dfrac{1}{2}\) × BC × AD

= \(\dfrac{1}{2}\) × 9 × 6

= 1 × 9 × 3

= 27 cm²

Area of ΔABC = \(\dfrac{1}{2}\) × base × height

27 = \(\dfrac{1}{2}\) × AB × CE

27 = \(\dfrac{1}{2}\) × 7.5 × CE

\(\dfrac{27 × 2}{7.5}\) = CE

CE = \(\dfrac{54}{7.5}\)

CE = 7.2 cm