Area and Perimeter NCERT Exercise 11.2

Part 2

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 = 12cm and QM = 7.6 cm. Find:
(a) The area of the parallelogram PQRS (b) QN, if PS = 8cm

(a) Given, Base = 12 cm, Height = 7.6 cm

Area of the parallelogram PQRS = Base `xx` Height

`= 12 cm \xx 7.6 cm = 91.2` sq cm

(b) Given; Base PS = 8cm, Area of the parallelogram(as calculated in the above question) = 91.2 cm², Height QN= ?

Area = Base `xx` Height

`⇒ 91.2 cm^2 = 8cm xx\ QN`

Or, `QN = (91.2)/(8) = 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 = 35cm and AD = 49 cm. Find the length of BM and DL.

Solution: Given, Area = 1470 cm², AB = 35cm and AD = 49 cm, then BM =?, DL = ?.

Considering AB = Base;

Area = Base `xx` Height

`⇒ 1470 cm^2 = AB \xx\ DL`

`⇒ 1470 cm^2 = 35cm \xx\ DL`

Or, `DL = (1470)/(35) = 42` cm

Now considering AD = Base;

Area = Base `xx` Height

`⇒ 1470 cm^2 = AD \xx \BM`

`⇒ 1470 cm^2 = 49 cm \xx \BM`

Or, `BM = (1470)/(49) = 30` cm

Hence, DL = 42 cm and BM = 30 cm

Question 7: ∆ ABC is right angled at A. AD is perpendicular to BC. If AB = 5cm, BC= 13cm and AC = 12cm. Find the area of ∆ABC. Also find the length of AD.

Solution: Given, Base = AB = 5cm, Height = AC = 12 cm

Area of the triangle `= 1/2 xx \text(base)xx\text(height)`

`= 1/2 xx 5 xx 12 = 30` sq cm

Now, consider BC = Base, therefore, height AD = ?

Area of the triangle `= 1/2 xx\text(base)xx\text(height)`

Or, `30 = 1/2 xx 13 xx\AD`

Or, `AD = (30xx2)/(13) = 4.61` cm

Question 8: ∆ ABC is isosceles with AB = AC = 7.5 cm and BC = 9cm. The height AD from A to BC is 6cm. Find the area of ∆ABC. What will be the height from C to AB i.e. CE?

Solution: Given, AB = AC = 7.5 cm, BC = 9cm, AD = 6cm, Area = ?, CE=?

In this triangle When BC is considered as Base then AD will become height. And when AB is considered as Base then CE will become height.
Now, Base BC = 9cm, Height AD = 6cm

Area `∆ = 1/2 xx \text(base) xx \text(height)`

`= 1/2 xx 9 xx 6 = 27` sq cm

Using the base AB = 7.5 cm, height CE can be calculated as follows:

Area `∆ = 1/2 xx \text(base) xx \text(height)`

Or, `27 = 1/2 xx 7.5 xx \CE`

Or, `CE = (27xx2)/(7.5) = 7.2` cm