Ex 11.4, 10 (ii) In the following figures, find the area of the shaded portions:
Area of shaded portion = Area of PQRS − Area of ∆PTQ − Area of ∆STU
− Area of ∆QRU
Area of ∆PQRS
Here,
QR = 20 cm
and
SR = SU + UR
= 10 + 10
= 20 cm
Hence, All side of PQRS are same
∴ PQRS is a square with
Side = a = 20 cm
Area = a × a
= 20 × 20 = 400 cm2
Area of ∆PTQ
Here, Δ PTQ is a right angle triangle
with Base PQ and height PT
Base = b = PQ
= SQ
= 20 cm
Height = h = PT
= PS − TS
= 20 − 10
= 10 cm
Area of ∆PTQ = 1/2 × b × h
= 1/2 × 20 × 10
= 10 × 10
= 100 cm2
Area of ∆STU
Here, Δ STU is a right angle triangle
with Base SU and height ST
Base = b = SU = 10 cm
Height = h = ST = 10 cm
Area of ∆AEF = 1/2 × b × h
= 1/2 × 10 × 10 = 5 × 10 = 50 cm2
Area of ∆QRU
Here, Δ QRU is a right angle triangle
with Base UR and height QR
Base = b = UR = 10 cm
Height = h = QR = 20 cm
Area of ∆QRU = 1/2 × b × h
= 1/2 × 10 × 20
= 5 × 20 = 100 cm2
Now,
Area of shaded portion
= Area of PQRS − Area of ∆PTQ − Area of ∆STU − Area of ∆QRU
= 400 − 100 − 50 − 100
= 400 − (100 + 50 + 100)
= 400 − 250
= 150 cm2
∴ Area of shaded portion is 150 cm2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!