(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm
Area of shaded portion
= Area of rectangle ABCD – (Area of ∆FAE + area of ∆EBC)
= (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC)
= (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10)
= 180 – (30 + 40)
= 180 – 70
= 110 cm²

(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region
= Area of square PQRS – Area of ∆QPT – Area of ∆TSU – Area of ∆UQR
= (SR x QR) – 1/2 x PQ x PT – 1/2 x ST x SU – 1/2
= 20 x 20 – 1/2 x 20 x 10 – 1/2 x 10 x 10 – 1/2 x 20 x 10
= 400 – 100 – 50 – 100
= 150 cm²

Class 7 Maths Chapter 11 Exercise 11.4

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/