Q : 11     In Fig. \small 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i)  \small ar(ACB)=ar(ACF)
(ii)  \small ar(AEDF)=ar(ABCDE)

        

Answers (1)


We have a pentagon ABCDE in which BF || AC and CD  is produced to F.

(i) Since \DeltaACB and \DeltaACF are on the same base AC and between same parallels AC and FB.
\therefore  ar(\DeltaACB) = ar (\DeltaACF)..................(i)

(ii) Adding the ar (AEDC) on both sides in  equation (i),  we get

  ar(\DeltaACB) + ar(AEDC) =  ar (\DeltaACF) + ar(AEDC)
 \therefore \small ar(AEDF)=ar(ABCDE)

Hence proved.

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions