1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Areas and ratio of angle bisectors of two similar triangles are given. Match the correct ratio in each case.

A
9 : 7
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
7 : 9
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
1 : 2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
2 : 1
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Consider △ABC and △DEF with AP DQ as angular bisectors of ∠A and ∠D resp. ar(△ABC)ar(△DEF)=AB2DE2....(i) ∠A=∠D (∵△ABC∼△DEF) 12∠A=12∠D ∠BAP=∠EDQ In △ABP and △DEQ, ∠ABP=∠DEQ ∠BAP=∠EDQ ⇒ △ABP∼△DEQ (By AA similarity) ∴ ABDE=APDQ ⇒ AB2DE2=AP2DQ2 So we can say that ar(△ABC)ar(△DEF)=AP2DQ2 [From (i)] ∴ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding angle bisectors. Now AP2DQ2=ar(△ABC)ar(△DEF) 1. ar(△ABC)ar(△DEF)= 81 cm249 cm2 ⇒ AP2DQ2=√8140=97 2. ar(△ABC)ar(△DEF)= 49 cm281 cm2 ⇒ AP2DQ2=√4981=79 3. ar(△ABC)ar(△DEF)= x cm22x cm2 ⇒ AP2DQ2=√x2x=1√2 4. ar(△ABC)ar(△DEF)= 2x cm2x cm2 ⇒ AP2DQ2=√2xx=√21

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Similar Triangles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program