given triangle ABC similar to triangle PQR of AB/PQ is equal to 1/3 then find area of triangle ABC /area of triangle PQR
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Heya☺
We know that ,
The square of ratio of corresponding sides of similar triangles is equal to the ratio of area of those triangles .
Therefore ,
ar ( ∆ABC ) / ar ( ∆ PQR )= (AB /PQ)^2
= (1 /3 ) ^2
= 1/9
We know that ,
The square of ratio of corresponding sides of similar triangles is equal to the ratio of area of those triangles .
Therefore ,
ar ( ∆ABC ) / ar ( ∆ PQR )= (AB /PQ)^2
= (1 /3 ) ^2
= 1/9
siddhant2610:
bro
maybe u have apllied the theorm wrong
how when I check this question in cbse guide it's answer is 9
that's wrong
It is 1/9
OK
Yep
yaa that is wrong
Yeah...
as this was ques no 1 in the leaked paper of cbse class 10 ...which i gave
Answered by
0
we r given that ∆ABC is similar to ∆PQR.....
THEREFORE we can apply theorm of similarity of triangles......
hence .....which says .....
ar(ABC)/ar(PQR)=(AB/PQ)^2.....HENCE replacing AB and PQ by 1 and 3 respectively as they r sides of similar ∆......
ar(ABC)/ar(PQR)=(1/3)^2=>1/9
HENCE UR ANSWER IS 1/9....
HOPE THIS HELPS....
THEREFORE we can apply theorm of similarity of triangles......
hence .....which says .....
ar(ABC)/ar(PQR)=(AB/PQ)^2.....HENCE replacing AB and PQ by 1 and 3 respectively as they r sides of similar ∆......
ar(ABC)/ar(PQR)=(1/3)^2=>1/9
HENCE UR ANSWER IS 1/9....
HOPE THIS HELPS....
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