Prove that the ratio of areas of two similar triangles is equal to the square of their corresponding sides.

The question is to prove the ratio of the areas of two similar triangles is equal to the square of their corresponding sides not square of ratio of corresponding sides.

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Even me!!
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how to proof area theorem practically

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how are you guys area theorms so as i cant type much you can refer to text books i can give you some tips to win in exam first when teacher is away exchange your paper with someone sitting near you without seeing write the answer on your table with pencil so that you can refer while writing exam
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is this answer
 
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Please find this answer

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Hope it helps you

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Use RD Sharma book. You sill get all the answers to your questions.😍😍
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Hi
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In triangle ABC , D and E are points on sides AB and AC respectively ,such that DE parellel to BC and AD : DB = 3:1, if EA = 6.6 cm, then find AC
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You can check google for better answers
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Hj
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It's simple and easy

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same answer
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P and Q are points on sides of AB and AC resp of triangleABC right angled at B.if AP=3cm,PB=9cm,AQ=5cm and QC=15cm.show that BC=4PQ plz help me as fast as u can
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yes
 
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Some mistakes is there in this. Ad is not ac
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Hii evryone
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If ABC is an isosceles triangle and D is a point on BC such that AD perpendicular to the BC ,

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Consider two triangles ABC and DEF. AX and DY are the bisectors of the angles A and D respectively. Ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. so, Area (ΔABC) / Area (ΔDEF) = AB2/ DE2 (1) ΔABC ~ ΔDEF ⇒ ∠A = ∠D 1/ 2 ∠A = 1 / 2 ∠D ⇒ ∠BAX = ∠EDY Consider ΔABX and ΔEDY ∠BAX = ∠EDY ∠B = ∠E So, ΔABX ~ ΔEDY [By A-A Similarity] AB/DE = AX/DY ⇒ AB2/DE2 = AX2/DY2 (2) From equations (1) and (2), we get Area (ΔABC) / Area (ΔDEF) = AX2/ DY2 Hence proved.
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Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF. To Prove: area(ΔABC)/area(ΔDEF) = AM2/DN2 Proof: ΔABC ~ ΔDEF (Given) ∴ area(ΔABC)/area(ΔDEF) = (AB2/DE2) ...(i) and, AB/DE = BC/EF = CA/FD ...(ii) In ΔABM and ΔDEN, we have ∠B = ∠E [Since ΔABC ~ ΔDEF] AB/DE = BM/EN [Prove in (i)] ∴ ΔABC ~ ΔDEF [By SAS similarity criterion] ⇒ AB/DE = AM/DN ...(iii) ∴ ΔABM ~ ΔDEN As the areas of two similar triangles are proportional to the squares of the corresponding sides. ∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2
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thanks!!
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hyy
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Prove that in a triangle if a square of 1 side is equal to the sum of the squares of the other two sides and angle opposite to the first side is a right angle
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If ABC is an isosceles triangle and D is a point on BC such that AD perpendicular to the BC ,
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😁
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dont know
 
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This sum is already proved in book
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Please find this answer

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Please find this answer

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Please find this answer

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Please find this answer

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the first is correct
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hi
 
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Please find this answer

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Please find this answer

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Please find this answer

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Please find this answer

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same as aasha
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hhhhhhhh!!!
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Answer

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Please find this answer

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Theorem 1 in the picture

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Hello
I think this might help. Thanks

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Proof this therom :
Median. Line draw form midpoint of side in a triangle to it's opposite vertex .
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Corresponding angle is equal measure
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Corresponding angle is equal measure
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Please find this answer

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Please find this answer

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#theorem
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Hope it will help u friend.
Thank you!

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Hope it will help u mate.
Thank u

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