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

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Elsa answered this

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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Aasha Kotian answered this

Even me!!

Francis Dhar answered this

how to proof area theorem practically

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Alan answered this

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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Sanu answered this

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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Miss Ronica answered this

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Rahul answered this

Sanjay Sharma answered this

It's simple and easy

Ammu answered this

same answer

Rakash answered this

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

Soumik Mukherjee answered this

yes

Hari Surya Bharadhwaj answered this

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Sinu answered this

Some mistakes is there in this. Ad is not ac

Saloni answered this

Hii evryone

Anshika Gupta answered this

If ABC is an isosceles triangle and D is a point on BC such that AD perpendicular to the BC ,

Rajvjr answered this

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gx

Palak Verma answered this

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.

Palak Verma answered this

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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Ujvala answered this

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Great Godwin answered this

hyy

Vishak answered this

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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Vishav answered this

If ABC is an isosceles triangle and D is a point on BC such that AD perpendicular to the BC ,

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Veer Sharma answered this

z
z

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Gopi Kumar answered this

😁

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Derek answered this

dont know

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Vigneshwar answered this

This sum is already proved in book

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Thor answered this

the first is correct

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Abhishek Sharma answered this

same as aasha

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A answered this

Answer

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Aditya answered this

Theorem 1 in the picture

Gagan Jain answered this

Hello
I think this might help. Thanks

Amar Nath Thakur answered this

Proof this therom :
Median. Line draw form midpoint of side in a triangle to it's opposite vertex .

Shreyash Harangule answered this

Corresponding angle is equal measure

Shreyash Harangule answered this

Corresponding angle is equal measure

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Buddy ﹏✍ answered this

#theorem

Priyanshu Ranjan answered this

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Thank you!

Priyanshu Ranjan answered this

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Thank u

Motilal answered this

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P. Bavisha answered this

when the class start man

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._. .

Divya ❤ answered this

draw a perpendicular AD from A to BC and PS from P To QR. As two angles are equal so the third angle of both triangle should also be equal Hence , we can say the ratio of the area of two similar triangle is equal to the ratio of the square of thier crosseponding sides