Q. 234.5( 86 Votes )

In the given figu

Answer :

Given - DEFG is a square and BAC = 90⁰


To prove - FG2= BG x FC


Property – If one triangle is similar to the second triangle and the second triangle is similar to the third triangle then the first triangle is similar to the third one.


Answer –


DEFG is a square, therefore we can write,


DE = EF = FG = GD ………(1) (sides of a square)


As opposite sides of square are parallel.


DE || FG


ADE = DBG ………(2) (corresponding angles)


AED = ECF ………(3) (corresponding angles)


As given, BAC = 90⁰


Now, in ∆ADE & ∆GBD,


DAE = DGB ………angles of 90⁰


ADE = DBG ………from (2)


………by AA test of similarity


Now, in ∆ADE & ∆FEC,


DAE = EFC ………angles of 90⁰


AED = ECF ………from (3)


………by AA test of similarity


By property, if one triangle is similar to the second triangle and the second triangle is similar to the third triangle then the first triangle is similar to the third one.


As and



………corresponding sides of similar triangles


GD × EF = BG × FC


FG × FG = BG × FC ………from (1)


FG2 = BG × FC


Hence proved !!!


OR


Consider an equilateral triangle ABC with altitude CD drawn on side AB.



Given - ∆ABC is an equilateral triangle and CD is altitude drawn on side AB


To Prove – 3 BC2 = 4 CD2


Property – Measure of all angles of an equilateral triangle are of 60⁰.


Answer –


As CD AB.


BDC = 90⁰ ……….(1)


As measure of all angles of an equilateral triangle are of 60⁰.


CBD = 60⁰ ……….(2)


As sum of all angles of a triangle is 180⁰


for ∆BCD,


BDC + CBD + DCB = 180⁰


90⁰ + 60⁰ + DCB = 180⁰


150⁰ + DCB = 180⁰


DCB = 30⁰


Therefore, ∆BCD is a 30⁰-60⁰-90⁰.


Now, in ∆BCD,




Squaring on both sides,



3 AB2 = 4 AD2


Thus, three times the square of one side is equal to four times the square of one of its altitudes.


Hence proved !!!


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