ch 8.3-ssc - cbse

let the sides of the right angled ∆ be a,b,c let c be the hypotenuse such that a^2+b^2=c^2 

area of equilateral ∆ of side x is given as :area= {(x^2)*sqrt(3)}/4 

as c^2=a^2 + b^2..if we multiply both sides by sqrt(3)/4 

we get {(c^2)*sqrt(3)}/4 = {(a^2)*sqrt(3)}/4 + {(b^2)*sqrt(3)}/4 

hence proved area of the equilateral ∆ drawn on the hypotenuse = the sum of the areas of the equilateral ∆ drawn on the other two sides of the ∆.

Q. 8.3-2 )

Q. 8.3 - 3 )

Q. 8.3 - 4 ) In triangle ABC, XY || AC and XY divides the triangle into two parts of equal area. Find the ratio of AX/ XB .

Q. 8.3 - 5 ) Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given:   △ABC ~ △DEF. AP is the median to side BC of △ABC and DQ is the median to side EF of △DEF.ACDF=BCEF    {Corresponding sides of similar triangles are proportional}⇒ACDF=2PC2QF=PCQF      (1){P is the mid-point of BC and Q is the mid-point of EF} To Prove:   ar(△ABC)ar(△DEF)=AP2DQ2

 Proof:    ar(△ABC)ar(△DEF)=BC2EF2

{The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides}⇒ar(△ABC)ar(△DEF)=(2PC)2(2QF)2=PC2QF2              (2) In △APC and △DQFACDF=PCQF  from (1)And, ∠C=∠F    {Corresponding angles of similar triangles are equal} Therefore, by SAS similarity criterion, △APC ~ △DQFTherefore, APDQ=PCQF            (3) Putting (3) in (2), we getar(△ABC)ar(△DEF)=AP2DQ2

 Hence Proved

Q. 8.4 – 1) Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.Solution:

Given: ABCD is a rhombus. Diagonals AC and BD of rhombus intersect at point O.To Prove: AB2+BC2+CD2+DA2=AC2+BD2

 Proof:In △AOB∠AOB=90∘          {Diagonals of rhombus intersect at 90∘}Therefore, AO2+OB2=AB2        {By, pythagoras theorem}    (1)Similarly, AO2+OD2=AD2          {By, pythagoras theorem}    (2)Similarly, OD2+OC2=DC2          {By, pythagoras theorem}    (3)Similarly, OC2+OB2=BC2          {By, pythagoras theorem}    (4) Adding (1), (2), (3) and (4), we getAB2+BC2+CD2+DA2=2OA2+2OB2+2OD2+2OC2      (5) But, we have OA = OC and OB = OD       {Diagonals of rhombus bisect each other}    (6) Putting (6) in (5), we getAB2+BC2+CD2+DA2=4OA2+4OB2⇒AB2+BC2+CD2+DA2=(2OA)2+(2OB)2⇒AB2+BC2+CD2+DA2=AC2+BD2             {2OA = AC and 2OB= BD} 

Hence Proved

Q. 8.4 – 2

Q. 8.4 – 3)    

Q. 8.4 – 4)    

Q. 8.4 – 5)

Q. 8.4 – 6)

Q. 8.4 – 7)

Q. 8.4 – 8)

Q. 8.4 – 9)

Q. 8.4 – 10)

Q. 8.4 – 11)

Q. 8.4 – 12)