Ch 8.3-SSC - CBSE
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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 ∆.
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Ch 8.3-SSC - CBSE
Transcript of 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
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