Solution : Given : \(\triangle\) ABC, in which each side is of length 2a. To Find : AD (altitude) In \(\triangle\) ADB and \(\triangle\) ADC, AD = AD (common) \(\angle\) 1 = \(\angle\) 2 (90 each) AB = AC …
Prove that the altitude of an equilateral triangle of side 2a is \(\sqrt{3} a\). Read More »
Solution : Since ABC is an isosceles triangle with AC = BC and \({AB}^2\) = \(2{AC}^2\), therefore, \({AB}^2\) = \({AC}^2\) + \({AC}^2\) \(\implies\) \({AB}^2\) = \({AC}^2\) + \({BC}^2\) (because AC = BC, given) \(\therefore\) \(\triangle\) ABC is right angled at C.
Solution : Since ABC is an isosceles right triangle, right angled at C, therefore \({AB}^2\) = \({AC}^2\) + \({BC}^2\) Since given that the triangle is isosceles, \(\therefore\) BC = AC \(\implies\) \({AB}^2\) = \({AC}^2\) + \({AC}^2\) So, \({AB}^2\) = \(2{AC}^2\)
Solution : (i) Since, AC \(\perp\) BD, therefore, \(\triangle\) ABC ~ \(\triangle\) DBA and each triangle is similar to whole triangle ABD. \(\implies\) \(AB\over BD\) = \(BC\over AB\) So, \({AB}^2\) = BC.BD (ii) Since, \(\triangle\) ABC ~ \(\triangle\) DAC, therefore, \(\implies\) \(AC\over BC\) = \(DC\over AC\) So, \({AC}^2\) = BC.DC (iii) Since, \(\triangle\) ACD ~ \(\triangle\) …
Solution : Since, PM \(\perp\) QR, therefore, \(\triangle\) PQM ~ \(\triangle\) RPM \(\implies\) \(PM\over QM\) = \(MR\over PM\) So, \({PM}^2\) = QM.MR.
Solution : (i) 7 cm, 24 cm, 25 cm Let AB = 7 cm, BC = 24 cm, CA = 25 cm then \({AB}^2\) = 49, \({BC}^2\) = 576 and \({CA}^2\) = 625 \({AB}^2\) + \({BC}^2\) = 625 = \({CA}^2\) Yes, ABC is a right triangle and its hypotenuse is 25 cm. (ii) 3 cm, …
Solution : Since the ration of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides, therefore ratio of areas = \((4)^2\) : \((9)^2\) = 16 : 81 \(\therefore\) (d) is the correct answer,
Solution : Since \(\triangle\) ABC and BDE are equilateral triangles, they are equiangular and hence \(\triangle\) ABC ~ \(\triangle\) BDE So, \(area(\triangle ABC)\over area(\triangle BDE)\) = \({BC}^2\over {BD}^2\) or \(area(\triangle ABC)\over area(\triangle BDE)\) = \({2BD}^2\over {AC}^2\) \(\implies\) \(area(\triangle ABC)\over area(\triangle BDE)\) = \(4\over 1\) \(\therefore\) (d) is the correct answer.
Solution : Given : A square ABCD, equilateral triangles BCE and ACF have been drawn on side BC and the diagonal AC respectively. To Prove : \(area(\triangle BCE)\) = \(1\over 2\)\(area(\triangle DEF)\) Proof : Since all equilateral triangles are similar. \(\implies\) \(\triangle\) BCE ~ \(\triangle\) ACF \(area(\triangle BCE)\over area(\triangle ACF)\) = \({BC}^2\over {AC}^2\) Since, Diagonal …
Solution : Given : \(\triangle\) ABC ~ \(\triangle\) DEF and AP, DQ are their medians. To Prove : \(area(\triangle ABC)\over area(\triangle DEF)\) = \({AP}^2\over {DQ}^2\) Proof : Since the ratio of the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \(\therefore\) \(area (\triangle ABC)\over area …
