Similar Triangles Questions

Solution : Let the diagonals BD and AC of the rhombus ABCD intersect each other at O. Since the diagonals of the rhombus bisect each other at right angles. \(\therefore\)  \(\angle\) AOB = \(\angle\) BOC = \(\angle\) COD = \(\angle\) DOA = 90 and  AO = CO, BO = OD Since AOB is a right …

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Read More »

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 ~ \(\triangle\) …

In the figure, ABD is triangle right angled at A and AC \(\perp\) BD. Show that (i) \({AB}^2\) = BC.BD (ii) \({AC}^2\) = BC.DC (iii) \({AD}^2\) = BD.CD Read More »

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, …

Sides of some triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm Read More »

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 …

Prove that the area of an equilateral triangle described on one side of square is equal to half the area of the equilateral described on one of the diagonals. Read More »