mathemerize

Prove that the altitude of an equilateral triangle of side 2a is \(\sqrt{3} a\).

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 »

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

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 »

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

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 »

Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is (a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4

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.

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.

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 »

The areas of two similar triangles are in the ratio of the square of the corresponding medians.

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 …

The areas of two similar triangles are in the ratio of the square of the corresponding medians. Read More »