mathemerize

Solution : Since D and E are the mid-points of the sides BC and CA respectively of \(\triangle\) ABC. \(\therefore\)  DE || BA \(\implies\)  DE || FA           ……….(1) Since D and F are the mid-points of the sides BC and AB respectively of \(\triangle\) ABC. Therefore, DF || CA  \(\implies\)  DF …

D, E, F are the mid-points of the sides BC, CA and AB respectively of a \(\triangle\) ABC. Determine the ratio of the areas of \(\triangle\) DEF and \(\triangle\) ABC. Read More »

Solution : Given : \(\triangle\) ABC ~ \(\triangle\) DEF such that \(area(\triangle ABC)\) = \(area(\triangle DEF)\) To Prove : \(\triangle\) ABC \(\cong\) \(\triangle\)  DEF Proof : Since, the ratio of the area of two similar triangles is equal to the ratio of the square of  two corresponding sides. \(area(\triangle ABC)\over area(\triangle DEF)\) = \({AB}^2\over {DE}^2\) …

If the areas of two similar triangles are equal, then the triangles are congruent, i.e. equal and similar triangles are congruent. Read More »

Solution : Given : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC. To Prove : \(area(\triangle ABC)\over area(\triangle DBC)\) = \(AO\over DO\) Construction : Draw AE \(\perp\) BC and DF \(\perp\) BC. Proof : In triangles AOE and DOF, we have : \(\angle\) AEO …

In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(area(\triangle ABC)\over area(\triangle DBC)\) = \(AO\over DO\). Read More »

Solution : In triangles AOB and COD, \(\angle\) AOB = \(\angle\) COD       (vertically opposite angles) \(\angle\) OAB = \(\angle\) OCD        (corresponding angles) \(\therefore\) By AA similarity, \(\triangle\) AOB ~ \(\triangle\) COD \(\implies\)  \(area(\triangle AOB)\over area(\triangle COD)\) = \({AB}^2\over {DC}^2\) \(\implies\)  \(area(\triangle AOB)\over area(\triangle COD)\) = = \({2DC}^2\over {DC}^2\) = \(4\over …

Diagonals of trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD. Read More »

Solution : We have,    area of triangle ABC = 64 \(cm^2\) and area of triangle DEF = 121 \(cm^2\) We know that, \(ar(\triangle ABC)\over ar(\triangle DEF)\) = \(({BC\over EF})^2\) \(\implies\)  \(64\over 121\) = \(({BC\over 15.4})^2\) \(\implies\) \(8\over 11\) = \(BC\over 15.4\) \(\implies\) BC = 11.2 cm.

Solution : Given : AD and PM are the medians of triangles ABC and PQR respectively, where \(\triangle\) ABC ~ \(\triangle\) PQR. To Prove : \(AB\over PQ\) = \(AD\over PM\) Proof  : In triangles ABD and PQM, we have \(\angle\) B = \(\angle\) D        (because \(\triangle\) ABC ~ \(\triangle\) PQR) Since AD …

If AD and PM are medians of triangles ABC and PQR respectively, where \(\triangle\) ABC ~ \(\triangle\) PQR, prove that \(AB\over PQ\) = \(AD\over PM\). Read More »

Solution : Given : Let AC be a tower casts a shadow BC = 28 m and DF = 6 m be a pole casts a shadow EF = 4m To Find : Height of the tower Procedure : Now, In triangle ABC and DEF \(\angle\) ACB = \(\angle\) DEF          (each …

A vertical pole of length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of tower. Read More »

Solution : Given : Two triangles ABC and DEF, in which AP and DQ are the medians, such that \(AB\over DE\) = \(AC\over DF\) = \(AP\over DQ\) To prove : \(\triangle\) ABC ~ \(\triangle\) DEF Construction : Produce line AP to G, so that PG = AP. Join CG. And Produce line DQ to H, …

If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding side and the median of another triangle, the prove that the two triangles are similar. Read More »

Solution : Given : D is a point on the side BC of a triangle ABC, such that \(\angle\) ADC = \(\angle\) BAC To Prove : \({CA}^2\) = \(DC \times CB\) Proof : In triangles ABC and DAC, \(\angle\) BAC = \(\angle\) ADC       (given) \(\angle\) C = \(\angle\) C          …

D is a point on the side BC of \(\triangle\) ABC such that \(\angle\) ADC and \(\angle\) BAC are equal. Prove that \({CA}^2\) = \(DC \times CB\). Read More »

Solution : Given : \(\triangle\) ABC and \(\triangle\) PQR in which AD and PM are the medians, such that \(AB\over PQ\) = \(BC\over QR\) = \(AD\over PM\) To prove :  \(\triangle\) ABC ~ \(\triangle\) PQR Proof : \(AB\over PQ\) = \(BC\over QR\) = \(AD\over PM\) \(\implies\)  \(AB\over PQ\) = \({1\over 2}BC\over {1\over 2}QR\) = \(AD\over …

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \(\triangle\) ABC ~ \(\triangle\) PQR. Read More »