mathemerize

Solution : Let ABC be and equilateral triangle and let AD \(\perp\) BC. In \(\triangle\) ADB and ADC, we have : AB = AC          (given) AD = AD           (common side of triangle) and \(\angle\) ADB = \(\angle\) ADB        (each 90) By RHS criteria of …

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one its altitudes. Read More »

Solution : Let ABC be an equilateral triangle and let D be a point on BC such that BD = \(1\over 3\) BC. Draw AE \(\perp\) BC. Join AD. In \(\triangle\) AEB and AEC, we have : AB = AC            (ABC is equilateral) \(\angle\) AEB = \(\angle\) AEC and AE …

In an equilateral triangle ABC, D is a point on the side BC such that BD = \(1\over 3\) BC. Prove that \(9{AD}^2\) = \(7{AB}^2\). Read More »

Solution : We have : DB = 3CD Now, BC = DB + CD i.e. BC = 3CD + CD          [because BD = 3CD] BC = 4CD \(\therefore\)   CD = \(1\over 4\) BC  and  DB = 3CD = \(3\over 4\) BC             ……….(1) Since triangle ABD is …

The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that \(2{AB}^2\) = \(2{AC}^2\) + \({BC}^2\) Read More »

Solution : From triangle ACE, \({AE}^2\) = \({EC}^2\) + \({AC}^2\)       ……….(1)          (By Pythagoras Theorem) From triangle DCB, \({BD}^2\) = \({BC}^2\) + \({DC}^2\)        ………(2) Adding (1) and (2), we get \({AE}^2\) + \({BD}^2\) = \({EC}^2\) + \({AC}^2\) + \({BC}^2\) + \({DC}^2\) By Pythagoras Theorem in right triangle …

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that \({AE}^2\) + \({BC}^2\) = \({AB}^2\) + \({DE}^2\). Read More »

Solution : Let AB and CD be two poles. AB = 11 m, CD = 6 m. Draw a line CE || to BD. In \(\triangle\) AEC, \({AC}^2\) = \({CE}^2\) + \({AE}^2\) = \((12)^2\) + \((11 – 6)^2\) = 144 + 25 = 169 \(\implies\) AC = 13 m Hence, distance between their tops is …

Two poles of height 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m. find the distance between their tops. Read More »

Solution : Let the first plane starts from O and goes upto A towards north. Where OA = (\(1000 \times {3\over 2}\)) km = 1500 km Let the second plane starts from O at the same time and goes upto B towards west, where OB = (\(1200 \times {3\over 2}\)) km = 1800 km According …

An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after \(3\over 2\) hours. Read More »

Solution : Let AB = 24 m be guy wire attached to a vertical pole BC of height 18 m. To keep the wire taut, let it be fixed to stake at A. Then, ABC is a right angled triangle at C. \(\therefore\)  \({AB}^2\) = \({AC}^2\) + \({BC}^2\) So, \({24}^2\) = \({AC}^2\) + \({18}^2\) \(\implies\) …

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ? Read More »

Solution : Let ladder be AB, B be the window and CB be the wall. Then, ABC is a right triangle, angled at C. \(\therefore\)  \({AB}^2\) = \({AC}^2\) + \({BC}^2\) So,  \({10}^2\) = \({AC}^2\) + \(8^2\) or  \({AC}^2\) = 100 – 64 \(\implies\) \({AC}^2\) = 36 \(\implies\)  AC = 6 m

Question : In the figure, O is a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB. Show that (i)  \({OA}^2\) + \({OB}^2\) + \({OC}^2\) – \({OD}^2\) – \({OE}^2\) – \({OF}^2\) = \({AF}^2\) + \({BD}^2\) + \({CE}^2\) (ii)  \({AF}^2\) + \({BD}^2\) + \({CE}^2\) = \({AE}^2\) + …

In the figure, O is a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB. Read More »

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 »