Maths

Use ruler and compasses only for this question. (i) Construct the locus of points inside the triangle which are equidistant from B and C. (ii) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

Steps of construction: Draw line \[BC\text{ }=\text{ }6\text{ }cm\] and construct angle \[CBX\text{ }=\text{ }{{60}^{o}}\]. Cut off \[AB\text{ }=\text{ }3.5\]. Join \[AC\], triangle  \[ABC\]is the...

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Use ruler and compasses only for this question. (i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60degree (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.

Steps of construction: (i) Draw line \[BC\text{ }=\text{ }6\text{ }cm\] and construct angle \[CBX\text{ }=\text{ }{{60}^{o}}\]. Cut off \[AB\text{ }=\text{ }3.5\]. Join \[AC\], triangle  \[ABC\]is...

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Construct a triangle ABC in which angle ABC = 75o, AB = 5cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P; prove that P is equidistant from B and C; and also from AC and BC.

Steps of Construction: i) Draw a line segment \[BC\text{ }=\text{ }6.4\text{ }cm\] ii) At\[B\], draw a ray \[BX\]making an angle of \[{{75}^{o}}\] with \[BC\]and cut off \[BA\text{ }=\text{ }5\text{...

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In each of the given figures: PA = PB and QA = QB Prove in each case, that PQ (produced, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.

(i) (ii) SOLUTION: Construction: Join \[PQ\]which meets \[AB\text{ }in\text{ }D.\] Proof: \[As\text{ }P\text{ }is\text{ }equidistant\text{ }from\text{ }A\text{ }and\text{ }B.\] So, \[P\]lies on the...

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A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’.

Given, The young lady covers a distance of 6 km at a speed x km/hr. Along these lines, the time taken to cover initial 6 km \[=\text{ }6/x\text{ }hr\] [Since, Time = Distance/Speed] Likewise given,...

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The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr. (i) Find the time taken by each train to cover 300 km. (ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.

(i) Given, Speed of the conventional train \[=\text{ }x\text{ }km/hr\] Speed of the express train \[=\text{ }\left( x\text{ }+\text{ }25 \right)\text{ }km/hr\] Distance \[=\text{ }300\text{ }km\] We...

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