Construction of a line parallel to a given line

एक दी हुई रेखा के समांतर रेखा खींचना

हमें एक रेखा l और एक बिन्दु N जो दी गयी रेखा के बाहर स्थित है , दिया गया है। बिन्दु N से होते हुए रूलर और परकार की मदद से रेखा l के समांतर एक रेखा m की रचना करनी है। रचना के चरण इस प्रकार हैं :

चरण 1 - दी गयी रेखा l खींचिए और रेखा के बाहर एक बिन्दु N लीजिए।

चरण 2 – रेखा l पर एक बिन्दु A अंकित कीजिए और बिन्दु A से बिन्दु N को मिलाइए।

चरण 3 – बिन्दु A को केन्द्र मानकर और एक सुविधाजनक त्रिज्या लेकर एक चाप बनाइए। यह चाप रेखा l को B पर और AN को C पर काटता है।

चरण 4 – बिन्दु N को केन्द्र और चरण 3 की त्रिज्या लेकर एक चाप बनाइए जो NA को बिन्दु D पर काटता है।

चरण 5 – परकार को BC लंबाई के बराबर खोलिए।

चरण 6 – D को केन्द्र मानकर परकार का खुलाव चरण 5 के बराबर रखते हुए एक चाप खींचिए जो चरण 4 के चाप को बिन्दु E पर काटता है।

चरण 7 – EN को मिलाकर रेखा m खींचिए जो दी गयी रेखा l के समांतर है।

नीचे एपलेट में उपर दिए चरण दिखाए गए हैं। एपलेट में ∠BAN = ∠ENA है , जो अंत: एकांतर कोण हैं , अत: रेखा m || रेखा l है। बिन्दु K को माउस की मदद से खींचकर रेखा l की अलग अलग परिस्थितियों में समांतर रेखा की रचना देखी जा सकती है।

Construction of a line parallel to a given line

We are given a line l and a point N not on the given line. We need to construct a line m with the help of ruler and compass through N parallel to the given line l. Following are the steps of construction:

Step 1 – Draw a line l and take a point N, not on the given line.

Step 2 – Take a point A on the given line l and join A with N.

Step 3 – Taking A as a center and a suitable radius draw an arc. This arc cuts the line l at B and AN at C.

Step 4 – Taking N as a center and with the radius of step 3, draw an arc which cuts NA at point D.

Step 5 – Open the compass equal in length to BC.

Step 6 – Taking D as a center and opening of the compass as in step 5 draw an arc which intersects the arc of step 4 at point E.

Step 7 – Join points E and N to make line m which is parallel to the line l.

The steps of constructions can be seen in the applet below. We can see that ∠BAN = ∠ENA, these are internal alternate angles. So line m || line l. Drag point K to see the construction of the parallel line in different positions of given line l.

Median to Hypotenuse of a Right Triangle

Problem : Let us consider the right triangle PQR with the right angle P (Figure 1), and let PS be the median drawn from the vertex P to the hypotenuse QR. We need to find the relationship between the length of the median PS and the the length of the hypotenuse QR.

Solution :Draw a straight line passing through the midpoint S and parallel to the side PR intersecting the side PQ at the point T. (Figure 2).

The angle QPR is given as right angle. The angles QTS and QPR are equal to each other as as they are corresponding angles of the parallel lines PR and TS and the transversal PQ. Hence the angle QTS is a right angle.

As TS passes through the mid-point S and is parallel to PR , it divides the side PQ into two equal parts i.e. PT = TQ. So, the triangles PTS and QTS are right triangle triangles with equal sides PT and TQ , these triangles also have a common side TS. Hence, these triangles are congruent in as per the Side – Angle – Side (SAS) Rule. 

From this we can say that the other sides of these triangles are also equal to each other as they are the corresponding parts of the congruent triangles , thus PS = QS. Now QS is equal to half the length of the hypotenuse QR , we can say that the median PS is also equal to half the length of the hypotenuse.

Hence, we can conclude that in a right triangle , the length of median to hypotenuse is half the length of the hypotenuse.

Tangent Circles - 2

Given a circle with radius R. The circle is tangent to a line ‘a’ at point M. A number r is given. Construct a circle with radius r tangent to both the given circle and the given line ‘a’
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com