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Power Point Prepared By N K Srivastava KV NTPC Shaktinagar Class : VII Topic : Practical Geometry Construction of a line parallel to a given line , through a point not on the line. Step 1 : Take a line ‘l’ and a point ‘A’ not on ‘l’ .A l Step 2: Take any point B on ‘l’ and join B to A . A B l Step 3 : With B as centre and a convenient radius , draw an arc cutting l at C and BA at D. A D B C l Step 4 : Now with A as centre and the same radius as in Step 3,draw an arc EF cutting AB at G. Place the pointed tip of the compass at C and adjust the opening so that the pencil tip is at D F A E G D B C l Step 5 : With the same opening as in step 4 and with G as centre draw an arc cutting the arc EF at H. F A H E G D B C l Step 6 : Now , join AH to draw a line m . This line m is the required parallel line parallel to line ‘l’ and passing through A . F A H E m G D B C l Construction Of Triangles 1.Construction of a triangle when the measurements of all the three sides are given. ( SSS Criterion ) Example : Construct a triangle ABC , given that AB = 5 cm , BC = 6 cm and AC = 7 cm Solution Step 1 : First , we draw a rough sketch with given measures. A B 6 cm C Step 2 : Draw a line segment BC of length 6 cm . B 6 cm C Step 3 : From point B , point A is at a distance of 5 cm . So , with B as centre , draw an arc of radius 5 cm. B 6 cm C Step 4 : From point C , point A is at a distance of 7 cm . So , with C as centre , draw an arc of radius 7 cm. B 6 cm C Step 5 : Mark the point of intersection of arcs as A . Join AB and AC .Δ ABC is the required triangle. A B 6 cm C 2.Construction of a triangle when the lengths of two sides and the measurement of angle between them are given. ( SAS Criterion ) Example : Construct a triangle PQR , given that PQ = 3 cm , QR = 5.5 cm and ˂ PQR = 60⁰ Solution Step 1 : First , we draw a rough sketch with given measures. P 60⁰ Q 5.5 cm R Step 2 : Draw a line segment QR of length 5.5 cm . Q 5.5 cm R Step 3 : AT point Q , Draw QX making 60⁰ with QR. X 60⁰ Q 5.5 cm R Step 4 : With Q as centre draw an arc of radius 3 cm. IT cuts QX at point P. X P 60⁰ Q 5.5 cm R Step 5 : Join PR .ΔPQR is the required triangle. X P 60⁰ Q 5.5 cm R 3.Construction of a triangle when the measures of two of its angles and the length of side between them is given. ( ASA Criterion ) Example : Construct a triangle PQR , given that XY = 6 cm ,m YXZ = 30⁰ and m XYZ = 100⁰ Solution Step 1 : First , we draw a rough sketch with given measures.Z X 100⁰ 30⁰ 6 cm Y Step 2 : Draw a line segment XY of length 6 cm . X 6 cm Y Step 3 : AT point X , Draw XP making 30⁰ with XY and at point Y draw YQ making an angle 100⁰ with YX. P Q X 100⁰ 30⁰ 6 cm Y Step 4 : Z has to lie on both the rays XP and YQ. So the point of Intersection of these two rays is Z. Δ XYZ is the required triangle. Q Z X 100⁰ 30⁰ 6 cm Y P 4.Construction of a Right -angled triangle when the length of its one leg and its hypotenuse are given. ( RHS Criterion ) Example : Construct a triangle LMN , right – angled at M , given that LN = 5 cm and MN = 3 cm. Solution Step 1 : First , we draw a rough sketch with given measures. L 90⁰ M 3 cm N Step 2 : Draw a line segment MN of length 3 cm . M 3 cm N Step 3 : AT point M , Draw MX perpendicular to MN. X 90⁰ M 3 cm N Step 4 : With N as centre , draw an arc of radius 5 cm. X L 90⁰ M 3 cm N Step 5 : Now join LN.Δ MNL is the required right – angled triangle. X L 90⁰ M 3 cm N HOME WORK 1. 2. 3. 4. Take any line m and draw a line perpendicular to passing through it and not on it. Construct a triangle ABC in which AB = 6 cm , BC = 7 cm and AC = 4 cm Construct a triangle PQR in which PQ = 8 cm Q = 60⁰ and QR = 6 cm Construct a triangle XYZ in which YZ = 7 cm , Y = 30⁰ and Z = 100⁰ .