The document Facts That Matter, Ex 11.1 NCERT Solutions- Constructions Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

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**Facts that Matter ****We know that: **

I. To divide a line segment in a given ratio m:n, we divide this segment into (m + n) equal parts. Then we take m parts on one side and n on the other.

II. The idea of dividing a line segment in any ratio is used in construction of a triangle similar to a given triangle, whose sides are in a given ratio with the corresponding sides of the given triangle.

III. The scale-factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

**TANGENTS TO A CIRCLE**

Remember:I. If a point lies inside a circle, then there cannot be a tangent to the circle through this point.

II. If a point lies on the circle, then there is only one tangent to the circle at this point and it is perpendicular to the radius through that point.

III. If the point lies outside the circle, there will be two tangents to the circle from thispoint.

NOTE:

(i) For drawing a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this radius through this point.

(ii) The two tangents to a circle from an external point are equal.

**Construction of tangents to a circle from a point outside it.****Steps of construction:**

I. Let the centre of the circle be O and P be a point outside the circle.

II. Join O and P.

III. Bisect OP and let M be the midpoint of OP.

IV. Taking M as centre and MP or MO as radius, draw a circle intersecting the given circle at the points A and B.

V. Join PA and PB.

Thus, PA and PB are the required two tangents.

NOTE:In case, the centre of the circle is not known, then to locate its centre, we take any two non-parallel chords and then find the point of intersection of their perpendicular bisectors.

**Page No. 219 - 220****EXERCISE 11.1**

In each of the following, give the justification of the construction also:**Ques 1: Draw a line segment of length 7.6 cm and divide it in the ratio 5:8.****Measure the two parts.****Sol: **Steps of construction:

I. Draw a line segment AB = 7.6 cm.

II. Draw a ray AX making an acute angle with AB.

III. Mark 13 (8 + 5) equal points on AX, and mark them as X_{1}, X_{2}, X_{3}, ........, X_{13}.

IV. Join â€˜point X_{13}â€™ and B.

V. From â€˜point X_{5}â€™, draw X_{5}C â•‘ X_{13}B, which meets AB at C.

Thus, C divides AB in the ratio 5:8.

On measuring the two parts, we get: AC = 4.7 cm and BC = 2.9 cm.**Justification: **

In Î” ABX_{13} and Î” ACX_{5}, we have

âˆ´

â‡’ AC : CB = 5 : 8.**Ques 2: Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.****Sol: **Steps of construction:

I. Draw a âˆ†ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.

II. Draw a ray BX making an acute angle âˆ CBX.

III. Mark three points X_{1}, X_{2}, X_{3} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3}.

IV. Join X_{3}C.

V. Draw a line through X_{2} such that it is parallel to X_{3}C and meets BC at Câ€².

VI. Draw a line through Câ€² parallel to CA to intersect BA at Aâ€².

Thus, Aâ€²BCâ€² is the required triangle.**Justification: **

By construction, we have:

â‡’

But

â‡’

â‡’

Adding, 1 to both sides, we get

â‡’

â‡’

Now, in Î”BCâ€²Aâ€² and Î”BCA

we have CA â•‘ Câ€²Aâ€²

âˆ´ Using AA similarity, we have:

â‡’ [each equal to 2/3]**Ques 3: Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.****Sol: **Steps of construction:

I. Construct a Î” ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that âˆ CBX is an acute angle.

III. Mark 7 points of X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6} and X_{7} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4} = X_{4}X_{5} = X_{5}X_{6} = X_{6}X_{7}

IV. Join X_{5} to C.

V. Draw a line through X_{7} intersecting BC (produced) at Câ€² such that X_{5}C â•‘ X_{7}Câ€².

VI. Draw a line through Câ€² parallel to CA to intersect BA (produced) at Aâ€². Thus, Î”Aâ€²BCâ€² is the required triangle.**Justification: **

By construction, we have

Câ€²Aâ€² â•‘ CA

âˆ´ Using AA similarity, Î”ABC ~ Î”Aâ€²BCâ€²

Also X_{7}Câ€² â•‘ X_{5}C [By construction

âˆ´

âˆ´

âˆ´ **Ques 4: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.****Sol: **Steps of construction:

I. Draw BC = 8 cm

II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4 cm.

IV. Join AB and AC.

Thus, Î”ABC is the required isosceles triangle.

V. Now, draw a ray BX such that âˆ CBX is an acute angle.

VI. On BX, mark three points X_{1}, X_{2} and X_{3} such that: BX_{1} = X_{1}X_{2} = X_{2}X_{3}

VII. Join X_{2} to C.

VIII. Draw a line through X_{3} parallel to X_{2} C and intersecting BC (extended) to Câ€².

IX. Draw a line through Câ€² parallel to CA intersecting BA (extended) at Aâ€², thus, Î”Aâ€²BCâ€² is the required triangle.**Justification:**

We have Câ€²Aâ€² â•‘ CA [By construction]

âˆ´ Using AA similarity, Î” ABC ~ Î”Aâ€²BCâ€²

â‡’

Since, X_{3}Câ€² â•‘ X_{2}C [By construction]

â‡’

â‡’ [By BPT]

But

â‡’

Thus, **Ques 5: Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and âˆ ABC = 60Â°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.****Sol: **Steps of construction:

I. Construct a Î”ABC such that BC = 6 cm, AB = 5 cm and âˆ ABC = 60Â°.

II. Draw a ray such that âˆ CBX is an acute angle.

III. Mark four points X_{1}, X_{2}, X_{3} and X_{4} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}

IV. Join X_{4}C and draw X_{3}Câ€² â•‘ X_{4}C such that Câ€² is on BC.

V. Also, draw another line through Câ€² and parallel to CA to intersect BA at Aâ€².

Thus, Î”Aâ€²BCâ€² is the required triangle.**Justification:**

By construction, we have:

X_{4}C â•‘ X_{3}Câ€²

[By BPT]

But [By construction]

â‡’ ...(1)

Now, we also have

[By construction]

[using AA similarity]

â‡’ [From (1)]**Ques 6: Draw a triangle ABC with side BC = 7 cm, âˆ B = 45Â°, âˆ A = 105Â°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Î” ABC.****Sol: **Steps of construction:

I. Construct a Î” ABC such that BC = 7 cm, âˆ B = 45Â° and âˆ A = 105Â°.

II. Draw a ray BX making an acute angle âˆ CBX with BC.

III. On BX, mark four points X_{1}, X_{2}, X_{3} and X_{4} such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}.

IV. Join X_{3} to C.

V. Draw X_{4}Câ€² â•‘ X_{3}C such that Câ€² lies on BC (extended).

VI. Draw a line through Câ€² parallel to CA intersecting the extended line segment BA at Aâ€².

Thus, Î”Aâ€²BCâ€² is the required triangle.**Justification:**

By construction, we have:

Câ€²Aâ€² â•‘ CA

âˆ´ Î”ABC ~ Î”Aâ€²BCâ€² [AA similarity]

â‡’ ...(1)

Also, by construction,

X_{4}Câ€² â•‘ X_{3}C

Î” BX_{4}Câ€² ~ Î” BX_{3}C

â‡’

But

â‡’ ..(2)

From (1) and (2), we have:**Ques 7: Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.****Sol: **Steps of construction:** **

I. Construct the right triangle ABC such that âˆ B = 90Â°, BC = 4 cm and BA = 3 cm.

II. Draw a ray BX such that an acute angle âˆ CBX is formed.

III. Mark 5 points X_{1}, X_{2}, X_{3}, X_{4} and X_{5} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4} = X_{4}X_{5}.

IV. Join X_{3} to C.

V. Draw a line through X_{5} parallel to X_{3}C, intersecting the extended line segment BC at Câ€².

VI. Draw another line through Câ€² parallel to CA intersecting the extended line segment BA at Aâ€².

Thus, Î”Aâ€²BCâ€² is the required triangle.**Justification: **

By construction, we have:

Câ€²Aâ€² â•‘ CA

âˆ´ Î”ABC ~ Î” Aâ€²BCâ€² [By AA similarity ]

â‡’ ...(1)

Also, X_{5}Câ€² â•‘ X_{3}C [By construction]

âˆ´ Î” BX_{5}Câ€² ~ Î” BX_{3}C

â‡’

But ...(2)

From (1) and (2) we get

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