# Important Questions Class 10 Maths Chapter 11 Constructions

In general, the word construction means to build or to make something and in mathematics, we also do many constructions by using geometric shapes like lines, curves, circles, triangles, squares, trapezoids, cones, pentagons, etc. However, it is important that there should be a mathematical explanation attached to those constructions. For any method of drawing whether to divide a line in different ratios, make two similar triangles, tangents over a circle, or any polygon there must be mathematical reasoning attached to it.

The important question for class 10 maths Chapter 11 Constructions can be obtained from the following topics:

- Division of a line segmentÂ
- Alternative method of division
- Construction of similar triangles
- Construction of tangentsÂ
- Division of a line in different ratio
- Constructing pair of tangents form any external point

## Important Questions Chapter 11 – Constructions

**Q1. ****Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are ****3/5**** times the corresponding sides of the given triangle.Â **

**Solution:**

âˆ´ âˆ†ABâ€™Câ€™ is the required âˆ†.

**Q2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm.**

**Solution:**

Steps of Construction:

Draw two circles with radius OA = 4 cm and OP = 6 cm with O as centre. Draw âŠ¥ bisector of OP at M. Taking M as centre and OM as radius draw another circle intersecting the smaller circle at A and B and touching the bigger circle at P. Join PA and PB. PA and PB are the required tangents.

Verification:

In rt. âˆ†OAP,

OA^2 + AP^2= OP^2 â€¦ [Pythagorasâ€™ theorem

(4)^2+ (AP)^2= (6)^2

AP2 = 36 â€“ 16 = 20

AP = +âˆš20=âˆš(4×5)

= 2âˆš5 = 2(2.236) = 4.472 = 4.5 cm

By measurement, âˆ´ PA = PB = 4.5 cm

**Q3. Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45Â°**

**Solution:**

Draw âˆ AOB = 135Â°, âˆ OAP = 90Â°, âˆ OBP = 90Â°

âˆ´ PA and PB are the required tangents.

**Q4. Draw two tangents to a circle of radius 3.5 cm, from a point P at a distance of 6.2 cm from its center.**

**Solution:**

OP = OC + CP = 3.5 + 2.7 = 6.2 cm

Hence AP & PB are the required tangents.

**Q5. ****Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and âˆ B = 90Â°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.**

**Solution:**

Steps of Construction:

- Draw BC = 8 cm.
- From B draw an angle of 90Â°.
- Draw an arc BA = 6 cm cutting the angle at A.
- Join AC. âˆ´ âˆ†ABC is the required âˆ†.
- Draw âŠ¥ bisector of BC cutting BC at M.
- Take Mas centre and BM as radius, draw a circle.
- Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.

AB and AE are the required tangents.

Justification: âˆ ABC = 90Â° â€¦[Given

Since, OB is a radius of the circle.

âˆ´ AB is a tangent to the circle.

Also, AE is a tangent to the circle.

**Q6. Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then construct another triangle whose sides are ****2/3**** times the corresponding sides of the first triangle.**

**Solution:**

Here, AB = 5 cm, BC = 7 cm, AC = 6 cm and ratio is 2/3 times of corresponding sides.

âˆ´ âˆ†Aâ€™BCâ€™ is the required triangle.

**Q7. Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are ****3/4**** times the corresponding sides of the isosceles triangle.**

**Solution:**

âˆ´ âˆ†Aâ€™BCâ€™ is the required triangle.

**Q8. Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.**

**Solution:**

Draw two circles on A and B as asked.

Z is the mid-point of AB.

From Z, draw a circle taking ZA = ZB as radius,

so that the circle intersects the bigger circle at M and N and smaller circle at X and Y.

Join AX and AY, BM and BN.

BM, BN are the required tangents from external point B.

AX, AY are the required tangents from external point A.

Justification:

âˆ AMB = 90Â° â€¦[Angle in a semi-circle

Since, AM is a radius of the given circle.

âˆ´BM is a tangent to the circle

Similarly, BN, AX and AY are also tangents.

**Q9. ****Construct a âˆ†ABC in which AB = 6 cm, âˆ A = 30Â° and âˆ†B = 60Â°. Construct another âˆ†ABâ€™Câ€™ similar to âˆ†ABC with base ABâ€™ = 8 cm.**

**Solution:**

Steps of construction:

- Draw a âˆ†ABC with side AB = 6 cm, âˆ A = 30Â° and âˆ B = 60Â°.
- Draw a ray AX making an acute angle with AB on the opposite side of point C.
- Locate points A1, A2, A3 and A4 on AX.
- Join A3B. Draw a line through A4 parallel to A3B intersecting extended AB at Bâ€™.
- Draw a line parallel to BC intersecting ray AY at Câ€™.

Hence, âˆ†ABâ€™Câ€™ is the required triangle

**Q10. Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose sides are ****4/5**** of the corresponding sides of first triangle.**

**Solution:**

âˆ´ âˆ†Aâ€™BCâ€™ is the required âˆ†.

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