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Constructions

Short Answer Type

71. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides arc

2 over 3

of the corresponding sides of the first triangle.

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72. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides arc

7 over 5

of the corresponding sides of the first triangle.

2093 Views

73. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another

triangle whose sides are

1 begin inline style 1 half end style

times the corres ponding sides of the isosceles triangle.

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74. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are

3 over 4

of the corresponding sides of the triangle ABC.

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75. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ΔABC.

2325 Views

Long Answer Type

76. 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 over 3

times the corresponding sides of the given triangle.

513 Views

Short Answer Type

77. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

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78. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Fig. Steps of Construction :(i) Join PO and bisect it. Let M b

Fig.

Steps of Construction :
(i)    Join PO and bisect it. Let M be the midpoint of PO.
(ii)    Taking Mas centre and MO as radius, draw a circle. Let it intersect the given circle at the point Q and R.
(iii)    Join PQ.
By measurement PQ = 4.5 cm
Then PQ is the required tangent.
By actual calculation,

PQ space equals space square root of OP squared minus OQ squared end root space space space space space space space space space left square bracket space By space Pythogoras space Theorem right square bracket space space space space space space equals space square root of left parenthesis 6 right parenthesis squared minus left parenthesis 4 right parenthesis squared end root space space space space space space equals space square root of 36 minus 16 end root space equals space square root of 20 space space space space space space equals space 4.47 space cm

Justification :
Join OQ. Then ∠PQO an angle in the semi-circle and, therefore
∠PQO = 90
⇒ PQ ⊥ OQ
Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.

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79. Draw a circle of radius 3 cm. Take two points at a distance 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

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80. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

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