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Byju's Answer

Standard VII

Mathematics

Construction of Incircle of a Triangle

The steps of ...

Question

The steps of construction of an isosceles triangle PQR with angle PQR as 110^o are as follows:

(i) Draw the base QR.

(ii) Join P to R to obtain the required triangle PQR.

(iii) At point Q, draw a ray QX making an angle 110° with QR.

(iv) Taking Q as centre, draw an arc of radius equal to PQ. It intersects QX at the point P.

The correct order of steps is

(i), (iii), (iv), (ii)

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(i), (iii), (ii), (iv)

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(i), (ii), (iii), (iv)

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(i), (iv), (ii), (iii)

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Solution

The correct option is A
(i), (iii), (iv), (ii)

The steps of construction in order are:

(i) Draw the base QR.

(ii) At point Q, draw a ray QX making an angle 110° with QR.

(iii) Taking Q as centre, draw an arc of radius equal to PQ. It intersects QX at the point P.

(iv) Join P to R to obtain the required triangle PQR.

Suggest Corrections

Similar questions

Q. Find the correct order of steps to construct a tangent from an external point P to a circle (where O is the centre of the circle and P is the point from which tangents are drawn).

(i) Bisect the segment OP, at L.
(ii) Join OP.
(iii) From P, join wherever this circle intersects the original circle.
(iv) Draw a circle with centre as L and radius as LO.

Q. Construct a

△ P Q R

such that

∠ P = 30^{\circ}, ∠ Q = 60^{\circ}

and

P Q = 10 c m

Type 1 if following are the steps of construction and type 0 if following are not tue steps of construction.

Steps of Construction:

(i) Draw a line segment

P Q = 10 c m

(ii) At

P

, draw a ray making an angle of

30^{\circ}

(iii) At

Q

, draw another ray making an angle of

60^{\circ}

which intersects the first ray at

R

△ P Q R

is the required triangle.

The steps involved in dividing a line segment are given below. Arrange them in the correct order.

I. Draw a ray $P M$ making an acute angle with $P Q$

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IV. Draw a line parallel to $Q P_{m + n}$ at $P_{m}$ .

V. Join $Q P_{m + n}$ .

VI. Draw an angle equal to $∠$ $P P_{m + n} Q$ at $P_{m}$ .

The point Q' at which the line $P_{m} Q'$ intersects PQ divides the line PQ in the ratio $m : n$

Q. Ajay is constructing a square PQRS with PQ = 5 cm. The steps of construction are as follows.

(i) Draw a line segment PQ of length 5 cm. Construct 90° at the points P and Q. Extend the rays from P and Q.
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Q. By following these steps of construction,

(i) Draw a line segment PQ = 11 cm

(ii) At P construct an angle of 60^o and at Q, an angle of 45^o.

(iii) Bisect these angles. Let the bisectors of these angles intersect at a point A.

(iv) Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.

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