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Standard VIII

Mathematics

Construction of Parallelogram When Two Adjacent Sides and One Diagonal Are Given.

By following ...

Question

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.

(v). Join AB and AC. Triangle ABC has been obtained. In this ∆ABC,

AB + BC + CA > 11cm

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AB + BC = 11cm

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AB + BC + CA = 11cm

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BC + CA = 11cm

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Solution

The correct option is C
AB + BC + CA = 11cm

AB + BC + CA = 11cm

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Similar questions

In a right triangle, sum of the squares of two sides is equal to the square of the hypotenuse.

By following these steps of construction,

(i) Draw a line segment PQ = 11 cm

(ii) At P construct an angle of 60o 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.

(v). Join AB and AC. Triangle ABC has been obtained. In this ABC,

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.

(v). Join AB and AC. Triangle ABC has been obtained. In this ∆ABC,

Q. To construct a triangle ABC in which

∠ B = 60^{\circ}, ∠ C = 45^{\circ}

and AB + BC + AC = 12 cm

The steps of construction are:
Step 1: Draw a line segment XY=12 cm
Step 2: At X, construct an angle of $60^{\circ}$ and at y construct an angle of $45^{\circ}$
Step 3: Bisect these angles. Let bisector of these angles intersect at point A
Step 4 will be

The steps to construct a triangle with given base angles $∠$ B and $∠$ C and BC + CA + AB, are:.
A. Draw a line segment, say XY equal to BC + CA + AB.
B. Make $∠$ LXY equal to $∠$ B and MYX equal to $∠$ C.
C. Bisect $∠$ LXY and $∠$ MYX. Let these bisectors intersect at a point A.
D. Draw perpendicular bisectors PQ of AX and RS of AY.
E. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC.

Pallav got following question in his exam:

Given the base angles, say $∠$ B and $∠$ C and BC + CA + AB, construct triangle ABC.

He followed the following steps for the construction.

1. Draw a line segment, say XY equal to BC + CA + AB.

2. Make $∠$ LXY equal to $∠$ B and $∠$ MYX equal to $∠$ C.

3. Bisect $∠$ LXY and $∠$ MYX. Let these bisectors intersect at a point A

4. Draw perpendicular bisectors PQ of AX and RS of AY.

5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC

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The correct option is C AB + BC + CA = 11cm AB + BC + CA = 11cm

The correct option is C
AB + BC + CA = 11cm

AB + BC + CA = 11cm