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Construct an equilateral triangle, given its side 4 cm and justify the construction.
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Construction Procedure:
1. Let draw a line segment AB=4 cm .
2. With A and B as centres, draw two arcs on the line segment AB and note the point as D and E.
3. With D and E as centres, draw the arcs that intersect the previous arc respectively that forms an angle of 60° each.
4. Now, draw the lines from A and B that are extended to meet each other at point C.
5. Therefore, ABC is the required triangle.
Justification:
From construction, it is observed that
and
We know that, the sum of the interior angles of a triangle is equal to 180°
(Sides opposite to equal angles are equal)
Hence, justified.
Construct a triangle XYZ in which \(\angle Y=30^o,\angle Z=90^o \) and \(XY+YZ+ZX=11\ cm \).
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Construction Procedure:
The steps to draw the triangle of given measurement is as follows:
1. Draw a line segment AB which is equal to XY + YZ + ZX = 11 cm.
2. Make an angle Y = 30° from point A and the angle be LAB
3. Make an angle Z = 90° from point B and the angle be MAB
4. Bisect LAB and MAB at point X.
5. Now take the perpendicular bisector of the line XA and XB and the intersection point be Y and Z respectively.
6. Join XY and XZ
7. Therefore, XYZ is the required triangle
Construct a triangle ABC in which BC = 8 cm, \(\angle \)B = 45° and AB – AC = 3.5 cm.
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Construction Procedure:
The steps to draw the triangle of given measurement is as follows:
1. Draw a line segment of base BC = 8 cm
2. Measure and draw B = 45° and draw the ray BX
3. Take a compass and measure AB – AC = 3.5 cm.
4. With B as centre and draw an arc at the point be D on the ray BX
5. Join DC
6. Now draw the perpendicular bisector of the line CD and the intersection point is taken as A.
7. Now join AC
8. Therefore, ABC is the required triangle.
Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. Are these lines parallel?
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According to the question,
A line segment AB of length 4 cm.
To draw a perpendicular to AB through A and B, respectively.
Steps of construction:
1. Draw AB = 4 cm.
2. With A as centre, draw an arc, intersecting AB at P.
3. With P as centre and the same radius, draw an arc intersecting the arc drawn in step 2 at Q.
4. With Q as centre and the same radius, draw an arc, intersecting the arc drawn in step 3 at R.
5. With R as centre and the same radius, draw an arc, intersecting the arc drawn in step 5 at X.
6. Draw OX and produced it to C and D.
7. Now, repeat the steps from 2 to 7 to draw the line EF perpendicular through B.
Yes, these lines are parallel because the sum of the interior angles on the same side of the transversal is 180 degrees.
Draw a line segment AB = 8 cm. Draw 13 part of it. Measure the length of \(\dfrac{1}{3} \) part of AB.
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Steps of Construction :
1. Draw a line segment AB = 8 cm.
2. Draw its perpendicular bisector and let it intersect AB in M.
3. Draw the perpendicular bisector of MB and let it intersect AB in N. Thus, AN = 13 of AB = 6 cm.
Why we cannot construct a \(\triangle \)ABC, if \(\angle \)A = 60°, AB = 6 cm and AC + BC = 5 cm but construction of \(\triangle \)ABC is possible if \(\angle \)A = 60°, AB = 6 cm and AC – BC = 5 cm ?
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We know that, by triangle inequality property, construction of triangle is possible if sum of two sides of a triangle is greater than the third side. Here, AC + BC = 5 cm which is less than AB (6 cm) Thus, ABC is not possible.
Also, by triangle inequality property, construction of triangle is possible, if difference of two sides of a triangle is less than the third side
Here, AC – BC = 5 cm, which is less than AB (6 cm)
Thus, ABC is possible.
Construct an angle of 90° at the initial point of the given ray.
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Steps of Construction :
1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc, cutting OA at P.
3. With P as centre and same radius, draw an arc cutting the arc drawn in step 2 at Q.
4. With Q as centre and the same radius as in steps 2 and 3, draw an arc, cutting the arc drawn in step 2 at R.
5. With Q and R as centres and same radius, draw two arcs, cutting each other in S.
6. Join OS and produce to B. Thus, AOB is the required angle of 90°
Draw a straight angle. Using compass bisect it. Name the angles obtained.
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Steps of Construction :
1. Draw any straight angle (say AOC).
2. Bisect AOC and join BO.
3. AOB is the required bisector of straight angle AOC.
Draw any reflex angle. Bisect it using compass. Name the angles so obtained.
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Steps of Construction :
1. Let AOB be any reflex angle.
2. With O as centre and any convenient radius, draw an arc cutting OA in P and OB in Q.
3. With P and Q as centres, draw two arcs of radius little more than half of it and let they intersect each other in C. Join OC. Thus, OC is the required bisector. Angles so obtained are AOC and COB
Construct a triangle whose sides are in the ratio 2 : 3 : 4 and whose perimeter is 18 cm.
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Steps of Construction :
1. Draw a line segment .
2. At A, construct an acute angle .
3. Mark 9 points on , such that
Join .
5. From and , draw , intersecting in and respectively.
6. With M as centre and radius AM, draw an arc.
7. With N as centre and radius NB, draw another arc intersecting the previous arc at L.
8. Join LM and LN. Thus, LMN is the required triangle.
Construct a \(\triangle ABC\) with \(BC = 8\ cm, \angle B = 45^o\) and \(AB – AC = 3.1\ cm\).
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Steps of Construction :
1. Draw any line segment BC = 8 cm.
2. At B, construct an angle CBX = 45°.
3. From BX, cut off BD = 3.1 cm.
4. Join DC.
5. Draw the perpendicular bisector ‘p’ of DC and let it intersect BX in A.
6. Join AC. Thus, ABC is the required triangle.
Construct a triangle \(ABC\) in which \(BC = 4.7\ cm, AB + AC = 8.2\ cm\) and \(\angle C = 60^o\).
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Given : In and .
Required : To construct .
Steps of Construction :
1. Draw BC = 4.7 cm.
2. Draw
3. From ray CX, cut off CD = 8.2 cm.
4. Join BD.
Draw the perpendicular bisector of BD meeting CD at A.
6. Join AB to obtain the required triangle ABC.
Justification :
A lies on the perpendicular bisector of BD, therefore, AB = AD
Now, CD = 8.2 cm
AC + AD = 8.2 cm
AC + AB = 8.2 cm.
Construct \(\triangle \)XYZ, if its perimeter is 14 cm, one side of length 5 cm and \(\angle \)X = 45°.
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Here, perimeter of XYZ = 14 cm and one side XY = 5 cm
YZ + XZ = 14 – 5 = 9 cm and X = 45°.
Steps of Construction :
1. Draw a line segment XY = 5 cm.
2. Construct an YXA = 45° with the help of compass and ruler.
3. From ray XA, cut off XB = 9 cm.
4. Join BY.
5. Draw perpendicular bisector of BY and let it intersect XB in Z.
6. Join ZY. Thus, ∆XYZ is the required triangle.
To construct a triangle, with perimeter 10 cm and base angles 60° and 45°.
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Given : In ,
and .
Required : To construct .
Steps of Construction :
1. Draw DE = 10 cm.
2. At D, construct of and at E, construct DEQ = 1 of
3. Let DP and EQ meet at A.
4. Draw perpendicular bisector of AD to meet DE at B.
5. Draw perpendicular bisector of AE to meet DE at C.
6. Join AB and AC. Thus, ABC is the required triangle.
Construct an equilateral triangle whose altitude is 6 cm long.
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Steps of Construction :
1. Draw a line PQ and take any point S on it.
2. Construct the perpendicular SR on PQ.
3. From SR, cut a line segment SA = 6 cm.
4. At the initial point A of the line segment AS, construct SAB = 30° and SAC = 30°.
5. The arms AB and AC of the angles SAB and SAC meet PQ in B and C respectively. Then, ABC is the required equilateral triangle with altitude of length 6 cm.
Construct a rhombus whose diagonals are 8 cm and 6 cm long. Measure the length of each side of the rhombus.
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Steps of Construction :
1. Draw a line segment PR = 8 cm.
2. Draw the perpendicular bisector XY of the line segment PR. Let O be the point of intersection of PR and XY, so that O is the 8 cm midpoint of PR.
3. From OX, cut a line segment OS = 3 cm and from OY, cut a line segment OQ = 3 cm.
4. Join PS, SR, RQ and QP, then PQRS is the required rhombus.
5. Measure the length of segments PQ, QR, RS and SP, each is found to be 5 cm long.