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A triangle \( ABC\) can be constructed in which \(\angle B=60^o,\angle C=45^o \) and \(AB + BC + AC = 12\ cm\). Is this statement true ? Justify your answer
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can be constructed with the given conditions.
Can we construct a triangle \( ABC\) in which \(\angle A=105^o,\angle B=75^o \) and \( AB + BC + AC = 12\ cm\)?
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I.e. which is not possible
So cannot be constructed with the given conditions
Can we construct a triangle \(ABC \) in which \(\angle B=105^o,\angle C=90^o \) and \( AB+BC+AC=10\ cm\)?
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cannot be constructed with the given conditions
Construct \(\triangle ABC \) in which \(BC=6.4\ cm,\angle B=45^o \) and \(AB – AC = 2\ cm \).
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Step 1 : Draw a line segment BC of length 6.4 cm.
Step 2 : Draw an angle of 45° at point B on CB using protractor
Step 3 : Draw Ray BP
Step 4 : From B take BX = 2 cm
Step 5 : Join CX
Step 6 : Draw perpendicular bisector of CX intersect ray BP at A
Step 7 : Join A to C,
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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To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction.
1. Draw a line segment AB = 4 cm.
2.Taking 4 as centre and radius more than AB (i.e., 2 cm) draw an arc say it intersect AB at E.
3.Taking E as centre and with same radius as above draw an arc which intersect previous arc at F.
Again, taking F as centre and with same radius as above draw an arc which intersect previous arc (obtained in step ii) at G.
5.Taking G and F are centres, draw arcs which intersect each other at H.
6.Join AH . Thus, AX is perpendicular to AB at A. Similarly, draw BY AB at B.
Now, we know that if two lines are parallel, then the angle between them will be 0°
or
180°.
Here,
and
So, the lines XA and YS are parallel.
Construct \(\triangle ABC \) in which \(BC=4.5\ cm,\angle B=45^o \) and \(AB+AC=5.6\ cm \)
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(1) Draw a line segment
(2) draw angle
(3) Cut a point on line like
(4) Join
(5) Bisect the line which cuts at point Join
is the required triangle
Construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm. Bisect its smallest angle and measure each part.
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1. Draw a line segment BC of length 4.8 cm.
2. From B, point A is at a distance of 3.6 cm. So, having B as centre, draw an arc of radius 3.6 cm.
3. From C, point A is at a distance of 3 cm. So, having C as centre, draw an arc of radius 3 cm which intersect previous arc at A.
4. Join AB and AC. Thus, ABC is the required triangle.
Flere, angle B is smallest, as AC is the smallest side. To direct angle B, we use the following steps.
1.Taking B as centre, we draw an are intersecting AB and BC at D and E, respectively.
2.Taking D and E as centres we draw arcs intersecting at P.
3. Joining BP, we obtain angle bisector of B.
4. Flere,
Thus,
Construct a right angled triangle in which the base is 3 cm and the difference of hypotenuse and perpendicular is 1 cm.
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(1) Draw a line segment
(2) At , draw
(3) Produce to such that is a line
(4) From ray , cast off
(5) Join
(6) Draw the perpendicular bisector of they intermates at
(7) Join
Construct a triangle \(XYZ \) in which \(\angle Y=30^o,\angle Z=90^o \) and \(XY + YZ + ZX = 11\ cm \)
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(1) Draw a line segment
(2) Make from point .
(3) Make a angle from point
(4) Bisect and . let these bisectors intersect at .
(5) Make perpendicular bisector of . Let it intersect at .
(6) Make perpendicular bisector . Let it intersect at .
(7) Join and
is the required
Construct a triangle \(ABC \) in which \(BC=7\ cm,\angle B=75^o \) and \(AB+AC=13\ cm \).
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(1) Draw base
(2) Let the ray be
From as centre cut an arc on ray . Let it intersect at
(4) Join and draw perpendicular bisector of
(5) Mark point where perpendicular bisector intersects
(6) Join
is the required
Construct a triangle \( ABC\) in which \(BC=4\ cm \). \(\angle B=30^o \) and \(AB+AC=6\ cm \)
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When
then
Construct a \(\triangle \)PQR with its perimeter = 10.4 cm and base angle of 45° and 120°
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Let and let .
Steps.
1. Draw a line segment say XY and equal to perimeter i.e., .
2. Draw and .
3. Make the bisector and and let them meet at a know point .
4. Draw perpendicular bisectors and of and , respectively.
5. Let intersect at and intersect at .
Join and .
Construct \(\triangle XYZ \) in which \(\angle Y=30^o\ \angle Z=90^o \) and perimeter is 11 cm.
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Given: In triangle and .
Required: To construct the .
Steps of Construction:
1. Draw a line segment
2. Make and
3. Bisect and . Let these bisectors meet at a point .
4. Draw perpendicular bisectors of and of .
5. Let intersect at and intersect at .
6. Join and .
Then, is the required triangle.
c a right triangle whose base is 6 cm and the difference of its hypotenuse and the other side is 2 cm.
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Steps of Construction :
1. Draw the base BC = 6 cm and at point B make an angle of .
2. Cut a line segment BD equal to AC – AB = 2 cm from the line BX extended on opposite side of line segment BC.
3. Join DC and draw the perpendicular bisector, say PQ of DC.
4. Let PQ intersect BX at A. Join AC
Then, ABC is the required triangle.
Construct \(\triangle ABC \) such that \(\angle B=60^o,\ \angle C=45^o \) and \(AB + BC + CA = 10\ cm. \)
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Steps of Construction :
1. Draw a line segment, say XY equal to BC + CA + AB = 10 cm.
2. Make angle LXY equal to and angle MYX equal to .
3. Bisect LXY and MYX. Let these bisectors intersect at point A.
4. Draw perpendicular bisector PQ of AX and RS of AY.
5. Let PQ intersect XY at B and RS intersect XY at C.
6. Join AB and AC.
ABC is the required triangle.
Construct an equilateral triangle if its altitude is 4 cm. Give justification of your construction.
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Steps of Construction :
1. Draw any line segment .
2. Take a point on and at , construct perpendicular to . From , cut off .
A , construct on either side of . Let and meet at points and respectively. Then, is the required equilateral triangle.
Construct a triangle \( ABC\) in which \(\angle A=45^o,\ \angle B=120^o \) and \(AB+BC=10.4\ cm \)
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Steps of Contruction:
1. Draw a ray DX and cut off a line segment DE= 10.4 cm from it.
2. At point D, construct .
3. At point E, construct .
4. Extend the rays DX and EY to meet at a point and name it as C.
5. Draw the perpendicular bisector of CD intersecting DE at a point A.
6. Draw a perpendicular bisector of CE intersecting DE at a point B.
7. Join CA and CB. ABC is the required triangle.
Construct a right triangle in which one side is 3.5 cm and sum of other side and potenuse is 5.5 cm
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Let given right triangle be ABC.
Then, given BC = 3.5 cm, B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ABC use the following steps
1. Draw the base BC = 3.5 cm
2. Make an angle XBC = 90° at the point B of base BC.
3. Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB.
4. Join DC and make an DCY equal to BDC.
5. Let Y intersect BX at A.
Thus, ABC is the required triangle.
Justification
Base BC and B are drawn as given.
In ACD, ACD = ADC [by construction]
AD = AC …(i)
[sides opposite to equal angles are equal] Now, AB = BD – AD = BD – AC [from Eq. (i)]
BD = AB + AC
Thus, our construction is justified.
Construct a triangle ABC, given that perimeter is \(12.5\ cm,\ \angle B=60^o \) and \(\angle C=75^o \)
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Steps of construction
1. Draw a line segment and locate points such that .
2. Construct rays such that and such that .
3. Draw the bisectors and of and respectively. Let these intersect at A.
4. Draw the perpendicular bisector of AP and AQ and let these intersect PQ at B and C respectively.
5. Join AB and AC.
Then ABC is the required triangle.
Construct a rhombus whose diagonals are 4 cm and 6 cm in lengths.
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We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another.
So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps.
1. Draw the diagonal say AC = 4 cm
2. Taking A and C as centres and radius more than AC draw arcs on both sides of the line segment AC to intersect each other.
3. Cut both arcs intersect each other at P and Q, then join PQ.
4. Let PQ intersect AC at the point O. Thus, PQ is perpendicular bisector of AC.
5. Cut off 3 cm lengths from OP and OQ, then we get points B and D.
Now, join AB, BC, CD, and DA.
Thus, ABCD is the required rhombus.
Construct a triangle ABC such that AB = BC = 6 cm and median AD = 4 cm.
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Steps of construction:
Step 1: Draw a line segment BC of 6 cm.
Step 2: Take midpoint D of BC.
Step 3: with center B and D and radii 6 cm and 4 cm draw two arcs which intersects each other A
Step 4: Join AB, AD and AC
Therefore ABC is the required triangle
Construct a triangle with perimeter 10 cm and base angle 60º and 45º
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Steps of Construction :
1. Draw a line segment, say XY equal to BC + CA + AB = 10 cm.
2. Make angle LXY equal to and angle MYX equal to .
3. Bisect LXY and MYX. Let these bisectors intersect at point A.
4. Draw perpendicular bisector PQ of AX and RS of AY.
5. Let PQ intersect XY at B and RS intersect XY at C.
6. Join AB and AC.
ABC is the required triangle.
Construct a triangle \(ABC \) in which \(BC=7.5\ cm \). \(\angle B=45^o \) and \( ACAB=2.5\ cm\).
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Steps of Construction :
1. Draw the base BC 7.5 cm and at point B make an angle say XBC of 45 degree.
2. Cut line segment BD equal to AC – BC (2.5 cm) from line BX extended on opiate side of BC.
3. Join DC and draw perpendicular bisector, say PQ of DC.
4. Let PQ intersect BX at A. Join AC.
Thus, ABC is the required triangle.
Construct a \(\triangle ABC \) in which \(\angle B=60^o \) and \(\angle C=45^o \) and the perimeter of the triangle is 11 cm.
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Steps of Construction :
1. Draw a line segment PQ = 11 cm (= AB + BC + CA).
2. At P construct an angle of 60° and at Q, an angle of 45°.
3. Bisect these angles. Let the bisectors of these angles intersect at a point A.
4. Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.
5. Join AB and AC
ABC is the required triangle.