Construction of Triangle 2
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Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.
Here, AB > AC, i.e. the side containing the base angle B is greater than third side
Steps of construction:
Step 1: Draw the base BC = 5.7 cm and draw a ray BY making an ∠YBC=60∘
Step 2: Cut the line segment BD = 3 cm from the ray BY.
Step 3 will be
- Let PQ intersect BX at A.Then, join AC
- Join DC and draw PQ⊥BY
- Join DC and draw PQ⊥DC
- Join DC and draw PQ⊥BC
- greater than or equal to
- greater than
- lesser than
- equal to
To construct a triangle given its base, a base angle and the difference of the other two sides, there is only a single type of construction.
True
False
How will you prove that the construction for a triangle with the given conditions is right?
Given conditions: Base length BC is given, base angle B is given, and difference of the other two sides is given (AB-AC) where AB is greater than AC. For going about the construction, I drew the base length BC, drew the ray BX with angle XBC known to me. Taking B as centre and radius equal to (AB-AC) I cut an arc on the ray BX intersecting it at point D. I then joined D to C. Then drew the perpendicular bisector of the line segment DC and named the point of intersection of this perpendicular bisector and the ray BX as A. Joined A to C and the triangle ABC was ready
Which of the following statements gives the best explanation to this construction?
Since the triangles AMD and AMC are congruent, AD = AC and hence the location of A has been plotted correctly
the triangles DBC and CAD are congruent the location of A is justified
AM is the altitude for the triangle ADC and hence the location of A is justified
None of these
Construct a triangle ABC, in which ∠B = 53.13°, ∠C = 36.87° and AB + BC+ CA = 24 cm. What type of triangle is this? [4 MARKS]
- 1
- 3
- 2
- 4
In the figure, if AM is the perpendicular bisector of CD,
- AD
- BD
- DM
- BM
Construct a triangle ABC in which BC is 5 cm, Angle B is 60∘ and AB-AC=0 cm. What type of triangle is this?
(i) △ABD≅△ACD
(ii) △ABP≅△ACP
(iii) AP bisects ∠A as well as △D.
(iv) AP is the perpendicular bisector of BC.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.
- SAS property
- SSS property
- RHS property
- AAA property
Construct a triangle ABC, in which ∠B = 53.13°, ∠C = 36.87° and AB + BC+ CA = 24 cm. What type of triangle is this? [4 MARKS]
- equilateral
- scalene
- isosceles
- right-angled
In the given figure AD is the bisector of ∠A and AB=AC. Then △ACD and △ADB are congruent by which criterion?
ASA
None of these
SSS
SAS
- 2
- 4
- 7
- 6
Δ ABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that ∠BCD is right angle. [4 MARKS]
- 6.5 cm
- 6 cm
- 5.5 cm
- 7.6 cm
Steps of construction:
Step 1: Draw BC = 6 cm
Step 2: Construct ∠YBC=45∘
Step 3 will be -
- Join AC, ΔABC is the required triangle.
- Draw perpendicular bisector of CD intersecting BY at A.
- From ray BY, cut-off line segement BD = 2.5 cm
- Join CD
- 2
- 4
- 6
- 7
- 1
- 2
- 3
- 4
Here, PR > PQ,
i.e., the side containing base angle is less than the third side.
Steps of construction:
Step 1: Draw the base QR = 5.5 cm
Step 2: At the point Q, make an ∠XQR=60∘
Step 3 will be
- Cut line segment QS = PR - PQ = 2 cm from the line QX extended on opposite side of linesegment QR
- Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR
- Cut line segment QS = PR - PQ = 2.5 cm from the ray QX
- Draw the perpendicular bisector LM of SR.
- True
- False
- 2
- 4
- 7
- 6
- False
- True