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Class VII - Maths: CONSTRUCTION OF TRIANGLES
One Word Answer Questions:
Q) The similar triangles are called?
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Q) What are Similar Figures?
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Q) PS/SQ = PT/TR and ∠ PST = ∠ PRQ.
Prove that PQR is an isosceles triangle?
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Q) If PQ ||RS, prove that Δ POQ ~ ΔSOR?
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Q) Full formula of SSS?
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Short Answer Questions:
Q) Give two different examples of pair of : a) Similar Figures b) Non- Similar Figures.
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Q) What can you say about the similarity of two triangles?
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Q) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Explain
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Q) If a line intersects sides AB and AC of a Δ ABC at D and E respectively
and is parallel to BC, prove that AD/AB = AE/AC.
    
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Q) ABCD is a trapezium with AB ||DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig.). Show that AE/ED = BF/FC.
    
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Long Answer Questions:
Q) Draw two triangles ABC and DEF such that AB = 2 cm, ∠ A = 50° AC = 4 cm, DE = 3 cm, ∠ D = 50° and DF = 6 cm.
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Q) In Fig.CM and RN are respectively the medians of ΔABC and
ΔPQR. If Δ ABC ~ ΔPQR, prove that:
(i) ΔAMC ~ Δ PNR
(ii)CM/RN = AB/ PQ
(iii) ΔCMB ~ Δ RNQ?
    
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Q) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Explain.
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Q) In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Explain.
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Q) O is any point inside a rectangle ABCD (see Fig.). Prove that OB2 + OD2 = OA2+ OC2.
    
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Contents
  • Trianagle
  • SSS Construction of Δ
  • SAS triangle construction
  • ASA Triangle Construction
  • RHS triangle Construction
Trianagle

A triangle can be drawn if you know the elements that are required for two triangles to be congurent. Thus, a triangle can be drawn in any of the situations given below i.e.,

  1. Three sides of the triangle.
  2. Two sides and the angle included between them.
  3. Two angles and the side included between them
  4. Hypotenuse and one adjacent side of a right-angled triangle.

A triangle can also be drawn if two of its sides and a non-included angle are given. However, it is important to remember that this condition is not suffiecient to make two triangles, congruent.

In order to be able to construct a Δ from three segments, one of the segments must be shorter than the sum of the other two: a < b+c , or b < a+c, or c < a+b


Construction of triangle when three sides given

SSS Construction of Δ
Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6cm.
Step 1 Draw a line YZ of length 5 cm.
Step 2 From Y, point X is at a distance of 4.5 cm. So, with Y as center, draw an arc of radius 4.5 cm.
Step 3 From Z, point X is at a distance of 6 cm. So, with Z as center, draw an arc of radius 6 cm.
Step 4 X has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark the point of interaction of arcs as X. Join XY and XZ. ΔXYZ is the required triangle as shown in the figure.


SAS triangle construction

Construct a ΔABC, in which ∠B = 70°, AB = 4.8 cm and BC = 5.2 cm.

Before doing construction, draw a rough sketch, so that you will get an idea which side is taken as a base.Here after drawing a rough sketch, we are sure that the base is AB = 4.8 cm.

Step 1 : Draw AB = 4.8 cm.
Step 2 : Using protractor, draw ∠ABK = 70 ° .
Step 3 : On the line segment BK, cut off BC = 5.2 cm.
Step 4 : Join A and C.
ΔABC is the required triangle.


ASA Triangle Construction

To construct a triangle when two of its angles,say B and C, and the included side BC are given,we proceed as follows :

Step 1 : Draw a line segment BC.
Step 2 : Draw ∠CBX of measure equal to that of ∠B.
Step 3 : Draw ∠BCY with Y on the same side of BC as X, such that its measure is equal to that of ∠C. Let BX and CY intersect at A. Then, ΔABC is the required triangle.

Example

Construct a ΔABC in which BC = 6 cm, ∠B = 35° and ∠C = 100 ° .

Step 1 : Draw a line segment BC = 6 cm.
Step 2 : Draw ∠CBX, such that ∠CBX = 35°
Step 3 : Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 100 °
Step 4 : Let BX and CY intersect at A. ΔABC is the required triangle.


RHS triangle Construction

A triangle is said to be a right triangle, if one of its three angles is a right angle.


To construct a right triangle ABC right angled at C when its hypotenuse is AC and one side is BC is given, we follow the following steps :

Step 1 : Draw a line segment BC of given length.
Step 2: Draw ∠BCX of measure 90 ° .
Step 3 : With center B and radius equal to the hypotenuse AB, draw an arc of the circle to intersect ray CX at A.
Step 4 : Join BA to obtain the required triangle ABC.

Example
Construct a right triangle PQR in which ∠Q = 90° , PR = 6 cm and QR = 4 cm.
Step 1: Draw QR = 4 cm.
Step 2 : Using compass, at Q draw ∠RQK = 90° .
Step 3 : Cut off RP = 6 cm.
Step 4 : Join P and R.
Now we get the required ΔPQR.

Construction of a triangle when measurements of the three sides are given

Example
Construct a ΔPQR with sides PQ = 4 cm, QR = 5 cm and RP = 7 cm.
Step 1 : Draw a rough sketch of the triangle and label it with the given measureents.

Step 2 : Draw a line segment QR of length 5 cm.

Step 3 : With centre Q, draw an arc of radius 4 cm.

Step 4 : Since P is at a distance of 7 cm from R, draw another arc from R with radius 7 cm such that it intersects first arc at P.

Step 5 : Join Q,P and P,R. The required Δ PQR is constructed.


Construction of a triangle with two given sides and the included angle

Example:1

Construct ΔABC in which AB = 4 cm, BC = 5cm and ∠B = 50°.

Step 1:Draw a rough sketch of a triangle and label it with the given measurements.

Step 2 : Draw a line segment AB of length 4 cm

Step 3 : Draw a ray `bar(BX)` making an angle 50° with AB.

Step 4: Draw an arc of radius 5 cm from B, which cuts `bar(BX)` at C

Step 5 : Join C, A to get the required ΔABC.


Example:2

Construct Δ MAN with MA = 4 cm, ∠ M = 45° and ∠A = 100°

Step 1: Draw rough sketch of a triangle and label it with the given measurements.

Step 2 : Draw line segment MA of length 4 cm

Step 3 : Draw a ray `bar(MX)`, making an angle 45° at M.

Step 4 : Draw a ray `bar(AY)`, making an angle 100°at A.Extend the ray `bar(MX)` if necessary to intersect ray `bar(AY)`.

Step 5 : Mark the intersecting point of the two rays as N. You have the required ΔMAN


Construction of right-angled triangle when the hypotenuse and a side are given.

Example
Construct ΔABC, right-angled at A, and BC = 6 cm; AB = 5 cm.
Step 1 : Draw a rough sketch of right-angled triangle and label it with given information.

Note: side opposite to the right angle is called hypotenuse.
Step 2 : Draw a line segment AB of length 5 cm.

Step 3 : Construct a ray `bar(AX)` perpendicular to AB at A.

Step 4 : Draw an arc from B with radius 6cm to intersect `bar(AX)` at ‘C’.

Step 5 : Join B,C to get the required ΔABC.


Construction of triangle when two sides and the non-included angle are given

Example
Construct ΔABC such that AB = 5 cm, AC = 4 cm, ∠B = 40°
Step 1 : Draw rough sketch of Δ ABC and label it with the given measurements.

Step 2 : Draw a line segment AB of length 5 cm.

Step 3 : Draw a ray `bar(BX)` making an angle 40° at B.

Step 4 : With A as centre and radius 4 cm, draw an arc to cut ray `bar(BX).

Step 5 : Mark the intersecting point as C and join C, A to get the required ΔABC.

Can you cut the ray `bar(BX)` at any other point? You will see that as ∠B is acute, the arc from A of radius 4 cm cuts the ray `bar(BX)` twice.


So we may have two triangles as given below:



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