Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2
Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2
Question 1.
Given that ∆ABC = ∆DEF (i) List all the corresponding congruent sides
(ii) List all the corresponding congruent angles.
Solution:
Given ∆ABC ≅ DEF.
(iv) By ASA criterion both triangles are congruent.
(v) By ASA criterion both triangles are congruent. Since two angles in one triangle are equal to two corresponding angles of the other triangle. Again one side is common to both triangle and the side is the included side of the angles.
(vi) Two sides are equal. One angle is vertically opposite angles and one equal.
By SAS criterion both triangles are cogruent.
Question 7.
I. Construct a triangle XYZ with the given conditions.
(i) XY = 6.4 cm, ZY = 7.7 cm and XZ = 5 cm
Solution:
Construction:
Step 1: Draw a line. Marked Y and Z on the line such that YZ 7.7 cm.
Step 2: With Y as centre drawn an arc of radius 6.4 cm above the line YZ.
Step 3: With Z as centre, drwan an arc or radius 5 cm to intersect arc drawn in steps. Marked the point of intersection as X.
Step 3: Joined YX and ZX. Now XYZ is the required triangle.
(ii) An equilateral triangle of side 7.5 cm
Solution:
Construction:
Step 1: Drawn a line. Marked X and Y on the line such that XY = 7.5 cm.
Step 2: With X as centre, drawn an arc of radius 7,5 cm above the line XY.
Step 3: With Y as centre, drawn an arc of radius 7.5 cm to intersect arc drawn in steps. Marked the point of intersection as Z.
Step 4: Joined XZ and YZ. Now XYZ in the required triangle.
(iii) An isosceles triangle with equal sides 4.6 cm and third side 6.5 cm
Solution:
Construction: .
Step 1: Drawn a line. Marked X and Y on the line such that XY = 6.5 cm.
Step 2: With X as centre, drawn an arc of radius 4.6 cm above the line XY
Step 3: with Y as centre, drawn an arc of radius 4.6 cm to intersect arc drawn in steps. Marked the point of intersection as Z.
Step 4: Joined XZ and YZ. Now XYZ is the required triangle.
II. Construct a triangle ABC with given conditions.
(i) AB = 7 cm, AC = 6.5 cm and ∠A = 120°.
Solution:
Construction:
Step 1: Drawn a line. Marked A and B on the line such that AB = 7 cm.
Step 2: At A, drawn a ray AX making an angle of 120° with AB.
Step 3: With A as centre, drawn an arc of radius 6.5 cm to cut the ray AX. Marked the point of intersection as C.
Step 4: Joined BC.
ABC is the required triangle.
(ii) BC = 8 cm, AC = 6 cm and ∠C = 40°.
Solution:
Construction:
Step 1: Drawn a line. Marked B and C on the line such fhat BC = 8 Cm.
Step 2: At C, drawn a ray CY making an angle of 40° with BC.
Step 3: With C as centre, drawn an arc of radius 6 cm to cut the ray CY, marked the point of intersection as A.
Step 4: Joined AB.
AB is the required triangle.
(iii) An isosceles obtuse triangle with equal sides 5 cm
Solution:
Construction:
Step 1: Drawn a line. Marked B and C on the line such that BC = 5 cm.
Step 2: At B drawn a ray BY making on obtuse angle 110° with BC.
Step 3: With B as centre, drawn an arc of radius 5 cm to cut ray BY. Marked the point of intersection as C.
Step 4: Joined BC. ABC is the required triangle.
III. Construct a triangle PQR with given conditions.
(i) ∠P = 60°, ∠R = 35° and PR = 7.8 cm
Solution:
Construction:
Step 1: Drawn a line. Marked P and R on the line such that PR = 7.8 cm.
Step 2: At P, drawn a ray PX making an angle of 60° with PR.
Step 3: At R, drawn another ray RY making an angle of 35° with PR. Mark the point of intersection of the rays PX and RY as Q.
PQR is the required triangle.
(ii) ∠P = 115°, ∠Q = 40° and PQ = 6 cm
Solution:
Construction:
Step 1: Drawn a line. Marked P and Q on the line such that PQ = 6 cm.
Step 2: At P, drawn O ray PX making an angle of 115° with PQ.
Step 3: At Q, drawn another ray QY making an angle of 40° with PQ. Marked the point of intersection of the rays PX and Q Y as R.
PQR is the required triangle.
(iii) ∠Q = 90°, ∠R = 42° and QR = 5.5 cm
Solution:
Construction:
Step 1: Drawn a line. Marked Q and R on the line such that QR = 5.5 cm.
Step 2: At Q, drawn a ray QX making an angle of 90° with QR.
Step 3: At R, drawn another ray RY making an angle of 42° QR. Marked the point of intersection of the rays QX and RY as P.
PQR is the required triangle.
Objective Type Questions
Question 8.
If two plans figures are congruent then they have
(i) same size
(ii) same shape
(iii) same angle
(iv) same shape and same size
Answer:
(iv) same shape and same size
Question 9.
Which of the following methods are used to check the congruence of plane figures?
(i) translation method
(ii) superposition method
(iii) substitution method
(iv) transposition method
Answer:
(ii) superposition method
Question 10.
Which of the following rule is not sufficient to verify the congruency of two triangles.
(i) SSS rule
(ii) SAS rule
(iii) SSA rule
(iv) ASA rule
Answer:
(iii) SSA rule
Question 11.
Two students drew a line segment each. What is the condition for them to be congruent?
(i) They should be drawn with a scale.
(ii) They should be drawn on the same sheet of paper.
(iii) They should have different lengths.
(iv) They should have the same length.
Answer:
(iv) They should have the same length.
Question 12.
In the given figure, AD = CD and AB = CB. Identify the other three pairs that are equal.
(i) ∠ADB = ∠CDB, ∠ABD = ∠CBD, BD = BD
(ii) AD = AB, DC = CB, BD = BD
(iii) AB = CD, AD = BC, BD = BD
(iv) ∠ADB = ∠CDB, ∠ABD = ∠CBD, ∠DAB = ∠DBC
Answer:
(i) ∠ADB = ∠CDB, ∠ABD = ∠CBD, BD = BD
Question 13.
In ∆ABC and ∆PQR, ∠A = 50° = ∠P, PQ = AB, and PR = AC. By which property ∆ABC and ∆PQR are congruent?
(i) SSS property
(ii) SAS property
(iii) ASA property
(iv) RHS property
Answer:
(ii) SAS property
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