CBSE Class 7 Mathematics Geometric Twins Assignment Set 01

Short Answer Questions

Question 1. What are congruent figures? Explain the test of superimposition that is used to check whether two figures are congruent.
Answer:
Figures that are exact copies of each other, having the same shape and the same size, are called congruent figures.
Test of superimposition: trace the first figure on tracing paper and place the tracing over the second figure.
If the tracing covers the second figure exactly, with no part sticking out and no gap left, the figures are congruent.
If some part does not cover, the figures are not congruent.
In simple words: Congruent figures are exact copies, and a tracing of one fits exactly on the other.

Teacher's Note:
1. Superimposition means placing one figure on top of the other.
2. The tracing may be turned or flipped.

Question 2. While checking congruence, a figure may be rotated or flipped before placing it on the other figure. Explain why this is allowed, and give one everyday example of a congruent pair in which one piece has to be turned around.
Answer:
Rotating or flipping changes only the position of a figure or the way it faces; it does not change its shape or size.
Congruence is only about shape and size, so we are allowed to turn or flip the tracing before superimposing it.
Example: our left and right hands (or two pieces cut together from a folded paper); one must be flipped over to fit on the other.
In simple words: Turning or flipping does not change shape or size, so it is allowed; our two hands are an example.

Teacher's Note:
1. Position does not matter in congruence.
2. Mirror-image pieces are still congruent.

Question 3. State the measurements you would take in each case, and give a reason:
(i) to draw a circle congruent to a given circle
(ii) to draw a rectangle congruent to a given rectangle
Answer:
(i) Only the radius (or the diameter). All circles have the same round shape, and the radius fixes the size completely.
(ii) The length and the breadth. All four angles of a rectangle are right angles already, so the two side lengths fix its shape and size.
In simple words: A circle needs only its radius; a rectangle needs its length and breadth.

Teacher's Note:
1. Ask which measurements can still change.
2. Fix exactly those measurements.

Question 4. A symbol on a signboard is made of two straight arms meeting at a point. The arm lengths are AB = 5 cm and BC = 9 cm. Is this information enough to redraw the symbol exactly? What extra measurement is needed and why?
Answer:
No, the two arm lengths alone are not enough.
With AB = 5 cm and BC = 9 cm, the arms can be opened out by different amounts at B, so many different-looking symbols can be drawn.
The extra measurement needed is the angle ∠ABC between the two arms.
The two lengths together with this angle fix both shape and size, so an exact copy can be drawn.
In simple words: Two arm lengths leave the opening free; measuring the angle at B fixes the symbol.

Teacher's Note:
1. Think of a pair of compasses opened by different amounts.
2. Lengths plus the angle between them fix the figure.

Question 5. State the SSS condition for the congruence of two triangles. Using a triangle with sides 5 cm, 7 cm and 9 cm, explain what the condition promises.
Answer:
SSS condition: if the three sides of one triangle are equal to the three sides of another triangle, the two triangles are congruent.
So every triangle drawn with sides 5 cm, 7 cm and 9 cm will be an exact copy of the given one.
Its angles need not be measured at all; the three sides already fix the shape and the size.
In simple words: Any triangle with sides 5, 7 and 9 cm is an exact copy of any other such triangle.

Teacher's Note:
1. Three sides fix a triangle completely.
2. The angles come out the same automatically.

Question 6. State the SAS condition for congruence. Explain why the angle used must be the one included between the two measured sides.
Answer:
SAS condition: if two sides and the angle included between them in one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent.
The included angle tells us how far the two known sides are opened out at their common vertex, so the third side is then fixed.
An angle at some other vertex does not fix this opening, and two different triangles may then be possible (the SSA case).
In simple words: The angle between the two sides fixes their opening; any other angle may allow two triangles.

Teacher's Note:
1. Included means lying between the two sides.
2. A non-included angle leads to SSA.

Question 7. It is given that △ PQR ≅ △ STU. Write:
(i) the three pairs of corresponding sides
(ii) the three pairs of corresponding angles
Answer:
The vertices match in the order written: P ↔ S, Q ↔ T, R ↔ U.
(i) PQ and ST, QR and TU, PR and SU.
(ii) ∠P and ∠S, ∠Q and ∠T, ∠R and ∠U.
In simple words: Match the letters by position and then read off the sides and angles.

Teacher's Note:
1. First, second and third letters go together.
2. A side uses two matched letters.

Question 8. In △ ABC, AB = AC and ∠B = 65°. Find ∠C and ∠A, giving a reason for each step.
Answer:
AB = AC. Angles opposite equal sides are equal; ∠C is opposite AB and ∠B is opposite AC.
So ∠C = ∠B = 65°.
The angles of a triangle add up to 180°, so ∠A = 180° − 65° − 65° = 50°.
In simple words: The base angles are both 65°, and the top angle is 180° − 130° = 50°.

Teacher's Note:
1. Find the equal angle first.
2. Then use the angle sum property.

Question 9. Explain why each angle of an equilateral triangle must measure 60°. Write the two facts you use in your reasoning.
Answer:
Facts used: angles opposite equal sides are equal; the three angles of a triangle add up to 180°.
In equilateral △ABC, AB = AC gives ∠B = ∠C, and AB = BC gives ∠A = ∠C. So all three angles are equal.
Hence 3 × (each angle) = 180°, so each angle = 60°.
In simple words: All three angles are equal and add to 180°, so each is 60°.

Teacher's Note:
1. Use two pairs of equal sides.
2. Then divide 180° by 3.

Question 10. In the figure, PQ = 7 cm, ∠QPS = 35° and an arc of radius 5 cm drawn from Q cuts the ray l at two points R and S.
Explain why this construction shows that the SSA condition does not guarantee congruence.
Answer:
The data PQ = 7 cm, the 35° angle at P and a side of 5 cm opposite that angle is two sides and a non-included angle (SSA).
The arc of radius 5 cm drawn from Q cuts the ray l at two points, R and S.
Both △ PQR and △ PQS have PQ = 7 cm, ∠P = 35° and a 5 cm side from Q, yet they have different shapes and sizes.
So the same measurements give two non-congruent triangles: SSA does not guarantee congruence.
In simple words: The 5 cm arc cuts the ray twice, so two different triangles fit the same data.

Teacher's Note:
1. Two crossing points mean two possible triangles.
2. That is why SSA is not a congruence condition.

Question 11. The angles of one triangle are 45°, 55° and 80°, and the angles of a second triangle are also 45°, 55° and 80°. Are the two triangles necessarily congruent? Give a reason and say what else must be known.
Answer:
No, they need not be congruent.
Equal angles make the triangles the same in shape only; many triangles of different sizes have angles 45°, 55° and 80°.
At least one pair of corresponding sides must also be equal. Then the ASA (or AAS) condition holds and the triangles are congruent.
In simple words: Same angles give the same shape; one equal side is needed to fix the size.

Teacher's Note:
1. AAA is not a congruence condition.
2. One side length fixes the size.

Question 12. It is given that △ CAT ≅ △ DOG. Write any three other correct ways of expressing the same congruence, and state the rule you followed while writing them.
Answer:
Here C ↔ D, A ↔ O, T ↔ G. Any three of these are correct:
△ ACT ≅ △ ODG, △ ATC ≅ △ OGD, △ TCA ≅ △ GDO, △ TAC ≅ △ GOD, △ CTA ≅ △ DGO.
Rule: the letters of both names must be rearranged in exactly the same way, so first matches first, second matches second and third matches third.
In simple words: Reorder both names together so that every matched pair stays in the same place.

Teacher's Note:
1. Never reorder only one triangle.
2. Check: C always above D, A above O, T above G.

Question 13. PQRS is a rectangle and PR is one of its diagonals. Show that △ PQR ≅ △ RSP, naming the condition you used and listing the three pairs of equal parts.
Answer:
In △ PQR and △ RSP:
PQ = RS (opposite sides of the rectangle)
QR = SP (opposite sides of the rectangle)
PR = RP (the diagonal is a common side)
All three pairs of sides are equal, so by the SSS condition, △ PQR ≅ △ RSP.
In simple words: Two pairs of opposite sides and the shared diagonal give SSS.

Teacher's Note:
1. Opposite sides of a rectangle are equal.
2. A common side counts as a pair of equal sides.

Question 14. In △ ABC and △ DEF, ∠A = ∠D = 40°, ∠C = ∠F = 70° and AB = DE = 6 cm. Show that the two triangles are congruent and express the congruence.
Answer:
By the angle sum, ∠B = 180° − 40° − 70° = 70° and ∠E = 180° − 40° − 70° = 70°. So ∠B = ∠E.
Now ∠A = ∠D, AB = DE and ∠B = ∠E, and the side AB lies between the two equal angles.
So the ASA condition holds and △ ABC ≅ △ DEF. (The given data itself is AAS.)
In simple words: Find the third angles, then the equal side lies between two equal angles: ASA.

Teacher's Note:
1. AAS can be changed into ASA using the angle sum.
2. Keep the matching A↔D, B↔E, C↔F.

Question 15. In the figure, O is the midpoint of both LN and MP.
Show that △ LOM ≅ △ NOP and hence write a pair of equal sides of the figure.
Answer:
In △ LOM and △ NOP:
LO = NO (O is the midpoint of LN)
MO = PO (O is the midpoint of MP)
∠LOM = ∠NOP (vertically opposite angles)
The equal angles lie between the equal sides, so by the SAS condition, △ LOM ≅ △ NOP.
Hence the corresponding sides are equal: LM = NP.
In simple words: Two halves and the vertically opposite angles give SAS, so LM = NP.

Teacher's Note:
1. Midpoint gives two equal halves.
2. Vertically opposite angles are equal.

Question 16. Two right-angled triangles have a hypotenuse of 13 cm each, and one more side of 5 cm each. Are they congruent? Name the condition used and explain which parts are matched.
Answer:
Yes, they are congruent.
Matched parts: the right angles, the hypotenuses (13 cm each, the sides opposite the right angles) and one more side (5 cm each).
This is the RHS condition, which guarantees congruence.
In simple words: Right angle, equal hypotenuse and one equal side: RHS.

Teacher's Note:
1. The hypotenuse must be one of the matched sides.
2. RHS works only for right-angled triangles.

Question 17. In the figure, PQ = PS and RQ = RS, and the diagonal PR is drawn.
Show that △ PQR ≅ △ PSR and explain why PR divides ∠QPS into two equal parts.
Answer:
In △ PQR and △ PSR:
PQ = PS (given)
RQ = RS (given)
PR = PR (common side)
By the SSS condition, △ PQR ≅ △ PSR.
So the corresponding angles ∠QPR and ∠SPR are equal.
These two angles together make ∠QPS, so PR divides ∠QPS into two equal parts.
In simple words: SSS gives the congruence, and its corresponding angles at P are equal halves of ∠QPS.

Teacher's Note:
1. Congruent triangles have equal corresponding angles.
2. This is the kite property.

Question 18. The SSA arrangement does not guarantee congruence, yet the RHS condition also uses two sides and an angle that is not between them. Explain why RHS still works.
Answer:
In SSA the given angle is not between the two given sides, so the arc drawn with the second side can cut the ray at two different points, giving two different triangles.
In RHS the angle is a right angle and the given side opposite it is the hypotenuse, the longest side.
The arc of the hypotenuse then meets the perpendicular line at only one point on each side of the base, and those two triangles are mirror images, so they are congruent.
So RHS always fixes the triangle.
In simple words: With a right angle and the hypotenuse, the arc gives only one triangle (up to flipping).

Teacher's Note:
1. RHS is the special case of SSA that works.
2. The hypotenuse is opposite the right angle.

Question 19. In △ ABC, AB = AC and ∠A = 60°.
(i) Find ∠B and ∠C, showing each step.
(ii) Compare the three angles of △ ABC with the angles of an equilateral triangle and say what you notice.
Answer:
(i) Step 1: AB = AC, so the angles opposite them are equal: ∠B = ∠C.
Step 2: ∠B + ∠C = 180° − 60° = 120°.
Step 3: Since they are equal, ∠B = ∠C = 60°. So ∠A = ∠B = ∠C = 60°.
(ii) An equilateral triangle also has all three angles equal to 60°. So △ABC has exactly the same angles as an equilateral triangle: an isosceles triangle whose angle between the equal sides is 60° has all its angles 60°.
In simple words: With AB = AC and ∠A = 60°, all three angles become 60°, just like an equilateral triangle.

Teacher's Note:
1. Base angles share 120° equally.
2. Such a triangle is in fact equilateral.

Question 20. In △ ABC, AB = 5 cm, BC = 6 cm, CA = 7 cm; in △ PQR, PQ = 6 cm, QR = 7 cm, RP = 5 cm. Are the triangles congruent? Write the congruence with the correct order of vertices and justify the order.
Answer:
Yes. Both triangles have sides 5 cm, 6 cm and 7 cm, so the SSS condition holds.
Match the equal sides: AB = RP (5 cm), BC = PQ (6 cm), CA = QR (7 cm).
B is the common vertex of AB and BC; P is the common vertex of RP and PQ. So B ↔ P. In the same way A ↔ R and C ↔ Q.
Hence △ ABC ≅ △ RPQ.
In simple words: Pair the equal sides, then the shared corners match: △ABC ≅ △RPQ.

Teacher's Note:
1. Match sides first, then vertices.
2. A vertex is shared by its two sides.