CMDLearn

Learn

Class 7 Maths: Geometric Twins, 8 practice questions with answers

CMDLearn ·

These are 8 practice questions on Chapter 1, Geometric Twins, of the new NCERT Class 7 Maths textbook, Ganita Prakash, Part 2. Each has a worked answer and the textbook page it comes from. They cover matching the vertices of two congruent triangles, the extra measurement that makes two brackets match, a boat's course and a buoy's range, a midpoint in an isosceles triangle, the angles of isosceles triangles, two tiles with the same angles that still do not match, and congruence in a real setting. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the distinctive numbers is taken from the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 7, Part 2, Chapter 1, printed pages 1 to 23):

Matching and conditions

Question 1. In ∆PQR, PQ = 11 cm, QR = 9 cm and RP = 6 cm. In ∆LMN, LM = 9 cm, MN = 6 cm and NL = 11 cm. Are the triangles congruent? If so, write the congruence with the vertices in the right order.

Answer: yes, by SSS: ∆PQR ≅ ∆NLM. The three sides of each triangle are 11, 9 and 6 cm, so the SSS condition holds. To match the vertices, look at which two sides meet at each one. Q is where the 11 cm and 9 cm sides meet; in ∆LMN those are NL and LM, which meet at L. P is where the 11 cm and 6 cm sides meet, and in ∆LMN that is N. R is where the 9 cm and 6 cm sides meet, which is M. So P matches N, Q matches L and R matches M, and we write ∆PQR ≅ ∆NLM. Writing ∆PQR ≅ ∆LMN would wrongly match side PQ (11 cm) with LM (9 cm). (The SSS condition and writing congruence, pages 5 to 8.)

Question 2. A carpenter is checking whether two triangular shelf brackets are exactly the same. Each bracket has a wall edge, a shelf edge and a slanting edge. For each case, name ONE more measurement that would make her sure they match, and the condition it gives. (a) The wall edges are equal and the shelf edges are equal. (b) Both have a right angle between the wall edge and the shelf edge, and their slanting edges are equal. (c) The angle at the top of the wall edge is the same in both, and so is the angle at the far end of the shelf edge.

Answer: (a) the angle between the wall and shelf edges (SAS), or the slanting edges (SSS); (b) one of the other two edges, say the wall edges (RHS); (c) any one pair of matching edges (ASA or AAS). (a) Two equal sides are not enough on their own. If the angle between those two edges is also equal, the condition is SAS; if instead the third edges are equal, it is SSS. (b) The slanting edge is the hypotenuse, opposite the right angle. One more pair of equal edges, for example the wall edges, gives RHS. (c) Equal angles alone only fix the shape, not the size. Adding one pair of matching edges settles it: the edge between the two angles (the slanting edge) gives ASA, and an edge that is not between them gives AAS. (Which measurements guarantee congruence, pages 9 to 17.)

Question 3. A boat leaves a jetty on a straight course that makes an angle of 33° with a straight shore. A buoy stands on the shore 8 km from the jetty, and its radio can be heard up to 6 km away. Explain why there are two points on the boat's course that are exactly 6 km from the buoy, and find roughly how far from the jetty each one is.

Answer: at about 2.6 km and about 10.8 km from the jetty. Draw the shore line with the jetty and the buoy 8 km apart, and the course from the jetty at 33° to the shore. A circle of radius 6 km round the buoy crosses the course at two points, because the course passes closer to the buoy than 6 km. The boat starts hearing the radio at the first crossing and loses it at the second. A careful scale drawing (or a calculation) gives about 2.6 km and about 10.8 km. Each point, with the jetty and the buoy, makes a triangle with sides 8 km and 6 km and the 33° angle at the jetty, so the same three measurements fit two different triangles: two sides with an angle that is not between them do not fix a triangle. (Two sides and a non-included angle, pages 10 to 12.)

Question 4. In ∆PQR, PQ = PR, and S is the midpoint of QR. (a) Show that ∆PQS ≅ ∆PRS. (b) Use this to show that PS is perpendicular to QR.

Answer: (a) by SSS; (b) ∠PSQ = ∠PSR and together they make 180°, so each is 90°. (a) PQ = PR is given; QS = RS because S is the midpoint of QR; and PS is common to both triangles. All three sides match, so ∆PQS ≅ ∆PRS by SSS. (b) Matching parts of congruent triangles are equal, so ∠PSQ = ∠PSR. These two angles sit side by side on the straight line QR, so together they make 180°, and each is 90°. So PS is perpendicular to QR. The same congruence also shows ∠Q = ∠R. (SSS and matching parts, pages 6 to 8; angles of an isosceles triangle, pages 17 and 18.)

Angles and real twins

Question 5. (a) An isosceles triangle has the angle between its equal sides equal to 54°. Find the other two angles. (b) In another isosceles triangle, each of the two equal angles is 52°. Find the third angle.

Answer: (a) 63° each; (b) 76°. The angles opposite the equal sides are equal. (a) The two equal angles share 180° − 54° = 126°, so each is 63°. (b) The third angle is 180° − 52° − 52° = 76°. (Angles of isosceles triangles, pages 17 and 18.)

Question 6. One angle of an isosceles triangle is 104°. Find the other two angles, and explain why 104° cannot be one of the two equal angles.

Answer: 38° and 38°. If 104° were one of the two equal angles, the triangle would have two angles of 104°, which add to 208°, more than 180°: impossible. So 104° is the angle between the equal sides, and the other two angles are equal and share 180° − 104° = 76°, so each is 38°. (Angles of isosceles triangles, pages 17 and 18; angle sum, page 15.)

Question 7. Tile A is a triangle with angles of 50° and 70°, and its 5 cm side is opposite the 50° angle. Tile B is a triangle with angles of 70° and 60°, and its 5 cm side is opposite the 60° angle. Are the two tiles congruent? Explain.

Answer: no. By the angle sum, Tile A's third angle is 180° − 50° − 70° = 60°, and Tile B's third angle is 180° − 70° − 60° = 50°. So both tiles have angles of 50°, 60° and 70°. But for AAS (or ASA, once the third angles are known) the equal sides must sit in the same place: in Tile A the 5 cm side faces the 50° angle, while in Tile B it faces the 60° angle. The matching sides are not equal, so the tiles are not congruent; one is a different size from the other. (The angle sum, page 15; two angles and a side, pages 14 to 16.)

Question 8. Two poles stand straight up on level ground. From the top of each pole a 10 m rope is pulled tight to a peg in the ground 6 m from the foot of that pole. Explain, using congruent triangles, why the two poles must be the same height.

Answer: the two right triangles are congruent by RHS, so the poles, which match, are equal. Each pole, its rope and the ground between the foot and the peg form a triangle with a right angle at the foot of the pole. In both triangles the rope, opposite the right angle, is the hypotenuse, and it is 10 m in each; the ground side is 6 m in each. Right angle, equal hypotenuse and one equal side: by RHS the two triangles are congruent. The poles are matching sides, so they are equal in height. (Checking with coordinates gives a height of 8 m for each, but the congruence argument does not need the number.) (The RHS condition, pages 16 and 17.)

Practise more, and check your marks

Where CMDLearn fits

CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.

Sources

All opened and read on 1 November 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page for this chapter.

The rules are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.