Figure it Out — Rectangles and Squares — Quadrilaterals

1. 14 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 94

   Find all the other angles inside the following rectangles. (i) Rectangle ABCD in which the diagonal AC makes an angle of 30° with the side AB. (ii) Rectangle PQRS whose diagonals meet at O, with ∠QOR = 110°.

   Hint. The diagonals of a rectangle are equal and bisect each other, so all four triangles they cut the rectangle into are isosceles.

   The single fact doing all the work here: the diagonals of a rectangle are equal and bisect each other, so if O is their intersection then OA = OB = OC = OD. Every one of the four triangles around O is therefore isosceles, and its two base angles are equal.

   (i) Rectangle ABCD, ∠CAB = 30°

   In △ABD, the angle at A is 90° (angle of a rectangle) and ∠ABD is what we want: ∠ABD = 180° − 90° − ∠ADB. Taking the diagonals through O: since OA = OB, triangle AOB is isosceles, so ∠OBA = ∠OAB = 30°. And ∠ABD is exactly ∠OBA, because D, O and B are collinear. ∠ABD = 30°

   ∠CAD is the rest of the right angle at A: ∠CAD = ∠DAB − ∠CAB = 90° − 30° = 60°

   In △ABD, angles sum to 180° with the right angle at A: ∠ADB = 180° − 90° − 30° = 60°

   ∠BDC is the rest of the right angle at D: ∠BDC = ∠ADC − ∠ADB = 90° − 60° = 30°

   Since AB ∥ DC and AC is a transversal, ∠ACD and ∠CAB are alternate angles: ∠ACD = 30°

   ∠ACB = ∠BCD − ∠ACD = 90° − 30° = 60°

   (ii) Rectangle PQRS, ∠QOR = 110°

   Vertically opposite angles at O: ∠POS = 110°

   Linear pairs along each diagonal: ∠QOP = 180° − 110° = 70°, and ∠ROS = 70°

   Now use the isosceles triangles. In △QOR, OQ = OR, so its base angles are equal: ∠OQR = ∠ORQ = (180° − 110°) ÷ 2 = 35°

   In △POQ, OP = OQ, so: ∠OQP = ∠OPQ = (180° − 70°) ÷ 2 = 55°

   In △ROS, OR = OS, so: ∠ORS = ∠OSR = (180° − 70°) ÷ 2 = 55°

   Check: at vertex Q the two parts must rebuild the right angle — 35° + 55° = 90° ✓

   ✦ (i) ∠ABD = 30°, ∠CAD = 60°, ∠ADB = 60°, ∠BDC = 30°, ∠ACD = 30°, ∠ACB = 60°. (ii) ∠POS = 110°, ∠QOP = ∠ROS = 70°, ∠OQR = ∠ORQ = 35°, ∠OQP = ∠OPQ = ∠ORS = ∠OSR = 55°.

2. 24 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 94

   Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of (i) 30° (ii) 40° (iii) 90° (iv) 140°.

   Hint. The chapter proved that equal diagonals bisecting each other force a rectangle, whatever the angle between them.

   Construction (the same steps for every part, only the angle changes):

   1. Draw a line segment AB = 8 cm — this will be one diagonal.
   2. Mark its midpoint O, so AO = OB = 4 cm.
   3. At O, draw a ray making the required angle with OB.
   4. On that ray, cut off OC = OD = 4 cm on either side of O, so the second diagonal CD is also 8 cm and is also bisected at O.
   5. Join AD, DB, BC and CA.

   ADBC is the required quadrilateral. Parts (ii), (iii) and (iv) follow exactly the same process with 40°, 90° and 140° at step 3.

   What you should notice about the results — this is the real point of the question.

   The chapter's Deduction proved that if the diagonals are equal and bisect each other, then all four angles of the quadrilateral come out as 90°, no matter what angle the diagonals make. Working it through: if the angle between the diagonals is x, the four angles at O are x, x, 180° − x, 180° − x. Each of the four triangles round O is isosceles, so their base angles are (180° − x)/2 = 90° − x/2 and x/2 respectively. Each corner of the quadrilateral is made of one of each: (90° − x/2) + (x/2) = 90°.

   So every one of the four figures is a rectangle, since a quadrilateral with all angles 90° is by definition a rectangle.

   The special case is (iii): when the diagonals cross at 90°, the four triangles become congruent by SSS, so all four sides come out equal too — and the rectangle is a square.

   ✦ All four constructions give rectangles, since equal diagonals bisecting each other force every angle to be 90° whatever the angle between them; the 90° case (iii) gives a square.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 94

   Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.

   Hint. PL and AM are the diagonals of the quadrilateral APML. Check the three things the chapter says decide a square.

   Step 1 — Identify what PL and AM are for the quadrilateral. The quadrilateral APML has vertices A, P, M, L. Its diagonals are the segments joining opposite vertices, which are AM and PL — exactly the two diameters.

   Step 2 — Check the three conditions.

   They are equal. Both are diameters of the same circle, and every diameter of a circle has the same length, namely twice the radius. So AM = PL.

   They bisect each other. Every diameter passes through the centre O, and the centre is the midpoint of any diameter since OA = OM = OP = OL = the radius. So both diagonals are cut in half at O.

   They are perpendicular. This is given.

   Step 3 — Apply the chapter's results. Equal diagonals that bisect each other already force all four angles to be 90°, so APML is at least a rectangle. Adding that they meet at 90°, the four triangles AOP, POM, MOL and LOA are congruent by SSS (two radii and the included right angle in each), so all four sides are equal as corresponding parts.

   A quadrilateral with all angles 90° and all sides equal is a square.

   ✦ APML is a square, because PL and AM are its diagonals and they are equal (both diameters), bisect each other (both pass through the centre) and are perpendicular (given).

4. 43 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 94

   We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these?

   Hint. This is the Carpenter's Problem in reverse — build a rectangle, and its corners are your right angles.

   Step 1 — Lay the two sticks as diagonals. Let the sticks be AB and CD, equal in length. Place them across each other so that their midpoints coincide at a point O. Finding the midpoint of a stick needs no measuring instrument — fold a length of thread against the stick and halve it, or balance the stick.

   Step 2 — Join the four endpoints with the thread. Run the thread from A to C, C to B, B to D and D back to A, forming the quadrilateral ACBD.

   Step 3 — Why the corners are exactly 90°. The two sticks are now the diagonals of ACBD, and by construction they are equal in length and bisect each other. That is precisely the condition the chapter proved forces every angle of the quadrilateral to be 90°: if the angle between the diagonals is x, each corner works out to (90° − x/2) + (x/2) = 90°.

   So ACBD is a rectangle, and each of ∠A, ∠C, ∠B and ∠D is an exact right angle — whatever angle you happened to cross the sticks at.

   Why this matters. This is not a textbook trick. Carpenters in Europe use exactly this method to square up a frame, and farmers in Mozambique use it to lay out a rectangular house base — no protractor required.

   A second method exists using the properties of an isosceles triangle; the book invites you to find it.

   ✦ Cross the two equal sticks so their midpoints meet, then join the four endpoints with the thread — the diagonals are equal and bisect each other, so the resulting quadrilateral is a rectangle and each of its corners is exactly 90°.

5. 53 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 94

   We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle?

   Hint. Try to draw a quadrilateral with opposite sides parallel and equal but with no right angle.

   No — this cannot be taken as a definition of a rectangle.

   Step 1 — See what the proposed definition actually captures. A quadrilateral whose opposite sides are parallel is, by the chapter's own definition, a parallelogram. So the proposed definition would let in every parallelogram, not just the rectangles.

   Step 2 — Produce a counter-example. Take the parallelogram the chapter constructs with adjacent sides 4 cm and 5 cm and an angle of 30° between them. Its opposite sides are parallel and equal — yet its angles are 30°, 150°, 30°, 150°, not one of them a right angle. It is plainly not a rectangle.

   Step 3 — Say what is missing. A definition has to state conditions that a shape satisfies if and only if it belongs to the class. 'Opposite sides parallel and equal' is a property every rectangle has, but it is not enough on its own, since parallelograms have it too. The missing ingredient is the angle condition.

   What does work as a definition (each of these three is shown in the chapter to pick out exactly the rectangles): • all four angles are 90°; • all four angles are 90° and opposite sides are equal; • the diagonals are equal and bisect each other.

   The general lesson is worth keeping: a property of a shape is not automatically a definition of it. A definition must exclude everything that is not in the class.

   ✦ No. Opposite sides parallel and equal defines a parallelogram, not a rectangle — a 4 cm by 5 cm parallelogram with a 30° angle satisfies it and has no right angle at all.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp104.pdf). The chapter builds every property by deduction (congruence and transversal arguments) rather than assertion, and its three 'Figure it Out' blocks sit on pages 94, 102 and 107-109. Every angle answer here was independently recomputed from the stated configuration and then checked against the book's own printed answer key. TWO DEVIATIONS ARE FLAGGED IN PLACE: (1) the key's answer to the Venn-diagram question 4(ii) contradicts its own answer to 4(i) and its own printed diagram — the mathematically correct answer under the book's stated definition of a kite is given, with the discrepancy explained; (2) question 5 (∠IOD in rectangles PAIR and RODS) has an answer in the key but its derivation depends on the exact printed figure, which cannot be recovered from the text, so the answer is cited rather than derived and this is stated openly in the solution.. Questions are referenced from the NCERT textbook for identification.