Chapter 1 exercise answers: Orienting Yourself: The Use of Coordinates

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(i) how far is the door from the left wall

1. In Fig. 1.3 one unit stands for 1 foot (the room is 12 ft by 10 ft, and B is (12, 10)).
2. The door D₁R₁ lies along the bottom wall, which is the x-axis. D₁ is the red point at the mark 8, and R₁ is at 11.5.
3. The distance of the door from the left wall (the y-axis) is the x-coordinate of its nearer end, D₁: 8 units.
4. Both ends of the door have y-coordinate 0, so the door lies on the x-axis itself: its distance from the x-axis is 0 units.

AnswerThe door is 8 units (8 ft) from the left wall (the y-axis) and 0 units from the x-axis — it lies on the x-axis.

(ii) What are the coordinates of D₁?

1. D₁ is on the x-axis, 8 units to the right of O, so its x-coordinate is 8 and its y-coordinate is 0.

AnswerD₁ = (8, 0).

(iii) If R₁ is the point (11.5, 0), how wide is the door?

1. D₁ (8, 0) and R₁ (11.5, 0) have the same y-coordinate, so the width of the door is the difference of their x-coordinates: 11.5 − 8 = 3.5 units.
2. One unit is 1 foot, so the door is 3.5 ft wide, which is about 1.07 m.
3. A typical room door is about 3 ft (about 90 cm) wide, so 3.5 ft is a comfortable width.
4. Doors meant for wheelchair users are usually expected to give at least about 3 ft (90 cm) of clear width. A 3.5 ft door gives more than that, so a person in a wheelchair should be able to enter easily.

AnswerThe door is 3.5 units (3.5 ft) wide. That is a comfortable width, and a person in a wheelchair should be able to enter easily.

(iv) is the bathroom door narrower or wider than the room door?

1. B₁ (0, 1.5) and B₂ (0, 4) have the same x-coordinate, so the bathroom door's width is the difference of their y-coordinates: 4 − 1.5 = 2.5 units.
2. The room door is 3.5 units wide (part iii), and 2.5 is less than 3.5.

AnswerThe bathroom door (2.5 ft) is narrower than the room door (3.5 ft).

Watch this explained “The room, from a corner”, 7:18 into Two axes, an origin, and why the order of the pair matters · हिंदी में देखें

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(i) Where will the fourth foot of the table be?

1. Call the three feet A (8, 9), B (11, 9) and C (11, 7).
2. A and B have the same y-coordinate (9), so AB is a horizontal side. B and C have the same x-coordinate (11), so BC is a vertical side. They meet at a right angle at B.
3. In a rectangle the fourth corner D is straight below A and straight to the left of C, so D takes A's x-coordinate (8) and C's y-coordinate (7).

AnswerThe fourth foot is at (8, 7).

(ii) Is this a good spot for the table?

1. The table covers x from 8 to 11 and y from 7 to 9.
2. In Fig. 1.5 the bed ends at about x = 6.5, the wardrobe is at x = 3 to 7 and y = 0 to 2, and the room door is on the bottom wall. The table overlaps none of these.
3. It is 1 ft from the right wall (x = 12) and 1 ft from the top wall (y = 10), and the open floor below it (y less than 7) leaves room for a chair.
4. Whether a spot is 'good' is a matter of judgement, so other answers are possible.

AnswerYes, it is a reasonable spot: the table does not overlap the bed, the wardrobe or the door, and there is open floor in front of it for a chair.

(iii) Can you make out the height of the table?

1. AB, from (8, 9) to (11, 9): same y-coordinate, so its length is 11 − 8 = 3 units.
2. BC, from (11, 9) to (11, 7): same x-coordinate, so its length is 9 − 7 = 2 units.
3. So the table top is 3 ft by 2 ft. The longer side is the length and the shorter side is the width (the book does the same for the dining room: length 18 ft, width 15 ft).
4. The coordinates give only positions on the floor plan — how far across and how far up the plan. The height of the table is measured straight up from the floor, and nothing in the plan records it.

AnswerWidth = 2 units (2 ft), length = 3 units (3 ft). No — the height cannot be found from these coordinates.

Watch this explained “The missing foot, and the shower”, 6:42 into The four quadrants, and reading a point's signs off its position · हिंदी में देखें

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1. The door is hinged at B₁ (0, 1.5) and its leaf is the segment B₁B₂, of length 4 − 1.5 = 2.5 units (from Exercise Set 1.1, part iv).
2. As the door swings open, its free tip stays exactly 2.5 units from the hinge B₁, tracing part of a circle of radius 2.5 units.
3. The wardrobe's corners are W₁ (3, 0), W₂ (7, 0), W₃ (7, 2), W₄ (3, 2); its nearest point to the hinge is its near edge at x = 3 (between y = 0 and y = 2), and since B₁'s y-coordinate 1.5 already falls in that range, the closest point on the wardrobe is (3, 1.5).
4. Distance from hinge to that nearest point = 3 − 0 = 3 units.
5. 3 units is more than the door's reach of 2.5 units, so the swinging door cannot touch the wardrobe — there is a 0.5-unit gap to spare.
6. If the door were made wider, its reach would grow past 2.5 units; once it reached 3 units the door would just touch the wardrobe, and any wider than that it would hit it.

AnswerNo, the door will not hit the wardrobe — it falls 0.5 unit short of the wardrobe's nearest edge. But there isn't much room to spare: making the door 0.5 unit (half a foot) wider would make it just graze the wardrobe, and any wider than that it would strike it. So if a wider door is wanted, either the wardrobe should be moved further away, or the door should be re-hung on its other edge (hinged at B₂ instead, swinging the other way).

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(i) the coordinates of the four corners O, F, R, and P

1. In Fig. 1.5 the bathroom is to the left of the y-axis. O is the origin, (0, 0).
2. F is on the y-axis at the mark 9, so F = (0, 9).
3. P is on the x-axis at the mark −6, so P = (−6, 0).
4. R is the top-left corner: straight above P and level with F, so R = (−6, 9).
5. Check: the bathroom is 0 − (−6) = 6 ft wide and 9 − 0 = 9 ft long, matching '6 ft × 9 ft' in Fig. 1.1.

AnswerO (0, 0), F (0, 9), R (−6, 9), P (−6, 0).

(ii) What is the shape of the showering area SHWR

1. Reading Fig. 1.5: S (−6, 6), H (−3, 6), W (−2, 9), R (−6, 9).
2. SH: both y-coordinates are 6, so SH is horizontal, of length −3 − (−6) = 3 units. WR: both y-coordinates are 9, so WR is horizontal, of length −2 − (−6) = 4 units. So SH and WR are parallel, with different lengths.
3. RS: both x-coordinates are −6, so RS is vertical. It meets SH and WR at right angles (at S and at R).
4. HW goes from (−3, 6) to (−2, 9): both coordinates change, so it is slanted.
5. A quadrilateral with exactly one pair of parallel sides is a trapezium. This one also has two right angles, so it is a right trapezium.

AnswerSHWR is a trapezium (a right trapezium). Its corners are S (−6, 6), H (−3, 6), W (−2, 9) and R (−6, 9).

(iii) Mark off a 3 ft × 2 ft space for the washbasin …

1. The free floor is below the shower: x from −6 to 0 and y from 0 to 6. The bathroom door is on the y-axis between B₁ (0, 1.5) and B₂ (0, 4), so keep the floor just inside the door clear.
2. Washbasin, 3 ft × 2 ft, in the bottom-left corner: (−6, 0), (−3, 0), (−3, 2), (−6, 2). Its sides are −3 − (−6) = 3 ft and 2 − 0 = 2 ft.
3. Toilet, 2 ft × 3 ft, against the left wall above the washbasin: (−6, 3), (−4, 3), (−4, 6), (−6, 6). Its sides are −4 − (−6) = 2 ft and 6 − 3 = 3 ft.
4. The two spaces do not overlap each other or the shower, and the way in from the door stays free. This is one possible answer; many other placements are also correct.

AnswerOne possible answer: washbasin space (−6, 0), (−3, 0), (−3, 2), (−6, 2); toilet space (−6, 3), (−4, 3), (−4, 6), (−6, 6). Other placements are also correct.

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(i) length of the dining room extends from point P to point A

1. From Fig. 1.5, P = (−6, 0) and A = (12, 0), both on the x-axis. PA = 12 − (−6) = 18 units, which is the given length, 18 ft.
2. Reiaan's room door D₁R₁ is on the x-axis and leads into the dining room, and the bedroom and bathroom are above the x-axis. So the dining room is on the other side of that wall: below the x-axis.
3. The width is 15 ft, so the room runs from y = 0 down to y = −15. Its other two corners are straight below A and P.
4. Sketch: a rectangle 18 units across and 15 units down, with its top side along the x-axis from P to A.

AnswerThe corners of the dining room are P (−6, 0), A (12, 0), (12, −15) and (−6, −15).

(ii) a rectangular 5 ft × 3 ft dining table precisely in the centre

1. The centre of the room is halfway across and halfway down: x = (−6 + 12) ÷ 2 = 3 and y = (0 + (−15)) ÷ 2 = −7.5. Centre = (3, −7.5).
2. Turn the table so that its 5 ft side runs along the length of the room (across) and its 3 ft side runs up and down. Its feet are at the corners of the table top.
3. Across: 5 ÷ 2 = 2.5 ft each side of x = 3, giving x = 0.5 and x = 5.5. Up and down: 3 ÷ 2 = 1.5 ft each side of y = −7.5, giving y = −6 and y = −9.
4. The table could also be turned the other way (3 ft across, 5 ft up and down); then its feet would be at (1.5, −5), (4.5, −5), (4.5, −10) and (1.5, −10).

AnswerWith the 5 ft side along the length of the room, the feet of the table are at (0.5, −6), (5.5, −6), (5.5, −9) and (0.5, −9).

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1. The point where the two axes cross is called the origin, O.
2. Every point on the x-axis has y-coordinate 0, and every point on the y-axis has x-coordinate 0.
3. The origin lies on both axes, so both its coordinates are 0.

Answerx-coordinate = 0, y-coordinate = 0; the point of intersection is O (0, 0).

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1. A line through W parallel to the y-axis is a vertical line; every point on it shares W's x-coordinate.
2. So H must have x-coordinate −5 — but its y-coordinate can be anything, since moving up or down stays on the same vertical line.
3. A point with x-coordinate −5 (negative) and y-coordinate positive lies in Quadrant II; with y-coordinate negative it lies in Quadrant III.
4. If H's y-coordinate were 0, H would sit on the x-axis itself, which is not inside any quadrant.

AnswerH's x-coordinate must be −5 (its y-coordinate cannot be predicted). H can lie in Quadrant II (if its y-coordinate is positive) or Quadrant III (if its y-coordinate is negative).

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(i) Two sides of RAMP that are perpendicular to each other.

1. Side AM runs from A (0,−2) to M (−5,−2): the y-coordinate does not change, so AM is horizontal.
2. Side MP runs from M (−5,−2) to P (−5,2): the x-coordinate does not change, so MP is vertical.
3. A horizontal side and a vertical side always meet at a right angle.

AnswerSides AM and MP are perpendicular to each other.

(ii) One side of RAMP that is parallel to one of the axes.

1. AM is horizontal (both its points have y = −2), so AM is parallel to the x-axis.
2. MP is vertical (both its points have x = −5), so MP is parallel to the y-axis too — there are actually two axis-parallel sides here.

AnswerAM is parallel to the x-axis (and MP is parallel to the y-axis as well).

(iii) Two points that are mirror images of each other in one axis.

1. Reflecting a point in the x-axis keeps its x-coordinate and flips the sign of its y-coordinate.
2. M = (−5, −2); flipping the sign of its y-coordinate gives (−5, 2), which is exactly P.
3. So M and P are mirror images of each other in the x-axis.

AnswerM and P are mirror images of each other, and the mirror is the x-axis.

Watch this explained “Manufacturing the right angle”, 1:06 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें

Open NCERT p. 12One way to think about it

1. Plot Z at (5, −6).
2. Pick any second point N so that the segment ZN is easy to measure — for instance, make ZN horizontal or vertical by giving N the same y-coordinate or the same x-coordinate as Z. Take N = (5, −2): it shares Z's x-coordinate, so ZN is vertical, of length |−2 − (−6)| = 4.
3. Pick a third point I that shares a coordinate with N, so that NI is easy to measure and meets ZN at a right angle. Take I = (9, −2): it shares N's y-coordinate, so NI is horizontal, of length |9 − 5| = 4, and NI ⊥ ZN (one horizontal, one vertical), giving a right angle at N.
4. The third side, IZ, is a slanted segment, so its length needs the distance formula (built from the Baudhāyana–Pythagoras theorem): IZ = √((9−5)² + (−2−(−6))²) = √(4² + 4²) = √32 = 4√2 units.

In shortOne valid triangle: I (9, −2), Z (5, −6), N (5, −2), right-angled at N, with ZN = 4 units, NI = 4 units and IZ = 4√2 units. (Any other choice of I and N that keeps the right angle works equally well — the book itself says answers will differ from person to person.)

Watch this explained “Manufacturing the right angle”, 1:06 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें

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1. Without negative numbers, every coordinate would have to be 0 or positive.
2. Such a system could only reach points with x ≥ 0 and y ≥ 0 — that is, Quadrant I together with the two positive half-axes and the origin.
3. Any point with a negative x or a negative y (such as a point in Quadrant II, III or IV) would have no pair of non-negative numbers that names it.

AnswerWithout negative numbers, the coordinate system could only locate points in one quarter of the plane (Quadrant I and its bordering half-axes). No, it would not be able to locate every point on the 2-D plane — three whole quadrants would be left with no address.

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1. Find the three distances using the distance formula: MA = √((0−(−3))²+(0−(−4))²) = √(9+16) = √25 = 5.
2. AG = √((6−0)²+(8−0)²) = √(36+64) = √100 = 10.
3. MG = √((6−(−3))²+(8−(−4))²) = √(81+144) = √225 = 15.
4. Three points are on one straight line exactly when the two shorter distances add up to the longest one: MA + AG = 5 + 10 = 15 = MG.

AnswerYes, M, A and G lie on the same straight line, because MA + AG = MG (5 + 10 = 15). Method: compute all three pairwise distances with the distance formula, and check whether the two smaller ones add up to the largest — no plotting needed.

Watch this explained “The near miss”, 7:22 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें

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1. RB = √((−2−(−5))²+(−5−(−1))²) = √(9+16) = √25 = 5.
2. BC = √((4−(−2))²+(−12−(−5))²) = √(36+49) = √85 ≈ 9.22.
3. RC = √((4−(−5))²+(−12−(−1))²) = √(81+121) = √202 ≈ 14.21.
4. Check: RB + BC ≈ 5 + 9.22 = 14.22, which is very close to RC ≈ 14.21 but not exactly equal — 5 + √85 ≠ √202 exactly, since (5+√85)² = 25+10√85+85 ≠ 202.

AnswerNo — R, B and C are not exactly on the same straight line, even though RB + BC comes extremely close to RC (about 14.22 vs 14.21). The near-equality is a coincidence of rounding, not an exact match, so the three points are not collinear.

Watch this explained “The near miss”, 7:22 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें

Open NCERT p. 12Checked by computerAnswers can differ: one example

(i) A right-angled isosceles triangle.

1. Take the origin O (0, 0) and two more points, one on each axis, at the same distance from O — for example (4, 0) and (0, 4).
2. The side from O to (4, 0) lies along the x-axis and the side from O to (0, 4) lies along the y-axis, so they meet at a right angle at O.
3. Both of these sides are 4 units long, so the triangle is right-angled and isosceles.
4. This is one example; any two points on the two axes at the same distance from O, such as (2, 0) and (0, 2), also work.

AnswerOne example: O (0, 0), (4, 0), (0, 4) — the right angle is at the origin and the two equal sides are 4 units each. Many other answers are possible.

(ii) one vertex in Quadrant III and the other in Quadrant IV

1. Take a point in Quadrant III (both coordinates negative), for example (−3, −4), and its mirror image in the y-axis, (3, −4), which is in Quadrant IV (x positive, y negative).
2. Distance from O to (−3, −4) = √((−3)² + (−4)²) = √25 = 5. Distance from O to (3, −4) = √(3² + (−4)²) = √25 = 5.
3. Two sides are equal, so the triangle is isosceles. (Its third side, from (−3, −4) to (3, −4), is 3 − (−3) = 6 units.)
4. This is one example; many other pairs of points work.

AnswerOne example: O (0, 0), (−3, −4), (3, −4) — the two sides meeting at the origin are both 5 units. Many other answers are possible.

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1. The midpoint of S (x₁, y₁) and T (x₂, y₂) is halfway between them in both directions: ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2). Work it out for each row and compare it with M.

2. Row	S	M	T	Midpoint of ST	Is M the midpoint?
   1	(−3, 0)	(0, 0)	(3, 0)	(0, 0)	Yes
   2	(2, 3)	(3, 4)	(4, 5)	(3, 4)	Yes
   3	(0, 0)	(0, 5)	(0, −10)	(0, −5)	No
   4	(−8, 7)	(0, −2)	(6, −3)	(−1, 2)	No

3. Row 1: ((−3 + 3) ÷ 2, (0 + 0) ÷ 2) = (0, 0), which is M. Yes.
4. Row 2: ((2 + 4) ÷ 2, (3 + 5) ÷ 2) = (3, 4), which is M. Yes.
5. Row 3: ((0 + 0) ÷ 2, (0 + (−10)) ÷ 2) = (0, −5), but M is (0, 5). No — M is not even between S and T.
6. Row 4: ((−8 + 6) ÷ 2, (7 + (−3)) ÷ 2) = (−1, 2), but M is (0, −2). No.
7. Connection: when M is the midpoint of ST, the x-coordinate of M is the average of the x-coordinates of S and T, and the y-coordinate of M is the average of their y-coordinates.

AnswerRow 1: Yes. Row 2: Yes. Row 3: No (the midpoint of ST is (0, −5)). Row 4: No (the midpoint of ST is (−1, 2)). Connection: M = ((x-coordinate of S + x-coordinate of T) ÷ 2, (y-coordinate of S + y-coordinate of T) ÷ 2).

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1. Since M is the midpoint of A and B: Mₓ = (Aₓ+Bₓ)/2 and Mᵧ = (Aᵧ+Bᵧ)/2, so Bₓ = 2Mₓ − Aₓ and Bᵧ = 2Mᵧ − Aᵧ.
2. Bₓ = 2(−7) − 3 = −14 − 3 = −17.
3. Bᵧ = 2(1) − (−4) = 2 + 4 = 6.

AnswerB = (−17, 6).

Open NCERT p. 13Checked by computer

1. P and Q cut AB into three equal parts, so A, P, Q and B are equally spaced along the segment.
2. Midpoint idea: the midpoint is the average of the two ends, each counted once. A point of trisection is a weighted average — for P, which is closer to A, count A twice and B once, then divide by 3; for Q, count A once and B twice.
3. P = ((2 × 4 + 16) ÷ 3, (2 × 7 + (−2)) ÷ 3) = (24 ÷ 3, 12 ÷ 3) = (8, 4).
4. Q = ((4 + 2 × 16) ÷ 3, (7 + 2 × (−2)) ÷ 3) = (36 ÷ 3, 3 ÷ 3) = (12, 1).
5. Check with equal steps: from A to B, x goes up by 16 − 4 = 12 and y goes down by 7 − (−2) = 9. One third of that is 4 across and 3 down: A (4, 7) → P (8, 4) → Q (12, 1) → B (16, −2).
6. Check with midpoints: the midpoint of A (4, 7) and Q (12, 1) is ((4 + 12) ÷ 2, (7 + 1) ÷ 2) = (8, 4) = P, and the midpoint of P (8, 4) and B (16, −2) is ((8 + 16) ÷ 2, (4 + (−2)) ÷ 2) = (12, 1) = Q.

AnswerP = (8, 4) and Q = (12, 1).

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(i) What is the radius of circle K?

1. A point lies on a circle centred at O exactly when its distance from O equals the radius.
2. OA = √(1²+(−8)²) = √(1+64) = √65.
3. OB = √((−4)²+7²) = √(16+49) = √65.
4. OC = √((−7)²+(−4)²) = √(49+16) = √65.
5. All three distances are equal, so A, B, C do lie on one circle centred at O, and its radius is √65.

AnswerRadius of circle K = √65 units (≈ 8.06 units).

(ii) D and E lie within the circle, on the circle, or outside

1. OD = √((−5)²+6²) = √(25+36) = √61 ≈ 7.81, which is less than √65 ≈ 8.06, so D is inside the circle.
2. OE = √(0²+9²) = √81 = 9, which is more than √65 ≈ 8.06, so E is outside the circle.

AnswerD lies within circle K, and E lies outside circle K.

Watch this explained “One old theorem, read in coordinates”, 8:27 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें

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1. Following the usual naming, D is the midpoint of BC, E is the midpoint of CA, and F is the midpoint of AB.
2. Each vertex is 'opposite' one midpoint and can be recovered as: A = F + E − D, B = D + F − E, C = D + E − F.
3. A = (0,3) + (6,5) − (5,1) = (0+6−5, 3+5−1) = (1, 7).
4. B = (5,1) + (0,3) − (6,5) = (5+0−6, 1+3−5) = (−1, −1).
5. C = (5,1) + (6,5) − (0,3) = (5+6−0, 1+5−3) = (11, 3).
6. Check: midpoint of B(−1,−1) and C(11,3) = (5,1) = D ✓; midpoint of C(11,3) and A(1,7) = (6,5) = E ✓; midpoint of A(1,7) and B(−1,−1) = (0,3) = F ✓.

AnswerA = (1, 7), B = (−1, −1), C = (11, 3).

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(ii)(a)

1. This numbering is really just a coordinate pair: the first number picks out one specific N–S street, and the second picks out one specific E–W street.
2. One specific street from each family cross at exactly one point, so a label like (4, 3) can only ever mean one crossing — the 4th N–S street meeting the 3rd E–W street.

AnswerExactly 1 street intersection can be referred to as (4, 3).

(ii)(b)

1. By the same reasoning, (3, 4) names the crossing of the 3rd N–S street with the 4th E–W street — again exactly one point.
2. Since 3 ≠ 4, (3, 4) is a different crossing from (4, 3), 200 m away from it in each direction.

AnswerExactly 1 street intersection can be referred to as (3, 4), and it is a different crossing from (4, 3).

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(i) whether any part of either circle lies outside the screen.

1. The screen covers x from 0 to 800 and y from 0 to 600.
2. Circle A (centre (100, 150), radius 80) reaches from x = 100 − 80 = 20 to x = 100 + 80 = 180, and from y = 150 − 80 = 70 to y = 150 + 80 = 230. All of this is on the screen.
3. Circle B (centre (250, 230), radius 100) reaches from x = 250 − 100 = 150 to x = 250 + 100 = 350, and from y = 230 − 100 = 130 to y = 230 + 100 = 330. All of this is on the screen too.

AnswerNo — no part of either circle lies outside the screen.

(ii) whether the two circles intersect each other.

1. Distance between the centres: AB = √((250 − 100)² + (230 − 150)²) = √(150² + 80²) = √(22500 + 6400) = √28900 = 170 pixels.
2. Sum of the radii = 80 + 100 = 180 pixels. 170 is less than 180, so the circles are close enough to overlap.
3. Difference of the radii = 100 − 80 = 20 pixels. 170 is more than 20, so the smaller circle is not sitting completely inside the larger one.
4. So the two circles cross each other.

AnswerYes — the two circles intersect: the distance between their centres (170 pixels) is less than the sum of their radii (180 pixels) and more than the difference of their radii (20 pixels).

Watch this explained “Letters instead of numbers”, 4:46 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें

Open NCERT p. 14Checked by computer

1. Find the four sides with the distance formula. AB = √((−1 − 2)² + (2 − 1)²) = √(9 + 1) = √10.
2. BC = √((−2 − (−1))² + (−1 − 2)²) = √(1 + 9) = √10. CD = √((1 − (−2))² + (−2 − (−1))²) = √(9 + 1) = √10. DA = √((2 − 1)² + (1 − (−2))²) = √(1 + 9) = √10. All four sides are equal.
3. Diagonals: AC = √((−2 − 2)² + (−1 − 1)²) = √(16 + 4) = √20 and BD = √((1 − (−1))² + (−2 − 2)²) = √(4 + 16) = √20. The diagonals are equal.
4. Right angle: in triangle ABC, AB² + BC² = 10 + 10 = 20 = AC², so by the converse of the Baudhāyana–Pythagoras theorem the angle at B is 90°.
5. Four equal sides and a right angle (and equal diagonals) make ABCD a square.
6. Area = side × side = √10 × √10 = 10 square units.

AnswerYes, ABCD is a square: all four sides are √10 units, both diagonals are √20 units, and the angle at B is 90°. Its area is 10 square units.

Watch this explained “One old theorem, read in coordinates”, 8:27 into The distance formula: Baudhāyana–Pythagoras rewritten in coordinates · हिंदी में देखें