The Baudhayana-Pythagoras Theorem Class 8 Solutions

Construct the new square on a diagonal of the original square. If the old side is a, the diagonal has length a\sqrt2, so the new area is (a\sqrt2)^2=2a^2.

No. A square of side a has area a^2, while one of side 2a has area 4a^2. The area becomes four times as large.

The original square contains two congruent small right triangles. The dotted square contains four copies of the same triangle, so its area is twice the original area.

Each side of the original square bisects a right angle of the dotted square. In a square, an angle bisector through one vertex passes through the opposite vertex, so each extension reaches the required dotted-square vertex.

Each is a right isosceles triangle whose two legs equal half a diagonal of the original square. Thus corresponding legs and the included right angle are equal, so the triangles are congruent by SAS.

Each new square uses twice as many copies of the same small triangle as the preceding square. Therefore its area is twice the preceding area.

Place one triangular piece along each side of Square 1, with their hypotenuses forming the four outer sides. The four pieces have the area of one complete square, so together with Square 1 the new square has double the original area.

Join the midpoints of the four sides. The inner tilted quadrilateral is a square, and the four corner triangles together occupy half of the original square, leaving the other half inside.

The larger square splits into eight congruent small right triangles. Four form the inner square and four remain at the corners, so the inner square occupies half the total area.

No. The area factor is (1/2)^2=1/4. Four squares with half the side length fill the original square.

Its vertices are the side midpoints. The four corner triangles are congruent right triangles, so PQRS has four equal sides. Adjacent acute angles of those triangles add to 90 degrees, making every angle of PQRS a right angle. The four corner triangles occupy half the paper, so PQRS occupies the other half.

If the hypotenuse is c, the square on it has twice the area of a unit square. Hence c^2=2, so c=\sqrt2.

Squaring decimals gives 1.414^2=1.999396<2<2.002225=1.415^2, so 1.414<\sqrt2<1.415. A terminating decimal with a non-zero last digit cannot square to exactly 2.000\ldots. If \sqrt2=m/n in lowest terms, then m^2=2n^2, which gives incompatible parity for the exponent of prime 2. Thus \sqrt2 is irrational and non-terminating.

Rotate the four congruent pieces so their long sloping edges form the outer boundary of one square and their shorter edges meet at the centre. Since all four pieces from both original squares are used, the new square has the combined, hence doubled, area.

Use c=a\sqrt2. The results are: 3\sqrt2, with 4.2<c<4.3; 4\sqrt2, with 5.6<c<5.7; 6\sqrt2, with 8.4<c<8.5; 8\sqrt2, with 11.3<c<11.4; and 9\sqrt2, with 12.7<c<12.8.

For equal side a, 2a^2=10^2=100. Thus a^2=50 and a=5\sqrt2, approximately 7.071.

For perpendicular sides a,b and hypotenuse c, the rearrangement shows that the square on c has the combined areas a^2+b^2. When a=b, this becomes c^2=2a^2, exactly the earlier diagonal construction.

The four surrounding right triangles are congruent, so all four boundary sides are equal. At each vertex, the two acute angles come from a right triangle and sum to 90 degrees. The figure therefore has four equal sides and four right angles, so it is a square.

The two starting squares have total area a^2+b^2. Cutting and rearranging changes neither area nor overlap, and the pieces exactly cover a square of side c, the hypotenuse. Therefore a^2+b^2=c^2.

c=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13 cm.

b=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15 cm.

For original side a, first construct a diagonal a\sqrt2. Use perpendicular legs a\sqrt2 and a; their hypotenuse is a\sqrt3, giving area 3a^2. For five times, use perpendicular legs 2a and a; the hypotenuse is a\sqrt5, giving area 5a^2.

(i) c=\sqrt{5^2+7^2}=\sqrt{74}. (ii) c=\sqrt{8^2+12^2}=4\sqrt{13}. (iii) b=\sqrt{15^2-9^2}=12. (iv) c=\sqrt{7^2+12^2}=\sqrt{193}. (v) c=\sqrt{1.5^2+3.5^2}=\sqrt{14.5}.

The triples, with legs interchangeable, are (3,4,5), (5,12,13), (6,8,10), (8,15,17), (9,12,15), and (12,16,20).

Yes. They are 10(3,4,5) and 100(3,4,5). Scaling all three sides by the same positive number preserves a^2+b^2=c^2.

If a^2+b^2=c^2, then (ka)^2+(kb)^2=k^2(a^2+b^2)=k^2c^2=(kc)^2. Hence (ka,kb,kc) is also a triple.

Yes, because \gcd(5,12,13)=1. The primitive triples with every entry at most 20 are (3,4,5), (5,12,13), and (8,15,17).

Using scale factors 2 through 6: from (3,4,5): (6,8,10),(9,12,15),(12,16,20),(15,20,25),(18,24,30). From (5,12,13): (10,24,26),(15,36,39),(20,48,52),(25,60,65),(30,72,78). From (8,15,17): (16,30,34),(24,45,51),(32,60,68),(40,75,85),(48,90,102). None is primitive because the scale factor is a common factor.

Yes. Divide a^2+b^2=c^2 by f^2 to obtain (a/f)^2+(b/f)^2=(c/f)^2. For (9,12,15) with f=3, this gives (3,4,5).

The nth odd number is 2n-1. The sum of the first n-1 odd numbers is (n-1)^2, so (n-1)^2+(2n-1)=n^2.

For an odd m, set 2n-1=m^2. Then (n-1,m,n) is a triple. Taking m=7,9,11,13,15 gives (24,7,25), (40,9,41), (60,11,61), (84,13,85), and (112,15,113).

No. In every generated triple, one leg is exactly one less than the hypotenuse. Consecutive integers have no common factor greater than 1, so all three entries cannot share such a factor.

Yes. For example, (8,15,17) is primitive, but neither leg is 16, one less than the hypotenuse, so this method does not generate it.

The diagonal is the hypotenuse of a right triangle with legs 5 and 5: d=\sqrt{5^2+5^2}=5\sqrt2 cm.

Top left: \sqrt{7^2+9^2}=\sqrt{130}. Top right: \sqrt{4^2+10^2}=2\sqrt{29}. Middle left: \sqrt{41^2-40^2}=9. Middle right: \sqrt{200-10^2}=10. Bottom left: \sqrt{150-10^2}=5\sqrt2. Bottom right: \sqrt{45^2-27^2}=36.

Rhombus diagonals bisect each other at right angles. A side is therefore \sqrt{12^2+35^2}=\sqrt{1369}=37 units.

Yes. Since c^2=a^2+b^2, we have c^2>a^2 and c^2>b^2. All side lengths are positive, so c>a and c>b.

True. Divide all three entries by their greatest common divisor. The resulting integer triple has no common factor and is primitive; the original is its scaled version.

Use the legs and hypotenuse of any five triples: rectangles 3×4, 5×12, 8×15, 7×24, and 9×40 have diagonals 5, 13, 17, 25, and 41 respectively.

Construct a right triangle with hypotenuse 7 and one leg 5. Its other leg is \sqrt{7^2-5^2}=\sqrt{24}=2\sqrt6. A square on that leg has area 24=49-25 square units.

A grid square based on the vector (a,b) has side squared, and hence area, a^2+b^2. Area 2 is possible with (1,1); 3 is impossible; 4 is possible with (2,0); and 5 with (1,2). On an unlimited grid, the possible positive integer areas are exactly the integers expressible as a sum of two integer squares.

The altitude bisects the base, giving a right triangle with hypotenuse 6 and one leg 3. Its height is \sqrt{6^2-3^2}=3\sqrt3. Thus the area is \frac12\times6\times3\sqrt3=9\sqrt3 square units.

Open the box labelled RED. It cannot contain red. If the ball drawn is blue, that box is BLUE; the box labelled GREEN must then be RED, and the box labelled BLUE is GREEN. If the drawn ball is green, swap blue and green in that reasoning.