Important Questions Class 8 Maths Part 2 Chapter 2 The Baudhayana-Pythagoras Theorem
Important Questions Class 8 Maths Part 2 Chapter 2 The Baudhayana-Pythagoras Theorem
The Baudhayana-Pythagoras Theorem is Chapter 2 of the Class 8 Maths book Ganita Prakash Part 2. It is about the Baudhayana-Pythagoras theorem for right-angled triangles. Students find the missing side of a right triangle, check whether three lengths make a right triangle, and solve real-life problems.
This page gives 12 important questions from Class 8 Maths Part 2 Chapter 2 The Baudhayana-Pythagoras Theorem with step-by-step answers prepared by Extramarks experts. The questions cover finding the hypotenuse and other sides, Baudhayana (Pythagorean) triples, squares on the sides of a triangle and ladder problems. Solve each question on paper first and then check your working with the answer.
Q1. A child is flying a kite with a taut string of length 26 m. At a particular moment, the point vertically below the kite is 10 m away from the child.
(i) Using the Baudhāyana Theorem, find the height of the kite above the ground.
(ii) If the horizontal distance becomes 24 m while the kite string remains 26 m, find the new height of the kite.
Answer: (i) Let the height of the kite be h m.
Using the Baudhāyana Theorem:
Therefore, the height of the kite is 24 m.
(ii) When the horizontal distance becomes 24 m, let the new height be H m.
H = 10 m
Therefore, the new height of the kite is 10 m.
Q2. Using the generation pattern
the value 49 occurs as 2n – 1. A student claims that the corresponding Baudhāyana triple is (24, 49, 25).
(a) Find the value of n.
(b) Explain the error in the student’s claim and obtain the correct Baudhāyana triple.
(c) State whether the obtained triple is primitive. Justify your answer.
Answer: (a) 2n – 1 = 49 ⇒ 2n = 50 ⇒ n = 25. So, n – 1 = 24.
(b) The student has incorrectly taken 49 as a side length. In the generation pattern, 49 represents the square of one side.
Hence, the correct Baudhāyana triple is (7, 24, 25).
(c) The numbers 7, 24 and 25 have no common factor greater than 1, so (7, 24, 25) is a primitive triple.
Q3. A surveyor marks a small triangular safety zone using side lengths 7 m, 24 m and 25 m. For a larger zone, every length is multiplied by 3, giving 21 m, 72 m and 75 m.
(a) Verify that the small zone is right-angled and state whether its triple is primitive.
(b) Without squaring the larger numbers, explain why the larger zone is also right-angled and state the scale factor.
(c) Is the larger triple primitive? Justify.
Answer: (a) For the small triangular zone:
Hence, 7, 24, 25 form a right-angled triangle. Also, 7, 24 and 25 have no common factor greater than 1. Therefore, (7, 24, 25) is a primitive Baudhāyana triple.
(b) Each side of the larger triangle is 3 times the corresponding side of the smaller triangle.
Scaling all three sides of a right-angled triangle by the same factor keeps the relation true,
so the larger triangle is also right-angled. The scale factor is 3.
(c) No, the larger triple is not primitive because all three numbers have a common factor of 3.
Hence, (21, 72, 75) is a scaled triple, not a primitive triple.
Q4. The figure shows a right triangle with perpendicular sides 6 cm and 8 cm. Squares are constructed on all three sides.
(a) Find the area of the square constructed on the hypotenuse.
(b) The square on the hypotenuse is halved by joining the midpoints of its sides. Find the area of the halved square.
(c) A student claims that the side length of the halved square is 5 cm. Is the student correct? Justify.
Answer: (a) By the Baudhāyana–Pythagoras theorem, the area of the square on the hypotenuse equals the sum of the areas of the squares on the perpendicular sides.
(b) Halving the square gives half its area.
Hence, the student is not correct.
Q5. An exhibition uses two square display panels with side lengths 8 m and 15 m. The organisers want to replace them with one square panel having exactly the same total display area. An engineer suggests a different design: use the diagonal of the 15 m square as the side of the new square.
Based on the given information, answer the following questions.
(a) Find the total area that the replacement square must have.
(b) Find the side length of the required replacement square.
(c) Evaluate the engineer’s alternative design. Will it have the required area? If not, find by how much its area differs.
Answer: (a) The replacement square must have the same area as the two original square panels together.
(b) Let the side length of the replacement square be s m.
(c) The engineer proposes using the diagonal of the 15 m square as the side of the new square. For a square of side 15 m, the square of its diagonal is:
Required area:
Difference:
Q6. In an isosceles right triangle, the square constructed on the hypotenuse has area 72 sq. cm.
(a) Find the area of the square constructed on each equal side.
(b) Find the length of each equal side.
(c) Without evaluating a decimal, state between which two consecutive whole numbers the hypotenuse lies.
Answer: (a) Let the length of each equal side be a cm and the hypotenuse be c cm. The area of the square on the hypotenuse is 72 sq. cm. Therefore,
Since the triangle is an isosceles right triangle, the squares on the two equal sides have equal areas. By the Baudhāyana–Pythagoras theorem,
(b) Length of each equal side
(c) Whole-number bounds for the hypotenuse
Since 8² = 64 < 72 < 81 = 9², the hypotenuse lies between 8 cm and 9 cm.
Q7. The figure shows a square of side 12 cm. The midpoints of its four sides are joined to form the inner square.
Find the area of the inner square without finding its side length.
Answer: Joining the midpoints of the four sides forms a square whose area is half the area of the original square.
Area of the original square:
Therefore, area of the inner square:
Q8. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance of 'a' metre from the wall. Find the distance of the foot of the ladder from the wall.
Answer:
Thus, the foot of the ladder is 9 m away from the wall.
Q9. A 13 m long ladder reached a ventilator 12 m high from the ground on placing it against a wall at a distance. Find the distance of the foot of the ladder from the wall.
Answer: Let AB be the height of the ventilator, 12 m, and AC be the ladder of length 13 m.
By Pythagoras' theorem,
AC² = AB² + BC²
13² = 12² + BC²
BC² = 169 – 144 = 25
BC = 5 m
Thus, the distance of the foot of the ladder from the wall is 5 m.
Q10. Alex goes 40 m due North and then 50 m due West. Find the distance from the starting point.
Answer:
By Pythagoras Theorem,
Q11. ABC is a triangle, right-angled at B. M is a point on BC. Prove that: AM² + BC² = AC² + BM².
Answer:
By Pythagoras Theorem,
By Pythagoras Theorem,
Add equation (1) and equation (2), we get
Q12. If the area of the larger square is 50 square units, find the side length of the smaller square.
Answer: The area of the larger square is twice the area of the smaller square.
Area of the smaller square = 50 ÷ 2 = 25 square units
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Q.1 Using identity, find the value of (4.7)2.
Marks:2
Ans
Use the identity as (a b)2 = a2 2ab + b2.
Hence, the required value is 22.09.
Q.2 Simplify: (a + 2) (c 4) + (a 2) (c + 2) + 2 (ac + 17)
Marks:2
Ans
(a + 2) (c 4) + (a 2) (c + 2) + 2 (ac + 17)
= ac 4a + 2c 8 + ac + 2a 2c 4 + 2ac + 34
= 4ac 2a + 22
Q.3 The value of 1001 × 1004 is
A. 1006004
B. 1005004
C. 1004004
D. 1003004
Marks:1
Ans [ 2049078 ]
Q.4 If the length of a rectangle is 2x2 + 3 units and its breadth is 2x2 7 units. What is the area and perimeter of the rectangle
Marks:4
Ans
(x + a) (x + b) = x2 + x (a + b) + ab
Area = length × breadth
= (2x2 + 3) × (2x2 7)
= (2x2)2 + 2x2(3 7) + 3(7)
= 4x4 8x2 21 square units
Perimeter = 2(length + breadth)
= 2(2x2 + 3 + 2x2 7)
= 2(4x2 4)
= 8x2 8 units
Q.5
Marks:2
Ans
CBSE Class 8 Maths Important Questions
FAQs (Frequently Asked Questions)
The most important topics are finding missing side lengths, identifying Baudhayana triples, primitive and scaled triples, square diagonals, isosceles right triangle questions, and real-life applications like the lotus-in-a-lake problem.
Square the two smaller numbers and add them. If the sum equals the square of the largest number, the numbers form a Baudhayana triple. For example, 8² + 15² = 17², so 8, 15, 17 is a triple.
A primitive Baudhayana triple has no common factor greater than 1. For example, 3, 4, 5 is primitive. A scaled triple is made by multiplying a primitive triple by the same number, such as 6, 8, 10.
First identify the hypotenuse. If the hypotenuse is missing, add the squares of the two shorter sides. If one shorter side is missing, subtract the square of the known side from the square of the hypotenuse.
The lotus-in-a-lake problem is important because it turns a real situation into a right triangle. It tests whether students can assign variables, draw the triangle, identify the hypotenuse, and apply the theorem correctly.