WebFeb 28, 2024 · Ex 6.5 Class 7 Maths Question 1. PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR. Solution: In right angled triangle PQR, we have QR 2 = PQ 2 + PR 2 From Pythagoras property) = (10) 2 + (24) 2 = 100 + 576 = 676 ∴ QR = = 26 cm The, the required length of QR = 26 cm. Ex 6.5 Class 7 Maths Question 2. WebSep 19, 2024 · Ex 6.5 Class 12 Maths Question 14. Find two positive numbers x and y such that x + y = 60 and xy 3 is maximum. Solution: We have x + y = 60 ⇒ y = 60 – x …(i) …
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle …
WebOct 17, 2024 · Karnataka State Syllabus Class 6 Maths Chapter 7 Fractions Ex 7.5. Question 1. Write these fractions appropriately as additions or subtractions:-. Solution: … WebClass 7 Maths Chapter 6 Exercise 6.5 Solution. All the contents are modified and updated for academic session 2024-23 CBSE and state board. Class 7 math NCERT exercise … fireworks nova scotia
NCERT Solutions Class 7 Maths Chapter 6 The Triangle …
WebEx 6.5 Class 10 Maths Question 7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Solution: The figure given below shows a rhombus ABCD in which AB = BC = CD = DA. The diagonals AC and BD bisect each other at O. In ∆AOB, ∠AOB = 90° Ex 6.5 Class 10 Maths Question 8. WebJun 23, 2024 · 0:00 / 44:49 Maths Introduction: The Triangles And its Properties Class 7 Maths Chapter 6 Exercise 6.5 The Triangle & Its Properties NCERT Class 7 Maths Magnet Brains 8.79M... WebSep 19, 2024 · Ex 6.5 Class 12 Maths Question 7. Find both the maximum value and the minimum value of 3x 4 – 8x 3 + 12x 2 – 48x + 25 on the interval [0,3]. Solution: Let f (x) = 3x 4 – 8x 3 + 12x 2 – 48x + 25 ∴f' (x) = 12x 3 – 24x 2 + 24x – 48 = 12 (x 2 + 2) (x – 2) For maxima and minima, f' (x) = 0 ⇒ 12 (x 2 + 2) (x – 2) = 0 ⇒ x = 2 fireworks nps