Rs aggarwal class 9 chapter 14

RS Aggarwal solutions for class 9 chapter 14 statistics are made available here. The chapter statistics for class 9 deals with Collection of data, presentation of data, the graphical representation of data in the form of bar graphs, histograms, and frequency polygon. Along with these students will also learn what is Measures of central tendency. The RS Aggarwal Class 9 Solutions Chapter 14

ML Aggarwal Solutions for Class 9 Maths Chapter 14 – Theorems on Area are provided here to help students understand all the concepts clearly and develop a strong command over the subject. This chapter mainly deals with problems based on theorems. Chapter 14 – Theorems on Area contains one exercise, and the Download PDF carouselExampleControls111 Previous Next Access Answers to ML Aggarwal Solutions for Class 9 Maths Chapter 14 – Theorems on Area. EXERCISE 14 1. Prove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms. Solution: Let us consider ABCD be a parallelogram in which E and F are mid-points of AB and CD. Join EF. To prove: ar (|| AEFD) = ar (|| EBCF) Let us construct DG ⊥ AG and let DG = h where h is the altitude on side AB. Proof: ar (|| ABCD) = AB × h ar (|| AEFD) = AE × h = ½ AB × h ….. (1) [Since E is the mid-point of AB] ar (|| EBCF) = EF × h = ½ AB × h …… (2) [Since E is the mid-point of AB] From (1) and (2) ar (|| ABFD) = ar (|| EBCF) Hence proved. 2. Prove that the diagonals of a parallelogram divide it into four triangles of equal area. Solution: Let us consider in a parallelogram ABCD, the diagonals AC and BD are cut at point O. To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD) Proof: In parallelogram ABCD, the diagonals bisect each other. AO = OC In ∆ACD, O is the mid-point of AC. DO is the median. ar (∆AOD) = ar (COD) ….. (1) [Median of ∆ divides it into two triangle...

Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 14 Areas Of Triangles And Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Areas Of Triangles And Quadrilaterals are extremely popular among Class 9 students for Maths Areas Of Triangles And Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate. Answer: Let the height of the triangle be h m. ∴ Base = 3 h m Now, Area of the triangle = Total Cost Rate = 783 58 = 13 . 5 ha = 135000 m 2 We have: Area of triangle = 135000 m 2 ⇒ 1 2 × Base × Heigh t = 135000 ⇒ 1 2 × 3 h × h = 135000 ⇒ h 2 = 135000 × 2 3 ⇒ h 2 = 90000 ⇒ h = 300 m Thus, we have: Height = h = 300 m Base = 3 h = 900 m Page No 533: Answer: Let : a = 42 cm , b = 34 cm and c = 20 cm ∴ s = a + b + c 2 = 42 + 34 + 20 2 = 48 cm By Heron ' s formula , we have : Area of triangle = s ( s - a ) ( s - b ) ( s - c ) = 48 ( 48 - 42 ) ( 48 - 34 ) ( 48 - 20 ) = 48 × 6 × 14 × 28 = 4 × 2 × 6 × 6 × 7 × 2 × 7 × 4 = 4 × 2 × 6 × 7 = 336 cm 2 We know that the longest side is 42 cm. Thus, we can find out the height of the triangle corresponding to 42 cm. We have: Area of triangl...

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 14 Theorems on Area Question 1. Prove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms. Solution: Question 2. Prove that the diagonals of a parallelogram divide it into four triangles of equal area. Solution: Question 3. (a) In the figure (1) given below, AD is median of ∆ABC and P is any point on AD. Prove that (i) area of ∆PBD = area of ∆PDC. (ii) area of ∆ABP = area of ∆ACP. (b) In the figure (2) given below, DE || BC. prove that (i) area of ∆ACD = area of ∆ ABE. (ii) area of ∆OBD = area of ∆OCE. Solution: Question 4. (a) In the figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD. (b) In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that: (i) area of ∆OAB + area of ∆OCD = \(\frac \) area of ∆ ABC. Solution: Question 4. Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle. Solution: Question 5. If each diagonal of a quadrilateral’ divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram. Solution: Question 6. In the given figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFB = 3 cm², find the area of pa...