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Nootan solutions for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई chapter 13 - Theorems on Area [Latest edition]

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Nootan solutions for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई chapter 13 - Theorems on Area - Shaalaa.com
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Solutions for Chapter 13: Theorems on Area

Below listed, you can find solutions for Chapter 13 of CISCE Nootan for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई.


Exercise 13AExercise 13B
Exercise 13A [Pages 256 - 260]

Nootan solutions for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई 13 Theorems on Area Exercise 13A [Pages 256 - 260]

Exercise 13A | Q 1. | Page 256

In the adjoining figure; if the area of parallelogram ABCD is 90 cm2, find :

  1. area of parallelogram ABEF.
  2. area of ΔABE.

Exercise 13A | Q 2. | Page 256

In the adjoining figure; AB = 16 cm, DM = 4 cm and BC = 8 cm. Find the length of DN.

Exercise 13A | Q 3. | Page 256

In the adjoining figure, BD = 2CD. If area of ΔABC = 36 cm2, find the area of ΔABD.

Exercise 13A | Q 4. | Page 256

In the adjoining figure, ABCD is a rectangle in which AB = 10 cm, AD = 6 cm. If AP || BQ, find :

  1. ar (|| gm ABQP) 
  2. ar (ΔАВР)

Exercise 13A | Q 5. | Page 257

The area of a parallelogram is 150 cm2. The ratio of its base and corresponding altitude is 3 : 2. Find the base and corresponding altitude.

Exercise 13A | Q 6. | Page 257

In the adjoining figure; AN = NB and DM = MC in the parallelogram ABCD. If area (ABCD) = 44 cm2,

  1. find the area of ΔВСЕ. 
  2. find the area of parallelogram BNMC.

Exercise 13A | Q 7. | Page 257

In the adjoining figure; ABCD is a parallelogram. If AP = BP and CP meets the diagonal BD at Q and area of ΔBPQ = 20 cm2, find:

  1. PQ : QC.
  2. Area of ΔPBC.
  3. Area of parallelogram ABCD.

Exercise 13A | Q 8. | Page 257

In the adjoining figure; ABCD is a parallelogram whose area is 324 cm2. If P is a point on AB such that AP : PB = 1 : 2, find:

  1. Area of ΔAPD. 
  2. QP : QD.

Exercise 13A | Q 9. | Page 257

In a parallelogram ABCD, P is a point on DC such that DP : PC = 2 : 3. If area of ΔDPB = 40 cm2, find the area of ◻ABCD.

Exercise 13A | Q 10. | Page 257

In the adjoining figure; BD = DC and AE = ED. Prove that area of ΔACE = `1/4` × area of ΔАВС.

Exercise 13A | Q 11. | Page 257

In the adjoining figure, D is the mid-point of BC and E be any point on AD.

Prove that : 

  1. area (ΔEBD) = area (ΔECD).
  2. area (ΔABE) = area (ΔACE).

Exercise 13A | Q 12. | Page 258

In a parallelogram ABCD, M and N are the points on AB and BC respectively.

Prove that:

  1. ΔCMD and ΔAND are equal in area. 
  2. area (ΔAND) = area (ΔAMD) + area (ΔCMB).
Exercise 13A | Q 13. | Page 258

In a trapezium ABCD, AB || DC and the diagonals AC and BD intersect at point ‘O’. Prove that area (ΔАOD) = area (ΔBОС).

Exercise 13A | Q 14. | Page 258

In the adjoining figure, O is the mid-point of diagonal AC of a quadrilateral ABCD. Prove that area (◻ABOD) = area (◻BODC).

Exercise 13A | Q 15. | Page 258

In the adjoining figure, AD = BD and ◻BDEC is a parallelogram. Prove that area (ΔABC) = area (◻BDEC).

Exercise 13A | Q 16. | Page 258

In the adjoining figure, PQ || BC.

Prove that :

  1. area (ΔABQ) = area (ΔАСР). 
  2. area (ΔΒΟP) = area (ΔCOQ).

Exercise 13A | Q 17. | Page 258

In the adjoining figure, ABCD is a parallelogram. A line through A intersects DC at P and meets BC produced at Q.

Prove that : area (ΔBCP) = area (ΔDPQ).

Exercise 13A | Q 18. | Page 258

In the adjoining figure, D is the mid-point of AB. P is any point on BC and CQ || PD meets AB at Q. 

Prove that : area (ΔBPQ) = `1/2` × area (ΔABC)

Exercise 13A | Q 19. | Page 259

In a trapezium ABCD, AB || DC. A line parallel to AC cuts AB at P and BC at Q. Prove that area (ΔАPD) = area (ΔACQ).

Exercise 13A | Q 20. | Page 259

In the adjoining figure, ΔABC is a right-angled triangle, right-angled at B; ABDE and ACGF are the squares. If BH is perpendicular to FG.

Prove that:

  1. area (ΔΕAC) = area (ΔΒΑF).
  2. area (◻ABDE) = area (◻AIHF).

Exercise 13A | Q 21. | Page 259

The diagonals AC and BD of a parallelogram ABCD intersect at ‘O’. A straight line through ‘O’ meets AB at P and DC at Q. 

Prove that: area (◻APQD) = `1/2` × area (◻ABCD).

Exercise 13A | Q 22. | Page 259

In the adjoining figure, ABCDE is a pentagon in which EG is drawn parallel to DA meets BA produced at G and CF is drawn parallel to DB meets AB produced at F. Prove that area of ABCDE = area of GDF.

Exercise 13A | Q 23. | Page 259

In the adjoining figure, BD = CD and P is a point on AC such that area (ΔAPD) : area (ΔАBD) = 2 : 3. 

Find:

  1. AP : PC. 
  2. area (ΔDCP) : area (ΔАВС).

Exercise 13A | Q 24. | Page 259

In the adjoining figure, ABCD and BEFG are two parallelograms.

Prove that : area (◻ABCD) = area (◻BEFG).

Exercise 13A | Q 25. | Page 259

ABCD 8 is a parallelogram. E and F are the mid-points of the sides AB and AD respectively. Prove that area (ΔAEF) = `1/8` × area (◻ABCD).

Exercise 13A | Q 26. | Page 260

In ΔAВC; E and F are the mid-points of sides AB and AC respectively. If FB and CE intersect at point ‘O’, prove that area (ΔOBC) = area (◻AEOF).

Exercise 13A | Q 27. | Page 260

In the adjoining figure, ABCD and ABFE are two parallelograms.

Prove that: area (ABCD) + area (ABFE) = area (EFCD).

Exercise 13A | Q 28. | Page 260

In the adjoining figure, PQ || BC, BE || CA and CF || BA. Prove that area of ΔABE = area of ΔACF.

Exercise 13B [Pages 260 - 261]

Nootan solutions for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई 13 Theorems on Area Exercise 13B [Pages 260 - 261]

Multiple Choice Questions Choose the correct answer from the given four options in each of the following questions :

Exercise 13B | Q 1. | Page 260

Two parallelograms are on equal bases and between the same parallel lines. The ratio of their areas is ______.

  • 1 : 1

  • 1 : 2

  • 2 : 1

  • 1 : 3

Exercise 13B | Q 2. | Page 260

In the adjoining figure, a parallelogram ABCD is shown. If DM ⊥ AB, DN ⊥ BC, then area of ΔABD is equal to:

  • `1/2 xx AB xx DN`

  • `1/2 xx BC xx DN`

  • `1/2 xx BC xx DM`

  • `1/2 xx AD xx DM`

Exercise 13B | Q 3. | Page 260

In the adjoining figure, a trapezium ABCD is shown in which AB || DC then, ar (ΔAOD) is equal to:

  • ar (ΔOCD)

  • ar (ΔAOB)

  • ar (ΔBCD)

  • ar (ΔBOC)

Exercise 13B | Q 4. | Page 260

A triangle and a rectangle are on the same base and between the same parallel lines. The ratio of their areas is ______.

  • 1 : 1

  • 1 : 2

  • 1 : 3

  • 1 : 4

Exercise 13B | Q 5. | Page 261

In the adjoining figure, ABCD is a parallelogram and M is the mid-point of AВ. If ar (|| gm ABCD) = 40 cm2, then ar (ΔADM) is equal to:

  • 5 cm2

  • 20 cm2

  • 15 cm2

  • 10 cm2

Exercise 13B | Q 6. | Page 261

The median of a triangle divides it into two ______.

  • triangles of equal area

  • congruent triangles

  • right triangles

  • isosceles triangles

  • equilateral triangle

Exercise 13B | Q 7. | Page 261

In ΔАВC; the mid-points of the sides BC, CA and AB are D, E and F respectively. The ratio of ar (ΔDEF) and ar (ΔABC) is ______.

  • 1 : 4

  • 1 : 3

  • 1 : 2

  • 1 : 1

Exercise 13B | Q 8. | Page 261

In the adjoining figure, a trapezium ABCD is shown in which AB || DC, AB = x and DC = y. If E and F are the mid-points of the sides AD and BC respectively, then ar (ABFE) : ar (EFCD) is:

  • x : y

  • (3x + y) : (x + 3y)

  • (x + 3y) : (3x + y)

  • (2x + y) : (3x + y)

Exercise 13B | Q 9. | Page 261

In the adjoining figure, a parallelogram ABCD is shown in which P divides AB in the ratio 1 : 2. If the area of the parallelogram ABCD is 60 cm2, then area (BPDC) is:

  • 20 cm2

  • 40 cm2

  • 50 cm2

  • 30 cm2

Exercise 13B | Q 10. | Page 261

In the adjoining figure, BM = 2CM. If ar (ΔАBC) = 30 cm2, then ar (ΔABM) is:

  • 20 cm2

  • 15 cm2

  • 10 cm2

  • 5 cm2

Solutions for 13: Theorems on Area

Exercise 13AExercise 13B
Nootan solutions for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई chapter 13 - Theorems on Area - Shaalaa.com

Nootan solutions for मैथमैटिक्स [अंग्रेजी] कक्षा ९ आईसीएसई chapter 13 - Theorems on Area

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