## RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a.

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
- RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
- RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
- RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

**Question **1.

**Solution:
**

**Question **2.

**Solution:
**Let a = 42 cm, b = 34 cm and c = 20 cm

(ii)Let base = 42 cm and corresponding height = h cm

Hence, the height corresponding to the longest side = 16 cm

**Question **3.

**Solution:
**Let a = 18 cm, b = 24 cm, c = 30 cm

Then,2s = (18 + 24 + 30) cm = 72 cm

s = 36 cm

(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(ii)Let base = 18 cm and altitude = x cm

Hence, altitude corresponding to the smallest side = 24 cm

**Question **4.

**Solution:
**On dividing 150 m in the ratio 5 : 12 : 13, we get

**Question **5.

**Solution:
**On dividing 540 m in ratio 25 : 17 : 12, we get

**Question **6.

**Solution:
**Let the length of one side be x cm

Then the length of other side = {40 (17 + x)} cm = (23 – x) cm

Hypotenuse = 17 cm

Applying Pythagoras theorem, we get

**Question **7.

**Solution:
**Let the sides containing the right – angle be x cm and (x – 7) cm

∴ perimeter of triangle (15 + 8 + 17) cm = 40 cm

**Question **8.

**Solution:
**Let the sides containing the right angle be x and (x 2) cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

**Question **9.

**Solution:
**Side of an equilateral triangle = a = 10 cm

**Question **10.

**Solution:
**Let each side of the equilateral triangle be a cm

**Question **11.

**Solution:
**Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm

**Question **12.

**Solution:
**Let each side of the equilateral triangle be a cm

**Question **13.

**Solution:
**Base of right angled triangle = 48 cm

**Question **14.

**Solution:
**Let the hypotenuse of right – angle triangle = 6.5 m

Base = 6 cm

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm

^{2}

**Question **15.

**Solution:
**The circumcentre of a right – triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)

= (2 × 8) cm = 16 cm

Base = 16 cm, height = 6 cm

Area of right angled triangle=

Hence, area of the triangle= 48 cm

^{2}

**Question **16.

**Solution:
**Let each equal side be a cm in length.

Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

**Question **17.

**Solution:
**Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm

= (2 41 + 80) cm

= (82 + 80) cm = 162 cm

Hence, perimeter of the triangle = 162 cm

**Question **18.

**Solution:
**Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Hence, area of the triangle = 48 cm

^{2}.

**Question **19.

**Solution:
**Let ∆ABC is a isosceles triangle. Let AC, BC be the equal sides

Then AC = BC = 10cm. Let AB be the base of ∆ABC right angle at C.

Hence, area = 50 cm

^{2}and perimeter = 34.14 cm

**Question **20.

**Solution:
**

Area of shaded region = Area of ABC – Area of DBC

First we find area of ABC

Area of shaded region = Area of ∆ABC – Area of ∆DBC

= (43.30 – 24) cm

^{2}= 19. 30 cm

^{2 }Area of shaded region = 19.3 cm

^{2}

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a are helpful to complete your math homework.

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