## RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

**Question 1.**

**Solution:**

(i) Base of the triangle = 42 cm

Height = 25 cm

Area = \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x 42 x 25 = 525 cm²

(ii) Base of the triangle = 16.8 m

and height = 75 cm = 0.75 m

Area = \(\frac { 1 }{ 2 }\) x Base x height

= \(\frac { 1 }{ 2 }\) x 16.8 x 0.75 m2 = 6.3 m²

(iii) Base of a triangle (b) = 8 m = 80 cm

and height (h) = 35 cm

Area = \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) x 80 x 35 = 1400 cm²

**Question 2.**

**Solution:**

Base of triangle = 16 cm

area of the triangle = 72 cm²

**Question 3.**

**Solution:**

Area of triangular region = 224 m²

Base = 28 m

**Question 4.**

**Solution:**

Area of triangle = 90 cm²

and height (h) = 12 cm

**Question 5.**

**Solution:**

Let height of a triangular field = x m

Then base (b) = 3x m

and area = \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) x 3x x x

**Question 6.**

**Solution:**

Area of the right angled triangle = 129.5 cm²

**Question 7.**

**Solution:**

In right angled ∆ABC,

Base BC = 1.2 m

and hypotenuse AC = 3.7 m

But AC² = AB² + BC² (Pythagoras Theorem)

⇒ (3.7)² = AB² + (1.2)²

⇒ 13.69 = AB² + 1.44

⇒ AB² = 13.69 – 1.44

⇒ AB² = 12.25 = (3.5)²

⇒ AB = 3.5 m

Now, area of ∆ABC = \(\frac { 1 }{ 2 }\) x base x altitude

= \(\frac { 1 }{ 2 }\) x 1.2 x 3.5 m² = 2.1 m²

**Question 8.**

**Solution:**

Legs of a right angled triangle = 3 : 4

Let one leg (base) = 3x

Then second leg (altitude) = 4x

Area = \(\frac { 1 }{ 2 }\) x base x altitude

= \(\frac { 1 }{ 2 }\) x 3x x 4x = 6x²

6x² = 1014

⇒ x² = \(\frac { 1014 }{ 6 }\) = 169 = (13)²

x = 13

one leg'(Base) = 3x = 3 x 13 = 39 cm

and second leg (altitude) = 4x = 4 x 13 = 52 cm

**Question 9.**

**Solution:**

One side BC of a right triangular scarf = 80 cm

and longest side AC = 1 m = 100 cm

By Pythagoras Theorem,

AC² = AB² + BC²

⇒ (100)² = AB² + (80)²

⇒ 10000 = AB² + 6400

⇒ AB² = 10000 – 6400

⇒ AB² = 3600 = (60)²

⇒ AB = 60

Second side = 60 cm

Area of the scarf = \(\frac { 1 }{ 2 }\) x b x h

= \(\frac { 1 }{ 2 }\) x 80 x 60 cm2 = 2400 cm²

Rate of cost = Rs. 250 per m²

Total cost =\(\frac { 2400 }{ 100 x 100 }\) x 250 = Rs. 60

**Question 10.**

**Solution:**

(i) Side of the equilateral triangle (a) = 18 cm

**Question 11.**

**Solution:**

Area of equilateral triangle = 16√3 cm²

Let each side = a

then \(\frac { \surd 3 }{ 4 }\) a² = 16√3

⇒ a² = \(\frac { 16\surd 3\times 4 }{ \surd 3 }\)

⇒ a² = 64 = (8)²

a = 8 cm

Each side = 8 cm

**Question 12.**

**Solution:**

Each side of an equilateral triangle = 24cm

Length of altitude = \(\frac { \surd 3 }{ 2 }\) a = \(\frac { \surd 3 }{ 2 }\) x 24

= 12√3 cm = 12 (1.73) = 20.76 cm

**Question 13.**

**Solution:**

(i) a = 13 m, b = 14 m, c = 15 m

= 2 x 2 x 3 x 7 x 7 x 7 = 4116 m²

**Question 14.**

**Solution:**

Let a = 33 cm, b = 44 cm, c = 55 cm

**Question 15.**

**Solution:**

Perimeter of the triangle = 84 cm

Ratio in side = 13 : 14 : 15

Sum of ratios =13 + 14 + 15 = 42

Let then first side = \(\frac { 84 x 13 }{ 42 }\) = 26 cm

**Question 16.**

**Solution:**

Let a = 42 cm, b = 34 cm, c = 20 cm

**Question 17.**

**Solution:**

In isosceles ∆ABC

Base BC = 48 cm.

and AB = AC = 30cm.

Let AD ⊥ BC

**Question 18.**

**Solution:**

Perimeter of an isosceles triangle = 32 cm

Base = 12 cm

Sum of two equal sides = 32 – 12 = 20 cm

Length of each equal side = \(\frac { 20 }{ 2 }\) = 10cm

Let AD ⊥ BC

BD = DC = \(\frac { 12 }{ 2 }\) = 6 cm

**Question 19.**

**Solution:**

In quadrilateral ABCD,

diagonal AC = 26 cm.

and perpendiculars DL = 12.8cm, BM = 11.2 cm

Area of quadrilateral ABCD

= \(\frac { 1 }{ 2 }\) (Sum of perpendicular) x diagonal

= \(\frac { 1 }{ 2 }\) (12.8 + 11.2) x 26 cm²

= \(\frac { 1 }{ 2 }\) x 24 x 26 = 312 cm²

**Question 20.**

**Solution:**

In quad. ABCD,

AB = 28 cm, BC = 26 cm, CD = 50 cm, DA = 40 cm

and diagonal AC = 30 cm

In ∆ABC,

**Question 21.**

**Solution:**

ABCD is a rectangle in which AB = 36 m

and BC = 24m

In ∆AED,

EF = 15 m

AD = BC = 24 m.

Now area of rectangle ABCD = l x b = 36 x 24 cm² = 864 cm²

Area of ∆AED = \(\frac { 1 }{ 2 }\) x AD x EF

= \(\frac { 1 }{ 2 }\) x 24 x 15 cm² = 180 cm²

Area of shaded portion = 864 – 180 = 684 m²

**Question 22.**

**Solution:**

In the fig. ABCD is a rectangle in which AB = 40 cm, BC = 25 cm.

P, Q, R and S and the mid points of sides, PQ, QR, RS and SP respectively

Then PQRS is a rhombus.

Now, join PR and QS.

PR = BC = 25cm and QS = AB = 40cm

Area of PQRS = \(\frac { 1 }{ 2 }\) x PR x QS

= \(\frac { 1 }{ 2 }\) x 25 x 40 = 500 cm²

**Question 23.**

**Solution:**

(i) Length of rectangle (l) = 18 cm

and breadth (b) = 10 cm

Area = l x b = 18 x 10 = 180 cm²

Area of right ∆EBC = \(\frac { 1 }{ 2 }\) x 10 x 8 = 40 cm²

and area of right ∆EDF = \(\frac { 1 }{ 2 }\) x 10 x 6 = 30 cm²

Area of shaded region = 180 – (40 + 30) = 180 – 70 = 110 cm²

(ii) Side of square = 20 cm

Area of square = a² = (20)² = 400 cm²

Area of right ∆LPM = \(\frac { 1 }{ 2 }\) x 10 x 10 cm² = 50 cm²

Area of right ∆RMQ = \(\frac { 1 }{ 2 }\) x 10 x 20 = 100 cm²

and area of right ∆RSL = \(\frac { 1 }{ 2 }\) x 20 x 10 = 100 cm²

Area of shaded region = 400 – (50 + 100 + 100) cm2 = 400 – 250 = 150 cm²

**Question 24.**

**Solution:**

In the quadrilateral ABCD

BD = 24 cm

AL ⊥ BD and CM ⊥ BD

AL = 5 cm and CM = 8 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.