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Answer 1

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Step-by-step explanation:

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Answer 2

Answer:

2) Option (d) 18 cm²

3) Option (a) $\tt{\dfrac{base \times altitude}{2}}$

4) Option (d) 10000 m²

5) Option (c) $\tt{ \sqrt{ \dfrac{area}{\pi} } }$

6) No option is correct. Correct answer is (πr + 2r).

7) Option (b) 48 cm²

8) Option (d) 110

9) Option (c) 25 cm

10) Option (a) Base × height

Match the columns : (a)→(ii) , (b)→(iv) , (c)→(i) , (d)→(iii)

Step-by-step explanation:

2) Area of shaded region = (area of bigger triangle) - (area of smaller triangle)

w.k.t., area of right-angled triangle with base 'b' and height 'h' is $\tt{\dfrac{b \times h}{2}}$.

⇒ Area of shaded region = $\tt{\dfrac{8 \times 6}{2} - \dfrac{4 \times 3}{2} = 24 - 6 }$

⇒ Area of shaded region = 18 cm²

3) Area of right-angled triangle is $\tt{\dfrac{base \times altitude}{2}}$.

4) 1 hectare is equal to 10000 m².

5) w.k.t., Area of circle with radius 'r' is, area = πr²

⇒ Radius of circle, r = $\tt{ \sqrt{ \dfrac{area}{\pi} } }$

6) Perimeter of semicircle = $\tt{ \dfrac{Circumference}{2} + \text{Diameter of circle} }$

w.k.t., circumference of circle with radius 'r' is 2πr.

⇒ Perimeter of semicircle = $\tt{ \dfrac{2\pi r}{2} + 2r }$

⇒ Perimeter of semicircle = $\tt{ \pi r + 2r }$

7) Given: Length of rectangle = 8 cm and Length of diagonal = 10 cm

w.k.t., According to Pythagoras theorem, (Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (Diagonal)² = (Length)² + (Breadth)²

⇒ (10)² = (8)² + (b)²

⇒ (b)² = 100 - 64 = 36

⇒ b = 6 cm

w.k.t., area of rectangle with length 'l' and breadth 'b' is l×b.

⇒ Area of rectangle = 8×6

⇒ Area of rectangle = 48 cm²

8) Given: Radius of circle, r = 7 cm

w.k.t., area of circle with radius 'r' is πr² and circumference of circle with radius 'r' is 2πr.

⇒ Area of circle = π × 7² = 49 × π = (49 × 22) / 7

⇒ Area of circle = 154 cm²

Also, circumference of circle = 2π × 7 = 14 × π = (14 × 22) / 7

⇒ Circumference of circle = 44 cm

Now, numerical difference between the area and circumference of circle is 154 - 44 = 110.

9) w.k.t., perimeter of circle = 2πr

⇒ Perimeter of adjoining figure = (Perimeter of circle)/4 + 7 + 7

⇒ Perimeter = (2πr)/4 + 14 = 7π/2 + 14 = 11 + 14

⇒ Perimeter = 25 cm

10) Area of parallelogram is Base × height.

Match the columns :

(a) w.k.t., area of triangle is $\tt{\dfrac{base \times height}{2}}$.

⇒ Area of triangle = (3×8)/2 = 3×4

⇒ Area of triangle = 12 cm²

i.e., (ii) in Column B.

(b) w.k.t., area of parallelogram is Base × height.

⇒ Area of parallelogram = 4×6

⇒ Area of parallelogram = 24 cm²

i.e., (iv) in Column B.

(c) w.k.t., area of circle with radius 'r' is πr² .

The diameter is 14 cm.

⇒ Radius of circle = 14/2 = 7 cm.

⇒ Area of circle = π × 7² = 49π = (49×22) / 7

⇒ Area of circle = 154 cm²

i.e., (i) in Column B.

(d) w.k.t., area of triangle is $\tt{\dfrac{base \times height}{2}}$.

⇒ Area of triangle = $\tt{\dfrac{\dfrac{2}{\sqrt{3}} \times \sqrt{3}}{2} = \dfrac{2}{2}}$

⇒ Area of triangle = 1 cm²

i.e., (iii) in Column B.