Question

Exercise 17.1
Work out the areas of each triangle.

4.1 cm
2.1​

Answer 1

$\sf{Answer}$

Given triangle is a Right angle triangle

As we know that ,

Area of Right angle triangle is 1/2 b h

Where

• b is the base
• h is the height of triangle

Plugging values

base = QR = 4.1 cm

Height = PQ = 2.6 cm

Area = $\sf\dfrac{1}{2}$bh

Area = $\sf\dfrac{1}{2}$× 4.1 × 2.6

Area = $\sf\dfrac{1}{2}$ × 2.6 × 4.1

Area =1.3 cm × 4.1 cm

Area = 5.33 cm²

So, Area of triangle is 5.33 cm²

Know more :-

Area of square is s²

Where" s" is side of square

________________

Area of rectangle = lb

Where l = length

b = breadth

________________

Area of rhombus is 1/2 d1 d2

d1 , d2 are diagonals of rhombus

______________________

Area of circle is πr²

___________________

Perimeter of square is 4s

Where" s" is side of square

_____________

Perimeter of rectangle is 2 (l + b)

Where l = length

b = breadth

________________

Perimeter of rhombus is 4a

Where "a "is side of rhombus

_____________

Perimeter of circle is 2πr

___________________

Answer 2

Given

• In ΔPQR ,
• ∠Q = 90°
• PQ = 2.6 cm
• QR = 4.1 cm

To find

• Area of the triangle

Solution

• QR = 4.1 (base)
• PQ = 2.6 (height)

We know, area of triangle :-

• 1/2 × base × height

Substituting the values, we get :-

• 1/2 × 4.1 × 2.6
• 1/2 × 10.66
• 10.66/2
• 5.33

Hence, the area of ΔPQR is 5.33 cm²

Learn more

• Area of square = side × side
• Area of rectangle = Length × Breadth
• Area of trapezium = 1/2 h(sum of parallel sides)
• Area of parallelogram = base × height
• Area of circle = πr²