Math, asked by ishikabahri7, 9 months ago

Please anwer the 10th and 9th question from the given picture.

I’ll mark you as the brainliest.

Attachments:

Answers

Answered by shyamkhenwar5035
19

9.a)ar(ABCD)=2(ar(BCD))

=2(1/2*b*h)

=2(1/2*5*7)

=35cm^2

b)ar(trapezium)=1/2(a+b)h

=1/2(10+14)*7

=12*7

=84cm^2

10.Let a=15cm,b=8cm,c=17cm

s=15+8+17/2

=40/2

=20cm

By Heron's Formula,

√s(s-a)(s-b)(s-c)

=√20(20-15)(20-8)(20-17)

=√20*5*12*3

=√4*5*5*3*2*2*3

=2*5*3*2

=60cm^2

now to find altitude,

Take b=17cm

Then,

ar(triangle)=1/2*b*h

60=1/2*17*h

60=17/2*h

h=60*2/17

h=120/17

h=7.05cm approx

Please mark me as Brainliest

Answered by EliteSoul
71

i) Given

Side of rhombus = 5 cm

Diagonal of rhombus = 7 cm

To find

Area of rhombus

Solution

As the diagonal of rhombus divides the rhombus into two triangles of base 5 cm & height 7 cm

➛ Area of ∆ = ½ × b × h

➛ Area of rhombus = 2[½ (5 × 7)]

➛ Area of rhombus = 2[½ × 35]

Area of rhombus = 35 cm² (Ans.)

__________________________

ii) Given

Parallel sides of trapezium = 10 cm & 14 cm

Height of trapezium = 7 cm

To find

Area of trapezium.

Solution

➛ Area of trapezium = ½ (a + b) × h

➛ Area of trapezium = ½(10 + 14) × 7

➛ Area of trapezium = ½ × 24 × 7

➛ Area of trapezium = 12 × 7

Area of trapezium = 84 cm² (Ans.)

___________________________

10) Given

Sides of ∆ are 15 cm,8 cm & 17 cm.

To find

Area of triangle.

Altitude of on the side of 17 cm.

Solution

First finding semi-perimeter of :

⇒ Semi-perimeter(s) = (15 + 8 + 17)/2

= 40/2

= 20 cm

Using Heron's formula :

➛ Area of ∆ = √[s(s - a)(s - b)(s - c)]

➛Area of ∆= √[20(20 - 15)(20 - 8)(20 - 17)]

➛ Area of ∆ = √[20 × 5 × 12 × 3]

➛ Area of ∆ = √[3600]

Area of = 60 cm² (Ans.)

Again, base of will be 17 cm.

We have to find out altitude of

➛ Area of ∆ = ½ × b × h

Putting values :

➛ 60 = ½ × 17 × h

➛ 60 × 2 = 17h

➛ 120 = 17h

➛ h = 120/17

Altitude = 7.06 cm (Ans.)


Anonymous: Great ❤️❤️❤️
Similar questions