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