100 points catch it plz find the solution
Answers
Step-by-step explanation:
31) According to the laws, the length of the inner rectangle = 40 − 3 − 3 = 34 m and the breath of the inner rectangle = 15 − 2 − 2 = 11 m.
∴ Area of the inner rectangle PQRS = Length × Breath
= 34 × 11
= 374 m2
Hence, the largest area where house can be constructed is 374 m2.
32) = 374 m2
Hence, the largest area where house can be constructed is 374 m2.
Let the sides of rhombus be of length x cm.
Perimeter of rhombus = 4x
⇒ 40 = 4x
⇒ x = 10 cm
Now,
In ∆ABC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
s=
10+10+12
2
=
32
2
=16 cm
∴ By Heron's formula,
Area of ∆ABC=
√s(s-a)(s-b)(s-c)
=
√16(16-10)(16-10)(16-12)
=
√16(6)(6)(4)
=48 cm2 ...(1)
In a†ADC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
s=
10+10+12
2
=
32
2
=16 cm
∴ By Heron's formula,
Area of ∆ADC=
√s(s-a)(s-b)(s-c)
=
√16(16-10)(16-10)(16-12)
=
√16(6)(6)(4)
=48 cm2 ...(2)
∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC
= 48 + 48
= 96 cm2
The cost to paint per cm2 = Rs 5
The cost to paint 96 cm2 = Rs 5 × 96
= Rs 480
The cost to paint both sides of the sheet = Rs 2 × 480
= Rs 960
Hence, the total cost of painting is Rs 960.
Answer 33:
Let the semi-perimeter of the triangle be s.
Let the sides of the triangle be a, b and c.
Given: s − a = 8, s − b = 7 and s − c = 5 ....(1)
Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20
⇒ 3s − 2s = 20 (∵ s=
a+b+c)
2
⇒ s = 20 cm ...(2)
∴ By Heron's formula,
Area of ∆=
√s(s-a)(s-b)(s-c)
=
√20(8)(7)(5)
(from (1) and (2)) =20
√14
cm2
Hence, the area of the triangle is 20
√14
cm2.
Answer:
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31.
∴ Length of inner-rectangle, EF = 40 – 3 – 3 = 34 m and breadth of inner-rectangle, FG =15 – 2 – 2 = 11 m
∴ Another rectangle EFGH will be formed inside the rectangle ABCD
∴ Area of inner rectangle, EFGH = Length x Breadth
= EF x FG = 34 x 11 = 374 m2
[∴ area of a rectangle = length x breadth]
Hence, the largest area where the house can be constructed in 374 m2.
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32.
let ABCD be a rhombus,
perimeter = 40
so, AB=BC=CD=DA=10
let diagnols ACBD intersect at O
so let BD =12
or BO=6so in trangle AOB
AB2= AO2+OB2
AO2=102-62
AO2=64
AO=8
AC=16
so area of rhombus is 1/2*d1*d2
ar ABCD = 1/2*12*16
=96cm2
total area of both sides = 192 cm2
cost of painting 1 m2 = Rs 5
cost of painting 10000 cm2 = Rs 5 (1m sq = 10,000 cm sq)
cost of painting 1 cm2 = Rs 5/10000
cost of painting 192 cm2 = Rs 5/10000 * 192
cost of painting 192 cm2 = Re 0.096
it will cost 9.6 paise for painting both the sides of rhombus
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33.
see in the picture
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34.
see in the picture
