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The area of a triangle, whose two sides are 8 cm and 11cm and the perimeter is 32 cm, will be:
Let s be the semiperimeter and a, b and c are sides of a triangle
Using Heron's formula,
Area of triangle = √s(s – a)(s – b)(s – c)
a = 8 cm, b = 11 cm c = 32  (8 + 11) = 13 cm
s = (8 + 11 + 13)/2 = 16
Area of triangle = √16(16 – 8)(16 – 11)(16 – 13)
⇒ Area of triangle = √16 × 8 × 5 × 3
⇒ √2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 3
⇒ 2 × 2 × 2 × √30 = 8√30 cm^{2}
If the length of each side of a triangle is multiplied by 3, then the % increase in area will be:
Let the sides be a, b, c
∴ New sides = 3a, 3b, 3c
∴ s_{new} = 3s
New area =
= 9∆
∴ Increase in area = 9∆ – ∆ = 8∆
∴ % Increase in area =
The sides of a triangle are 11cm, 15cm and 16 cm. The altitude to the largest side is:
= 5 × 2 × 3√7
= √cm^{2}
1/2 × Altitude to the largest side × largest side = 30√7 cm^{2}
⇒ Altitude to the largest side
The length of median of an equilateral triangle is √3 cm. The area of triangle is:
Length of median of equilateral triangle
⇒ a = 2 cm
∴ Area of triangle
In the figure, PQ : QR = 3 : 2. If the area of ∆PRT = 40 cm^{2}, then area of ∆TQR is:
⇒ ar (∆PQT) = 24 cm^{2}
∴ area (∆TQR) = ar (∆PRT) – ar (∆PQT) = 40 – 24 = 16cm^{2}
The area of an equilateral triangle with side 2√3 cm is
Given, the side of an equilateral triangle = 2√3 cm
We have to find the area of the equilateral triangle.
Area of an equilateral triangle = √3/4 (side)²
= √3/4 (2√3)²
= √3/4 (4 × 3)
= 3√3
Consider √3 = 1.732
= 3 × 1.732
= 5.196 cm²
Therefore, the area of an equilateral triangle is 5.196 cm²
If a square and equilateral triangle have same perimeter and, square has area A_{1} and equilateral triangle has area A^{2}, then
Perimeter of square = Perimeter of equilateral ∆ = x
∴ length of side of square = x/4
Length of side of equilateral ∆ = x/3
Clearly, A_{1} > A_{2}
Two parallel sides of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. The area of the trapezium is
AE = x, FB = (17 – x) cm
From ∆AED, and ∆CFB
CF^{2} = DE^{2}
⇒ (26)^{2} – (17 – x)^{2} = (25)^{2} – (x)^{2}
⇒ (26)^{2} – (25)^{2} = (17 – x)^{2} – (x)^{2}
⇒ (26 – 25) (26 + 25) = (17 – x + x) (17 – x – x)
⇒ 51 = 17 (17 – 2x)
⇒ 17 – 2x = 3
⇒ 2x = 14 ⇒ x = 7cm
∴ area = 1/2 x (60 + 77) x 24 cm^{2}
= 137 × 12 cm^{2} = 1644 cm^{2}
The area of rhombus whose perimeter is 80m and one of the diagonal is 24m.
Side of rhombus = 80/4 = 20 cm
∴ Length of other diagonal
∴
= 16 × 24 cm^{2} = 384 cm^{2}
∴ For ∆DBC,
= 14 × 3 × 3
= 126 cm^{2}
Ar(∆ABD) = 1/2 x 9 x 40
= 180 cm^{2}
∴ Total area = (126 + 180) cm^{2}
= 306 cm^{2}
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