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The area of an isosceles triangles whose equal sides are 12cm and the other side is 6cm long, will be:
In ∆ABD,
AB^{2} + BD^{2} = AD^{2}
⇒ AD^{2} = AB^{2} – BD^{2}
⇒ AD = √135 = 3√15 cm
∴ area (∆ABC) = 1/2 x 6 x 3√15
= 9√15 cm^{2}
The sides of a triangle are 50 cm, 78 cm and 112 cm. the smallest altitude is:
Here S = = 120 cm
∴ Area
= 2 × 3 × 2 × 2 × 2 × 5 × 7
= 240 × 7 cm^{2} = 1680 cm^{2}
Area = 1/2 x base altitude = 1680 cm^{2}
= 30 cm
The base and hypotenuse of a right triangle are 5 cm, 13 cm long. The length of altitude from the vertex containing right angle to the hypotenuse will be:
BC =
⇒ AC × BD = AB × AC
⇒ 13 × BD = 5 × 12
⇒
In ∆DOC,
OD^{2} + OC^{2} = DC^{2}
⇒ OD =
∴ DB = 2 × OD = 2 × 60 = 120 cm.
Area of rhombus =
= 9600 cm^{2}
If a square and rhombus have same perimeter, and area of square is S and area of rhombus is R, then
Area of square = a^{2}
Area of rhombus = a^{2} sin θ
∵ sin θ < 1
∴ a^{2} > a^{2} sin θ
ar(square) > ar(rhombus)
⇒ S > R
The area of kite in the adjoining figure is: (AC = BD = 32 cm)
Area of square ABCD = 1/2 × AC × BD
= 1/2 x 32 x 32
= 16 × 32 cm^{2}
= 512 cm^{2}
Here
Area of parallelogram, in the adjoining figure will be:
Area of ∆ACD = area of ∆ACB
⇒ area of ∆ACD =
= 4 × 6 × 7 × 2
= 4 × 84 = 336 cm^{2}
∴ area of ∥gm = 2 × ar(∆ACD)
= 2 × 336 cm^{2}
= 672 cm^{2}
In ∆AOB,
AB =
= 20 cm.
∴ ar(∆AOB) = 1/2 × 12 × 16 = 96 cm^{2}
For ∆ABC,
∴ ar (∆ABC) =
= 480 cm^{2}
∴ Area of shaded region
= 480 – 96
= 384 cm^{2}
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2 cm, the area of the triangle is:
Length of equilateral triangle = x/3 cm
Length of side of square = x/4 cm
Length of side of square
= 12 cm
The third side of triangle whose two sides are 26 and 28 cm and area is 336 cm^{2}, is
Let the length of third side be x
∴
Area
⇒ x = 30 cm
∴ Third side = 30 cm
The length of sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm, then, the height corresponding to the length side is:
Lengths of sides of triangle
∴ Area
∴ Height corresponding to the longest side
ABCD is a parallelogram, where,
AL = 8 cm, CM = 10 cm, AD = 6 cm. Find AB.
Area of ∥gm = AL × DC = CM × AD
⇒ 8 × DC = 10 × 6
⇒ DC = 60/8 = 7.5 cm = AB.
Here S =
∴ Required area
If each side of ∆ is doubled, then the area will become how many times?
If the sides of ∆ are a, b and c
∴ New sides are 2a, 2b, 2c
∴ New area =
The diagonal of a parallelogram divide it in 2 parts, the area of the two parts:
The two parts of the parallelogram are congruent because they have equal area.
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