Comparing x³+ 5³ with a³+b³ we get a = x, b = 5
We know,
a³+ b³ = (a+b) (a²−ab+b²)
x³+ 5³ = (x+5) (x²−(x)(5)+5²)
x³+ 5³ = (x+5) (x²−5x+25)

Now, x² + 8x + 16
This can be written as x² + 8x + 42
Comparing this with a² + 2ab + b² = (a + b)² we get a = x; b = 4
(x²) + 2(x)(4) + (4)² = (x + 4)²
x² + 8x + 16 = (x + 4)²

Given:
MA = 4 cm, AT = 3.6 cm,
TH = 4.5 cm, MH = 5 cm and ∠A = 85°


Steps:
1. Draw a line segment MA = 4 cm.
2. Make ∠A = 85°.
3. With A as centre, draw an arc of radius 3.6 cm. Let it cut the ray AX at T.
4. With M and T as centres, draw arcs of radii 5 cm and 4.5 cm respectively and let them cut at H.
5. Join MH and TH.
6. MATH is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MATH = \(\frac12\) × d × (h₁+ h₂) sq.units
= \(\frac12\) x 5.1 x (3.9 + 2.8)
= 2.55 x 6.7 = 17.09 cm²

Given: DE = 6 cm, EA = 5 cm, AR = 5.5 cm,
RD = 5.2 cm and a diagonal DA = 10 cm


Steps:
1. Draw a line segment DE = 6 cm.
2. With D and E as centres, draw arcs of radii 10 cm and 5 cm respectively and let them cut at A.
3. Join DA and EA.
4. With D and A as centres, draw arcs of radii 5.2 cm and 5.5 cm respectively and let them cut at R.
5. Join DR and AR.
6. DEAR is the required quadrilateral
Calculation of Area:
Area of the quadrilateral DEAR = \(\frac12\) x d x (h₁+ h₂) sq. units
= \(\frac12\) x 10 x (1.9+ 2.3)
= 5 x 4.2 = 21 cm²

We have, (5y³ − 25y² + 8y) ÷ 5y
= \(\cfrac { { 5y }^{ 3 }-25{ y }^{ 2 }+8y }{ 5y } \)

 = \({ y }^{ 3-1 }-{ 5y }^{ 2-1 }+\cfrac { 8 }{ 5 } \)
= \({ y }^{ 2 }-5y+\cfrac { 8 }{ 5 } \)

The given field has four triangles (I, III, IV & V) and a trapezium (II).
Area of the triangle (I) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 5\times 6=15{ m }^{ 2 }\)
Area of the trapezium (II) \(=\frac { 1 }{ 2 } h(a+b)=\frac { 1 }{ 2 } \times 13\times (6+4)=65{ m }^{ 2 }\)
Area of the triangle (III) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 8\times 4=\frac { 32 }{ 2 } =16{ m }^{ 2 }\)
Area of the triangle (IV) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
Area of the triangle (V) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
∴ The total area of the field = 15 + 65 + 16 + 65 + 65 = 226 m²

This is one of the solutions. Try for more

This is one of the solutions. Try for more.

Side of the square = 5 cm
Diameter of the semi-circle = 5 cm
∴ Radius = 2.5 cm
Side of the equilateral triangle = 5 cm
∴  Area of the keychain = area of the semi circle + area of the square + area of the equilateral triangle
\(=\frac { 1 }{ 2 } { \pi r }^{ 2 }+{ a }^{ 2 }+\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
\(=\left( \frac { 1 }{ 2 } \times 3.14\times 2.5\times 2.5 \right) +\left( 5\times 5 \right) +\left( \frac { \sqrt { 3 } }{ 4 } \times 5\times 6 \right) \)
= 9.81 + 25 + 10.83
= 45.64cm² (approx.)

Radius of a circular quadrant, r = 3.5 cm and side of a square, a = 3.5 cm.
The given figure is formed by the joining of 4 quadrants of a circle with each side of a square. The boundary of the given figure consists of 4 arcs and 4 radii.
(i) Perimeter of the given combined shape
= 4 x length of the arcs of the quadrant of a circle + 4 x radius
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\pi r \right) +4r\)
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\times 3.5 \right) +(1\times 3.5)\)
= 22 + 14 = 36 cm (approximately)
(ii) Area of the given combined shape
= area of the square + 4 x area of the quadrants of the circle
\({ a }^{ 2 }=\left( 4\times \frac { 1 }{ 4 } \times \pi { r }^{ 2 } \right) \)
\(=(3.5\times 3.5)+\left( \frac { 22 }{ 7 } \times 3.5\times 3.5 \right) \)
A = 12.25 + 38.5 = 50.75 cm² (approximately)