Here a3 – 8 can be written as a3 – 23
Comparing this with a3−b3, we get a=a,b=2
\(\therefore\) a³-b³ = (a-b)(a²+ab+b²)
a³-2³=(a-2)(a²+a(2)+22)
a³-8=(a-2)(a²+2a+4)

This is in the form of ax² + bx + c
We get a = 7,b = 2,c = −5
Now, the product = a x c = 7 x (−5) = –35 and sum b = 2
= 7c² + 2c − 5
= (7c² − 5c) + (7c − 5) (the middle term 2c can be written as – 5c + 7c)
= c(7c − 5) + 1(7c−5) (taking out the common factor 7c – 5 )
= (7c − 5)(c + 1)
Therefore, (7c–5), (c+1) are the two factors.

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

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)