Question

Q) How much paper of each shade is needed to make a kite given in the figure ,in which ABCD is a square with diagonal 44 cm.

The base side of green triangle is 14cm

Answer 1

Solution :

First of all we'll find the area of square ABCD

★ Area of square ABCD = 2 × Area of ∆ ABC

→ Area of ABCD = 2(½ × 44 × 22)

→ Area of ABCD = 2 × 88

→ Area of ABCD = 968 cm²

From the figure, it's clear that square is divided into 4 equal parts.

• Paper of red shade needed = $\sf \: \dfrac{1}{4} \times 968$

→ Paper of red shade needed = 242 cm²

• Paper of yellow shade needed = (242 + 242) cm²

→ Paper of yellow shade needed = 484 cm²

Now, the green shade is compromised of ∆ AOD and that downwards triangle

Let's find the area of downwards triangle :

$\sf \: s = \dfrac{20 + 20 + 14}{2} = 27 \: cm$

Using Heron's formula :

$\sf \triangle = \sqrt{s(s - a)(s - b)(s - c)}$

$\sf \longrightarrow \: \sqrt{27(27 - 20)(27 - 20)(27 - 14)}$

$\sf \longrightarrow \: \sqrt{27 \times 7 \times 7 \times 13}$

$\sf \longrightarrow 21 \sqrt{39 } \: {cm}^{2}$

$\sf \longrightarrow 21 \times 6.245 \: {cm}^{2}$

$\sf \longrightarrow 131.15 \: {cm}^{2}$

• Paper of green shade needed = 131.15 + 242

→ Paper of green shade needed = 373.15 cm²

Hence,

• Paper of red shade needed is 242 cm².

• Paper of yellow shade needed is 484 cm².

• Paper of green shade needed is 373.15 cm².

Answer 2

Step-by-step explanation:

Solution :

First of all we'll find the area of square ABCD

★ Area of square ABCD = 2 × Area of ∆ ABC

→ Area of ABCD = 2(½ × 44 × 22)

→ Area of ABCD = 2 × 88

→ Area of ABCD = 968 cm²

From the figure, it's clear that square is divided into 4 equal parts.

• Paper of red shade needed = \sf \: \dfrac{1}{4} \times 968

4

1

×968

→ Paper of red shade needed = 242 cm²

• Paper of yellow shade needed = (242 + 242) cm²

→ Paper of yellow shade needed = 484 cm²

Now, the green shade is compromised of ∆ AOD and that downwards triangle

Let's find the area of downwards triangle :

\sf \: s = \dfrac{20 + 20 + 14}{2} = 27 \: cms=

2

20+20+14

=27cm

Using Heron's formula :

\sf \triangle = \sqrt{s(s - a)(s - b)(s - c)}△=

s(s−a)(s−b)(s−c)

\sf \longrightarrow \: \sqrt{27(27 - 20)(27 - 20)(27 - 14)}⟶

27(27−20)(27−20)(27−14)

\sf \longrightarrow \: \sqrt{27 \times 7 \times 7 \times 13}⟶

27×7×7×13

\sf \longrightarrow 21 \sqrt{39 } \: {cm}^{2}⟶21

39

cm

2

\sf \longrightarrow 21 \times 6.245 \: {cm}^{2}⟶21×6.245cm

2

\sf \longrightarrow 131.15 \: {cm}^{2}⟶131.15cm

2

• Paper of green shade needed = 131.15 + 242

→ Paper of green shade needed = 373.15 cm²

Hence,

• Paper of red shade needed is 242 cm².

• Paper of yellow shade needed is 484 cm².

• Paper of green shade needed is 373.15 cm².