Question

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Answer 1

Answer:

:By dividing the quadrilateral into two triangles and applying Heron’s formula, we can calculate the area of triangles.

Heron's formula for the area of a triangle is: Area = √s(s - a)(s - b)(s - c)

Where a, b, and c are the sides of the triangle, and s = Semi-perimeter = Half the Perimeter of the triangle

Now, ABCD is the park shown in the figure below

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A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

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Solution:

By dividing the quadrilateral into two triangles and applying Heron’s formula, we can calculate the area of triangles.

Heron's formula for the area of a triangle is: Area = √s(s - a)(s - b)(s - c)

Where a, b, and c are the sides of the triangle, and s = Semi-perimeter = Half the Perimeter of the triangle

Now, ABCD is the park shown in the figure below

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

We have ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.

Let's connect B and D, such that BCD is a right-angled triangle.

In ∆BDC, apply Pythagoras theorem in order to find the length of BD

BD2 = BC2 + CD2 [Pythagoras theorem]

BD2 = 122 + 52

BD2 = 144 + 25

BD = √169

BD = 13 m

Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD

Now, Area of ∆BCD = 1/2 × base × height

= 1/2 × 12 m × 5 m

= 30 m²

Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m

Semi Perimeter of ΔABD

s = (a + b + c)/2

= (9 + 8 + 13)/2

= 30/2

= 15 m

By using Heron’s formula,

Area of ΔABD = √s(s - a)(s - b)(s - c)

= √15(15 - 9)(15 - 8)(15 - 13)

= √15 × 6 × 7 × 2

= 6√35

= 35.5 m² (approx.)

Area of ΔABD = 35.5 m²

Therefore,

Area of park ABCD = 30 m² + 35.5 m² = 65.5 m²

Thus, the park ABCD occupies an area of 65.2 m².

Answer 2

$\huge\sf\fbox\purple{ ★ Solution ★}$

°°° Explanation °°°

To find zero of the polynomial, p(x)=0

(i) If p(x)=x+5=0 then x=−5, i.e. −5 is the zero.

(ii) If p(x)=x−5=0 then x=5, i.e. 5 is the zero.

(iii) If p(x)=2x+5=0 then x= 2

−5 , i.e. 2−5 is the zero.

(iv) If p(x)=3x−2=0 then x= 32

, i.e. 32 is the zero.

(v) If p(x)=3x=0 then x=0, i.e. 0 is the zero.

(vi) If p(x)=ax=0 then x=0, i.e. 0 is the zero.

(vii) If p(x)=cx+d=0 then x= c−d

, i.e. c −d

is the zero.