Area MCQ Quiz in मल्याळम - Objective Question with Answer for Area - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Area of the triangle:

Area of the triangle with vertices \(\left(x_1, y_1\right), \left(x_2, y_2\right)\mbox{ and }\left(x_3, y_3\right)\) is given by:

\(|A| = \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\\ \end{vmatrix}\)

Calculation:

It is given that the triangle has vertices as \((3, 0), (0,4)\mbox{ and }(3, 4)\).

Thus, area of the triangle is given by:

\(\begin{align*} A &= \dfrac{1}{2}\begin{vmatrix} 3 & 0 & 1\\ 0 & 4 & 1\\ 3 & 4 & 1 \end{vmatrix}\\ &= \dfrac{1}{2}\left(3(4 - 4) -0(0 -3) + 1(0 - 12)\right)\\ &= -6 \end{align*}\)

Thus the area of the triangle is 6 square units.

Concept:

The area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is \({1 \over 2} [x_1(y_2 - y_3) +x_2 (y_3 -y_1) + x_3 (y_1 -y_2)]\)

Calculation:

The third vertex lies on the line y = x + 3

Let the third vertex be (k, k + 3)

⇒ The area of a triangle with vertices (2, 1), (3, -2), (k, k + 3) is

\({1 \over 2} [2(-2 - (k+3)) +3 (k+3 -1) + k (1 -(-2))]\)

⇒ \({1 \over 2} [2(-5 - k) +3 (k+2 ) + k (3))] = 5\)

⇒ \([-10 - 2k +3 k+6 + 3k] = 5\times 2\)

⇒ \([-4 +4 k] = 10\)

⇒ k = 14/4 = 7/2

⇒ The third vertex = (k, k + 3) = \(\left(\frac{7}{2}, \frac{13}{2}\right)\)

∴ The correct answer is option (1).

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (1, 1), B (- 1, - 1) and C (- √3, √3) are the vertices of the triangle ABC

Let x1 = 1, y1 = 1, x2 = - 1, y2 = - 1, x3 = - √3 and y3 = √3

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the  vertices of a Δ ABC then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{1}}&{{1}}&1\\ {{-1}}&{{- 1}}&1\\ {{-√3}}&{{√3}}&1 \end{array}} \right|\)

⇒ (1/2) |1 (-1 - √3) - 1 (-1 + √3) + 1 (- √3 - √3) |

⇒ (1/2) | (-1 - √3 + 1 - √3 - √3 - √3) |

⇒ (1/2) |-4 √3 | 

⇒ A = 2√3

Hence, option C is the correct answer.

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (4,4), B(3,- 16) and C(3,-2) are the vertices of the triangle ABC

Let x1 = 4, y1 = 4, x2 = 3, y2 = - 16, x3 = 3 and y3 = - 2

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the  vertices of a Δ ABC then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{4}}&{{4}}&1\\ {{3}}&{{-16}}&1\\ {{3}}&{{-2}}&1 \end{array}} \right|\)

⇒ A = 7 sq units

Hence, option B is the correct answer.

Concept:

• Area of an equilateral triangle = \(√3 a^2 \over 4\)

           where, a is the side of the triangle.

• If equation of circle is x2 + y2 + 2gx + 2fy + c = 0, then radius of circle, R = \(√{g^2 + f^2 - c}\)
• For an isosceles triangle, two sides equal to l and one a,

           a = 2l sin \(θ \over 2\), where θ is th angle opposite to a.

Calculation:

Area of the equilateral triangle given =  \(27√{3}\) sq. units

Let the side of the triangle be a,

⇒ \(√3 a^2 \over 4\)= \(27√{3}\)

⇒ a² = 27 × 4

⇒ a = 6√3 

Now the side 'a' and two radius joining centre of circle and vertices of triangle forms an isosceles triangle.

⇒ a = 2r sin \(120 \over 2\)

⇒ \(6√3 = 2r {√3 \over2}\)

⇒ r = 6

Now for given equation of circle, x2 + y2 + 10x + 12y + c = 0 

g = 5, f = 6, r = 6

⇒ r = \(\sqrt{g^2 + f^2 - c}\)

⇒ \(6=\sqrt{5^2 + 6^2 - c}\)

⇒ c = 25

∴ The correct answer is option (2).

Concept:

Area of Triangle = 1/2 × base × height

Calculation:

Here, x = 0, y = 0 and x + y + 2 = 0

So when x = 0, y = -2 and  

when y = 0, x = -2

so we get points, (0, -2) and (-2, 0)

Now, area of triangle  = 1/2 × 2 × 2

= 2 square units.

Hence, option (3) is correct.

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

Note: If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

CALCULATION: 

Here, we have to find the value of a for which the points (a, 2a), (3, 1) and (- 2, 6) are collinear

Let A = (a, 2a), B = (3, 1) and C = (- 2, 6)

Let x1 = a, y1 = 2a, x2 = 3, y2 = 1, x3 = - 2 and y3 = 6

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the  vertices of a Δ ABC then area of Δ ABC = |A| where A = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{a}}&{{2a}}&1\\ {{3}}&{{1}}&1\\ {{-2}}&{{6}}&1 \end{array}} \right|\)

⇒ A = (20 - 15a)/2

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ A = (20 - 15a)/2 = 0

⇒ a = 4/3

Hence, option A is the correct answer.

Concept:

The perimeter of the triangle = sum of all sides

Area of triangle = (1/2)base × height

Calculation:

Since it is not mentioned, Δ ABC is which type of triangle. So, to make calculation easy

Let us assume, Δ ABC is a right-angle triangle where ∠A is 90°.

Let a = 5, b = 4, and c = 3 (Pythagorean triple)

Using the above concept, Perimeter

p = 3 + 4 + 5

⇒ p = 12  

Area of triangle

q = (1/2) 3 × 4

⇒ q = 6

Hence, the required value

\(\Rightarrow p(p-2a) \tan \frac{A}{2}\) = 12(12 - 2 × 5)tan (90°/2) [∵ ∠A = 90°]

⇒ p(p − 2a)tan(A/2)= 24        [∵ tan 45° = 1

From option 4

4q = 4 × 6 = 24

∴ 4q is the correct answer.

Alternate MethodConcept:

The perimeter of the triangle = sum of all sides = a + b + c

Area of scalene triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\), 

Half-angle formulae: \(\displaystyle tan(\frac{A}{2})=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)

Where 2s = a + b + c and a, b, c are the sides of the triangle.

Calculation:

If p be the perimeter of the triangle,

p = a + b + c                ------(1)

q = \(\sqrt{s(s-a)(s-b)(s-c)}\)        ------(2)

where a, b, c are the sides of the triangle.

Now, 

\(\displaystyle p(p-2a) \tan \left(\frac{A}{2}\right)\) = (a + b+ c) (a + b+ c - 2a) \(\displaystyle \tan \left(\frac{A}{2}\right) \)

⇒ (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)

⇒ (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{(s-b)(s-c)}{s(s-a)}×\frac{s(s-a)}{ s(s-a)}} \)

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC= |A| where A = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

Note: If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

CALCULATION: 

Here, we have to find the value of k for  which the points (7, - 2), (5, 1) and (3, k) are collinear

Let A = (7, - 2), B = (5, 1) and C = (3, k)

Let x1 = 7, y1 = - 2, x2 = 5, y2 = 1, x3 = 3 and y3 = k.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the  vertices of a Δ ABC then area of Δ ABC = |A| where A = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{7}}&{{- 2}}&1\\ {{5}}&{{1}}&1\\ {{3}}&{{k}}&1 \end{array}} \right|\)

⇒ |A| = 4 - k

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ |A| = 4 - k = 0

⇒ k = 4

Hence, option C is the correct answer.

Given:

The sides of a triangular field are 41 m, 40 m and 9 m.

Formula used:

Area of triangular field = √ s(s - a) (s - b) (s - c) 

S = a +b + c / 2 

Calculation:

S = 41 + 40 + 9 / 2 = 90 / 2 = 45m 

Area of triangular filed = √ 45(45 - 41) (45 - 9) (45 - 40)

⇒ 4 × 5 × 9 = 180m² = 180 × 10⁴ cm²

NOW 900 cm² area required for 1 rose bed 

⇒ 180 × 10⁴ cm² area required for 180 × 10⁴ / 900 rose 

∴  2000 rose bed