SATHEE: Properties of Triangles 2 Question 17

Properties of Triangles 2 Question 17

17. Let $A B C$ and $A B C^{'}$ be two non-congruent triangles with sides $A B = 4, A C = A C^{'} = 2 \sqrt{2}$ and angle $B = 30^{\circ}$ . The absolute value of the difference between the areas of these triangles is

(2009)

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Solution:

In $△ A B C$ , by sine rule, $\frac{a}{\sin A} = \frac{2 \sqrt{2}}{\sin 30^{\circ}} = \frac{4}{\sin C}$

$\Rightarrow C = 45^{\circ}, C^{'} = 135^{\circ}$

When, $C = 45^{\circ} \Rightarrow A = 180^{\circ} - (45^{\circ} + 30^{\circ}) = 105^{\circ}$

When, $C^{'} = 135^{\circ} \Rightarrow A = 180^{\circ} - (135^{\circ} + 30^{\circ}) = 15^{\circ}$

Area of $△ A B C = \frac{1}{2} A B \times A C \sin A$

$= \frac{1}{2} \times 4 \times 2 \sqrt{2} \sin (105^{\circ})$

$= 4 \sqrt{2} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$= 2 (\sqrt{3} + 1)$ sq. units

Area of $△ A B C^{'} = \frac{1}{2} A B \times A C \sin A$

$\begin{aligned} = \frac{1}{2} \times 4 \times 2 \sqrt{2} \sin (15^{\circ}) \\ = 2 (\sqrt{3} - 1) sq. units \end{aligned}$

Difference of areas of triangle

$= | 2 (\sqrt{3} + 1) - 2 (\sqrt{3} - 1) | = 4 sq units$

Alternate Solution

Here, $A D = 2, D C = 2$

Difference of areas of $△ A B C$ and $△ A B C^{'}$

$=$ Area of $△ A C C^{'}$

$= \frac{1}{2} A D \times C C^{'} = \frac{1}{2} \times 2 \times 4 = 4 s q$ units

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Properties of Triangles 2 Question 17

17. Let $A B C$ and $A B C^{'}$ be two non-congruent triangles with sides $A B = 4, A C = A C^{'} = 2 \sqrt{2}$ and angle $B = 30^{\circ}$ . The absolute value of the difference between the areas of these triangles is

Solution:

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Properties of Triangles 2 Question 17

17. Let ABC and ABC′ be two non-congruent triangles with sides AB=4,AC=AC′=22 and angle B=30∘. The absolute value of the difference between the areas of these triangles is

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

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परीक्षा मंच

नमूना मॉक टेस्ट

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17. Let $A B C$ and $A B C^{'}$ be two non-congruent triangles with sides $A B = 4, A C = A C^{'} = 2 \sqrt{2}$ and angle $B = 30^{\circ}$ . The absolute value of the difference between the areas of these triangles is