Kidding me?

Geometry Level 4

In the given figure, A B = 3 cm , B C = 3 cm , C D = 2 cm , A D = 2 cm AB=3 \text{ cm}, BC=3\text{ cm}, CD=2\text{ cm}, AD=2\text{ cm} . A B C = 6 0 \angle ABC= 60^\circ . Find the area of the quadrilateral A B C D ABCD .

No trigonometry to be used.

Round off your answer correct to 2 decimal places.
Find your answer in sq cm units.


The answer is 1.91.

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2 solutions

Join AC.

Now, In tr. ABC, AB = BC = 3 cm Hence angle BAC = angle BCA

Clearly, using angle sum property of triangles,

angle BAC = angle BCA = angle ABC = 60 degrees Clearly, tr. ABC is an equilateral triangle

Then, AB=BC=AC=3 cm

Now, area(quad. ABCD) = area(tr. ABC) - area(tr. ADC) = [rt(3)/4]3^2 - (3/2) rt[2^2 - (3^2)/4]

Simpilifying the above expression using rt(3) = 1.732 and any other irrational number upto 3 decimal places (we have to round it off correct to 2 decimal places), we come up to:

(15.588 - 7.937)/4 = 7.651/4 = 1.91275

Rounding it off correct to 2 decimal places, we get 1.91 as the final answer

its a relative easy problem i wonder why you put it level 4

Ols Muka - 4 years, 5 months ago

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Well, I didn't put it on Level 4. @Brilliant Mathematics did.

Arkajyoti Banerjee - 4 years, 5 months ago

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We do not manually change the levels. If more people can't solve this problem, then the level of this problem will naturally increase.

I've changed it back to level 3.

Brilliant Mathematics Staff - 4 years, 5 months ago

Clearly Isosceles triangle ABC with vertex angle 60 is equilateral. So AC=3. Altitude 0f ABC from B= 3 3 2 . I n i s o s c e l e s Δ A D C , a l t i t u d e f r o m D = A D 2 ( A C 2 ) 2 = 7 2 . r e q u i r e d a r e a = 1 2 3 ( 3 3 2 7 2 ) = 1.9128 3\dfrac{\sqrt3 } 2.\\ In\ isosceles \Delta ~ADC, ~altitude ~ from ~ D=\sqrt{AD^2-(\frac {AC} 2)^2}=\dfrac{\sqrt7} 2.\\ \therefore ~required~ area =\frac1 2 *3*( 3\dfrac{\sqrt3 } 2 - \dfrac{\sqrt7} 2)=1.9128

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