What is its Area?

Geometry Level 4

A circle with center point A A and radius 9 c m 9cm touches another circle of radius 5 c m 5cm and center point B B .

Two distinct direct common tangent lines P Q ‾ \overline{PQ} and R S ‾ \overline{RS} are drawn to the circles such that points P P and S S lies on the circle with center A A .

If the area of quadrilateral P Q R S PQRS is △ c m 2 \triangle cm^2 , then find ⌊ △ ⌋ \lfloor \triangle \rfloor .

Note:


The answer is 172.

Zakir Husain by Zakir Husain , 41, India

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2 solutions

Hongqi Wang
Dec 24, 2020

Let ∠ P A B = θ \angle PAB = \theta

P Q = R S = ( r A + r B ) 2 − ( r A − r B ) 2 = ( 9 + 5 ) 2 − ( 9 − 5 ) 2 = 6 5 sin ⁡ θ = P Q A B = 3 7 5 cos ⁡ θ = 1 − sin ⁡ 2 θ = 2 7 S P Q B A = 1 2 ⋅ ( r A + r B ) ⋅ P Q = 1 2 ⋅ ( 9 + 5 ) ⋅ 6 5 = 42 5 S △ P A S = 1 2 r A 2 sin ⁡ 2 θ = r A 2 sin ⁡ θ cos ⁡ θ = 9 2 ⋅ 3 7 5 ⋅ 2 7 = 486 49 5 S △ Q B R = 1 2 r B 2 sin ⁡ 2 θ = r B 2 sin ⁡ θ cos ⁡ θ = 5 2 ⋅ 3 7 5 ⋅ 2 7 = 150 49 5 S P Q R S = 2 S P Q B A − S △ P A S + S △ Q B R = 2 ⋅ 42 5 − 486 49 5 + 150 49 5 = 540 7 5 ≈ 172.497 ⌊ S P Q R S ⌋ = 172 \begin{aligned} PQ = RS &= \sqrt {(r_A+r_B)^2 - (r_A-r_B)^2} \\ &= \sqrt {(9+5)^2 - (9-5)^2} \\ &= 6\sqrt 5 \\ \sin \theta &= \dfrac {PQ}{AB} = \dfrac 37 \sqrt 5 \\ \cos \theta &= \sqrt {1- \sin^2 \theta} = \dfrac 27 \\ S_{PQBA} &= \dfrac 12 \cdot (r_A+r_B) \cdot PQ \\ &= \dfrac 12 \cdot (9+5) \cdot 6\sqrt 5 \\ &= 42\sqrt 5 \\ S_{△PAS} &= \dfrac 12 r_A^2 \sin {2\theta} = r_A^2 \sin {\theta} \cos {\theta} \\ &= 9^2 \cdot \dfrac 37 \sqrt 5\cdot \dfrac 27 \\ &= \dfrac {486}{49} \sqrt5 \\ S_{△QBR} &= \dfrac 12 r_B^2 \sin {2\theta} = r_B^2 \sin {\theta} \cos {\theta} \\ &= 5^2 \cdot \dfrac 37 \sqrt 5\cdot \dfrac 27 \\ &= \dfrac {150}{49} \sqrt5 \\ S_{PQRS} &= 2S_{PQBA} - S_{△PAS} + S_{△QBR} \\ &= 2 \cdot 42 \sqrt 5 - \dfrac {486}{49} \sqrt5 + \dfrac {150}{49} \sqrt5 \\ &= \dfrac {540}{7} \sqrt5 \\ & \approx 172.497 \\ \lfloor S_{PQRS} \rfloor &= \boxed {172} \end{aligned}

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Hosam Hajjir
Dec 24, 2020

The two circles are depicted in the figure, as well as the quadrilateral C C ′ D ′ D C C' D' D (shaded).

If the slope of C D = − tan ⁡ θ CD = - \tan \theta , then sin ⁡ θ = 9 − 5 9 + 5 = 2 7 \sin \theta = \dfrac{9 - 5}{9+5} = \dfrac{2}{7} and cos ⁡ θ = 45 7 \cos \theta = \dfrac{\sqrt{45}}{7}

The coordinates of point C = ( C x , C y ) = ( 9 sin ⁡ θ , 9 cos ⁡ θ ) C = (C_x, C_y) = (9 \sin \theta, 9 \cos \theta ) , and the coordinates of point D = ( D x , D y ) = ( 14 + 5 sin ⁡ θ , 5 cos ⁡ θ ) D = (D_x , D_y) = ( 14 + 5 \sin \theta, 5 \cos \theta)

Hence, the area of the trapezoid is

[ C C ′ D ′ D ] = ( 14 cos ⁡ θ ) ( 14 − 4 sin ⁡ θ ) = 2 45 ( 14 − 8 7 ) ≈ 172.49667 [C C' D' D ] = (14 \cos \theta) (14 - 4 \sin \theta) = 2 \sqrt{45} (14 - \dfrac{8}{7} ) \approx 172.49667

Hence the answer is 172 \boxed{172}

2 Helpful 2 Interesting 2 Brilliant 0 Confused

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