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Geometry Level 4

ABCD is rectangle. P is any point on AB, and the foot of the perpendiculars from P to the diagonals are Q and R. If AB=24 cm and BC=7 cm, then find the value of PQ + PR.


The answer is 6.72.

Gilang Syahya by Gilang Syahya , 29, Indonesia

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1 solution

Gilang Syahya
Sep 13, 2014

Area of AOB = Area AOP + Area of BOP

24 × 7 4 = A O × P R 2 + B O × P Q 2 \frac{24 \times 7}{4} = \frac{AO \times PR}{2} + \frac{BO \times PQ}{2}

42 = 25 2 × P R 2 + 25 2 × P Q 2 42 = \frac{\frac{25}{2} \times PR}{2} + \frac{\frac{25}{2} \times PQ}{2}

42 = 25 4 × ( P R + P Q ) 42 =\frac{25}{4} \times ({PR+PQ})

42 × 4 25 = P R + P Q 42 \times \frac{4}{25} = PR + PQ

6.72 = P R + P Q 6.72 = PR+PQ

4 Helpful 0 Interesting 0 Brilliant 2 Confused

I think in the second line there is a typo. PR and PQ are exchanged.

Niranjan Khanderia - 3 years, 9 months ago

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