Difference between revisions of "1967 AHSME Problems/Problem 32"
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− | + | In quadrilateral <math>ABCD</math> with diagonals <math>AC</math> and <math>BD</math>, intersecting at <math>O</math>, <math>BO=4</math>, <math>OD = 6</math>, <math>AO=8</math>, <math>OC=3</math>, and <math>AB=6</math>. The length of <math>AD</math> is: | |
− | In quadrilateral <math>ABCD</math> with diagonals <math>AC</math> and <math>BD</math>, intersecting at <math>O</math>, <math>BO=4</math>, <math>OD = 6</math>, <math>AO=8</math>, <math>OC=3</math>, and <math>AB=6</math>. | + | <math>\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}</math> |
− | <math> | + | <math>\sqrt{166}</math> |
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+ | After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of <math>AB, BD(BD = BO + OD)</math>, but we want to find the value of AD. We can apply stewart's theorem now, letting <math>m = 4, n = 6, AD = X, AB = 6</math>, and we have <math>10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6</math>, and we see that <math>x = \sqrt{166}</math>, <math>\boxed{E \sqrt{166}}</math> | ||
== See also == | == See also == | ||
{{AHSME box|year=1967|num-b=31|num-a=33}} | {{AHSME box|year=1967|num-b=31|num-a=33}} |
Revision as of 18:18, 10 September 2020
In quadrilateral with diagonals and , intersecting at , , , , , and . The length of is:
After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of , but we want to find the value of AD. We can apply stewart's theorem now, letting , and we have , and we see that ,
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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