Triangles ABC and DEF are such that A= D, AB=DE =17, BC=EF =10 and AC-DF=12 . What is AC+DF ?

Correct answer: 30 Let A coincides with D, B coincides with E . With E(B) as centre draw a circle with radius 10 intersecting the line, making angle θ= A with AB , at F \& C . Let N be the foot of perpendicular from B(E) to C F So, BM=8 Hence, AM= 17^2-8^2 = (25)(9) =15 Hence, A F=15-6=9 \& A C=15+6=21 So, AC+DF=30

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

Triangles ABC and DEF are such that A= D, AB=DE =17, BC=EF =10 and...

Triangles ABC and DEF are such that

\angle \mathrm{A}=\angle \mathrm{D},

\mathrm{AB}=\mathrm{DE}

=17,

\mathrm{BC}=\mathrm{EF}

=10

and

\mathrm{AC}-\mathrm{DF}=12

. What is

\mathrm{AC}+\mathrm{DF}

Correct answer

Step-by-step solution

Let

A

coincides with

D, B

coincides with

E

. With

E(B)

as centre draw a circle with radius 10 intersecting the line, making angle

\theta=\angle \mathrm{A}

with AB , at

\mathrm{F} \& \mathrm{C}

Let

N

be the foot of perpendicular from

B(E)

C F

So,

\mathrm{BM}=8

Hence,

\mathrm{AM}=\sqrt{17^2-8^2}=\sqrt{(25)(9)}=15

Hence,

A F=15-6=9

A C=15+6=21

So,

\mathrm{AC}+\mathrm{DF}=30

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