Exercise 10 B
(1) ABCD is a square, prove that AC2=2AB2
(2) In Fig. 10.26, AB=BC=CA=2a and segment AD⊥ side BC, Show that
(i) AD=a√3, (ii) area of ΔABC=a2√3
Sol:1 (IMAGE TO BE ADDED)
From, Right angle triangle , △ABC
We get, AC2=AB2+BC2 (By Pythagoras Theorem)
=AB2+AB2 (ABCD is a square)
=2AB2
Sol:2 (IMAGE TO BE ADDED)
(i) From right angle triangle △ABC
We get,
(2a)2=a2+AD2 (by Pythagoras Theorem)
AD2 =4a2 - a2
(ii) Area of △ABC = a2√3
Question 3
In Fig 10.27, prove that AB2−AD2=CD2−CB2,
(IMAGE TO BE ADDED)
Sol : From , right angle triangle , △ABC
We get, AC2=AB2+BC2(By Pythagoras theorem).......(i)
From right angle triangle
△ABC We get,
AC2=AD2+DC2(By Pythagoras theorem)........(ii)
From (i) and (ii) We get
AB2+BC2=AD2+DC2
AB2−AD2=CD2−CB2 Proved
Question 4
In a △ABC,AD⊥BC, Prove that: AB2+CD2=AC2+BD2
(IMAGE TO BE ADDED)
Sol: From right angle triangle ,
△ABD We get,
AB2=BD2+AD2(By Pythagoras theorem we get)
from △ADC we get,
AC2=AD2+DC2 .....(ii)
From (i) & (ii) We get,
AB2−BD2=AC2−DC2
∴AB2+CD2= AC2+BD2 (Proved)
Question 5
In a quadrilateral ABCD, the diagonals AC,BD intersect at right angles. Prove that:
AB2+CD2=BC2+DA2
(IMAGE TO BE ADDED)
Sol: From right angle triangle ,
△OAB We get,
AB2=OA2+OB2..........(i)
△ OBC we get,
BC2=OB2+OC2.........(ii)
△OCD we get,
DC2=OC2+OD2............(iii)
△OAD We get,
AD2=OA2+OD2
From (i) and (ii) we get.
AB2+CD2= OA2+OB2+OC2+OD2
From (iii) &(i) we get
BC2+DA2= OA2+OB2+OC2+OD2
AB2+CD2= BC2+DA2
Question 6
In △ABC,∠B=90∘ and D is the mid point of BC, Prove that
(i) AC2=AD2+3CD2 (ii) BC2=4(AD2−AB2)
(IMAGE TO BE ADDED)
From , right angle triangle △ABC We get,
AC2=AB2+BC2.........(i)
From right angle triangle
△ABD We get,
AD2=AB2 + B D^{2}.......(ii)
From (i) & (ii) We get,
AC2=AD2−BD2+BC2=AD2−BD2+(2CD)2=AD2−CD2+4CD2=AD2+3CD2 (proved)
(ii) (IMAGE TO BE ADDED)
AC2=AB2+BC2AD2=AB2+BD2
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