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# Short Answer Type Questions (Part 2) - Triangles Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions (Part 2) - Triangles Class 10 Notes | EduRev

The document Short Answer Type Questions (Part 2) - Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Q11. ABC is a right triangle, right angled at A, and D is the mid-point of AB. Prove that
BC2 = CD2 + 3 BD2

Sol. We have a right Î” ABC in which <A = 90Â°
âˆ´Using Pythagoras Theorem, we have:
BC2 = AB2 + AC2   ...(1)
Again, D ACD is right D, <A = 90Â°
...(2) [Using Pythagoras Theorem]
Subtracting (2) from (1), we get
BC2 - CD2 = AB2 - AD2 ...(3)
Since D is the mid-point of AB
âˆ´2 BD = AB  and  AD = BD ...(4)
From (3) and (4), we have:
âˆ´BC2 - CD2 = (2 BD)2 - (BD)2
= 4 BD2 - BD2
BC2 = CD2 + 3 BD2

Q12. In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm and OA = 5 cm. Find the length of OC.

Sol. Let us draw EOF y AB â‡’ OE âŠ¥ AD and OF âŠ¥ BC
In Î” OFB
Å’ <F = 90Â°
âˆ´Using Pythagoras theorem, we have:
CB2 = OF2 + BF2 ...(1)
In Î” OED
<E = 90Â°
âˆ´Using Pythagoras theorem, we have:
OD2 = OE2 + DE2 ...(2)
Adding (1) and (2), we get
OB2 + OD2 = OF2 + BF2 + OE2 + DE2
= OF2 + AE2 + OE2 + CF2
[Å’ BF = AE and CF = DE]
= (OF2 + CF2) + (OE2 + AE2)
= OC2 + OA2
= OC2 + 52
â‡’ 62 + 82 = OC2 + 52
â‡’ 36 + 64 = OC2 + 25
â‡’ OC2 = 36 + 64 - 25 = 75
â‡’
Thus OC = 5âˆš3cm.

Q13. In the figure, if AD âŠ¥ Bc,  then prove that:
AB
2 + CD2 = AC2 + BD2

âˆ´ AD2 = AC2 - CD2 .....(1) (Using Pythagoras Theorem)
Similarly, in D AbD,
â‡’ AD2 = AB2 - DB2.....(2)
From (1) and (2), we have
AB2 - DB2 = AC2 - CD2
â‡’ AB2 + CD2 = AC2 + BD2

Q14. In the given figure, AD âŠ¥ BC and BD =  CD. Prove that:

Sol. Å’ BD = 1/2 CD
âˆ´ 3 BD = CD
Since BD + DC = BC
âˆ´ BD + 3 BD = BC
â‡’ 4 BD = BC
â‡’ BD = 1/4 BC
â‡’ CD = 3/4 BC
Now, in right Î” ADC, < D = 90Â°
By Pythagoras theorem, we get
CA2 = AD2 + CD2  ...(1)
AD2 = AB2 - BD2  ...(2)
From (1) and (2),
CA2 = AB2 - BD2+ CD2

Q15. In the given figure, M is the mid-point of side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produce D at E. Prove that EL = 2 BL.

Sol. We have parallelogram ABCD in which M is the mid point of CD.
In Î” EMD and Î” BMC
MD = MC [Å’ M is mid-point of CD]
<EMD = <CMB
[Vertically opposite angles]
<MED = <MBC
[Alternate interior angles]
âˆ´  Î” BMC â‰Œ   Î” EMD     [AAS congruency]
â‡’ BC = ED  â‡’  AD = ED    ...(1)
[Å’ BC = AD,  opposite sides of parallelogram]
Now, in   Î” AEL and   Î” CBL
<AEL = <CBL [Alternate interior angles]
<ALE = <CLB
[Vertically opposite angles]
âˆ´ By AA similarity, we have:

Q16. In the given figure, D ABC is right angled at C and DE ^ AB. Prove that D ABC ~ D ADE and hence find the length of AE and DE.

Sol.In Î” ABC and Î” ADE, we have:
<A = <A [Common]
<C = <E [Each = 90Â°]
âˆ´ Î” ABC ~ Î” ADE [AA Similarity]
...(1)
In right Î” ABC, <C = 90Â°
Using Pythagoras theorem, we have:
AB2 = BC2 + AC2
= 122 + 52
= 144 + 25 = 169
â‡’AB = âˆš169 = 13 cm
Now, from (1), we get

â‡’ DE = = 2.77 cm
And =
â‡’ AE

Q17. In the given figure, DEFG is a square and <BAC = 90Â°. Show that DE2 = BD Ã— EC.

Sol. In Î” DBG and Î” ECF
<3 + <1 = 90Â° = <3 + <4
âˆ´<3 + <1 = <3 + <4
â‡’ <1 = <4
<D = E = 90Â°
âˆ´Using AA similarity, we have:
=
BD Ã— EC = EF Ã— DG
But DG = EF = DE
âˆ´BD Ã— EC = DE Ã— DE
â‡’ BD Ã— EC = DE2
Thus, DE2 = BD Ã— EC

Q18. In the figure, AD âŠ¥ BC and BD =  CD. Prove that 2 CA2 = 2 AB2 + BC2

Sol. Å’ BD = 1/3 CD
â‡’ 3 BD = CD
âˆ´BC = BD + DC
â‡’ BC = BD + 3 BD
â‡’ BC = 4 BD ...(1)
And
From (2) ...(4)
Using Pythagoras theorem,
=
From (3) ...(4)
Using Pythagoras theorem,
=

Q19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

Sol. We have a quadrilateral ABCD such that its diagonals intersect at O and
=
Å’ <AOB = <COD [Vertically opposite angles]
âˆ´ Using SAS similarity, we have
Î” AOB ~ Î” COD
â‡’ Their corresponding angles are equal.
i.e., <1 = <2
But they form a pair of int. alt. angles.
â‡’ AB y DC
â‡’ ABCD is a trapezium.

Q20. Two triangles ABC and DBC are on the same base BC and on the same side of BC in which
<A = <D = 90Â°. If CA and BD meet each other at E, show that
AEÂ· EC = BEÂ· ED
Sol.
We have right D ABC and right D DBC on the same base BC such that
<A = <D = 90Â°
In D ABE and D DCE
<A = <D = 90Â°

<1 = <2 [Vertically opp. angles]
âˆ´ Using AA similarity, we have:
Î” ABE ~ Î” DCE
â‡’ Their corresponding sides are proportional.
â‡’
â‡’ AEÂ· EC = BEÂ· ED

Q21. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
Î” ABE ~ Î” CFB
Sol.
We have parallelogram ABCD
In Î” ABE and Î” CFB, we have
<A = <C
[Opposite angles of parallelogram]
<AEB = <EBC
[Alternate angles, Å’ AD y BC]
âˆ´ Using AA similarity, we get
Î” ABE ~ Î” CFB

Q22. In Î” ABC, if AD is the median, then show that AB2 + AC2 = 2 [AD2 + BD2].

Sol. Å’ AD is a median,
âˆ´  BD = DC
Let us draw AE â‰Œ BC
Now, in rt. Ds AEB and AEC, we have
AB2 = BE2 + AE2 ...(1)
AC2 = CE2 + AE2 ...(2)
AB2 + AC2 = BE2 + AE2 + CE2 + AE2
= (BD - ED)2 + AE2 +
(CD + DE)2 + AE2
= 2AE2 + 2ED2 + BD2 + CD2
= 2 [AE2 + ED2] + BD2 + BD2
[Å’ BD = CD]
[Å’ AE2 + ED2 = AD2]
Thus,  AB2 + AC2 = 2 [AD2 + BD2]

Q23. Triangle ABC is right angled at B and D is the mid point of BC. Prove that:
AC2 = 4 AD- 3 AB2
Sol.
D is the mid-point of BC.
âˆ´ BC = 2 BD

Now, in Î” ABC
AC2 = BC2 + AB2
= (2 BD)2 + AB2
= 4 BD2 + AB2 ...(1)
In right Î” ABD
Using Pythagoras theorem,
â‡’ BD2 = AD2 - AB2 ...(2)
From (1) and (2), we get
AC2 = 4 [AD2 - AB2] + AB2
â‡’ AC2 = - 4 AB2 + 4 AD2 + AB2
â‡’ AC2 = - 3 AB2 + 4 AD2
or AC2 = 4 AD2 - 3 AB2

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